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IB Physics Standard Level · Topic D.2

D.2 Electric and magnetic fields

156 free, high-quality past-paper-style questions on Electric and magnetic fields for IB Physics SL, each with a full mark scheme. Covers: Coulomb's law; Uniform fields between plates; Electric field lines; and more. For when you have finished the past papers, or want to practise one topic at a time.D.2 at Higher Level →Read the D.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.776 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1D-1A-10Coulomb's law Paper 1AEasy1 mark

Point charges of +2.0 μC and −3.0 μC are 0.10 m apart in a vacuum.

Which row gives the magnitude of the electric force between the charges and describes the force?

Magnitude of force / NNature of force
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Marking pointMarkNotes
Step 1F = kq1q2/r² = 8.99 × 109 × 2.0 × 10−6 × 3.0 × 10−6/0.10² = 5.4 N; unlike charges attract.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AThe separation has not been squared.
  • BCorrect: F = 5.4 N, and charges of opposite sign attract.
  • CThe magnitude is right, but charges of opposite sign attract each other.
  • DOne of the charges has been left in microcoulombs; both must be converted with 1 μC = 10−6 C.
2D-1A-12Uniform fields between plates Paper 1AEasy1 mark

Two parallel plates 0.020 m apart are connected to a 240 V supply.

Which row gives the electric field strength between the plates and the magnitude of the force on an electron between them?

Two parallel plates with a uniform electric field between them, the field lines running from the positive to the negative plate.+−dVuniform electric field between the plates
Diagram NOT accurately drawn
Field strength / V m−1Force on the electron / N
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Marking pointMarkNotes
Step 1E = V/d = 240/0.020 = 1.2 × 104 V m−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = eE = 1.60 × 10−19 × 1.2 × 104 = 1.9 × 10−15 N.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe field has been found by multiplying V by d instead of dividing.
  • BThe separation has been read as 0.20 m instead of 0.020 m, giving a field ten times too small.
  • CCorrect: E = 240/0.020 = 12 000 V m−1 and F = eE = 1.9 × 10−15 N.
  • DThe field is right, but the force has been quoted without multiplying by the charge.
3D-1A-15Electric field lines Paper 1AEasy1 mark

Two parallel metal plates carry equal and opposite charges.

Which statement about the electric field lines is correct?

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Marking pointMarkNotes
Step 1Field lines show the direction of the force on a positive test charge, so they run from + to −; between the central parts of the plates the field is uniform, and at the edges the lines bulge outwards.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AField lines point in the direction of the force on a positive test charge, so they leave the positive plate.
  • BThis ignores edge effects: near the edges the lines curve outwards and spread apart, so the field there is weaker.
  • CBetween the central parts of the plates the field is uniform, so the line spacing does not change across the gap.
  • DCorrect: the central field is uniform and directed from the positive plate to the negative plate.
4D-1A-33Conservation of charge Paper 1AEasy1 mark

Two identical small metal spheres X and Y on insulating stands carry charges of +6.0 nC and −2.0 nC. The spheres are touched together and then separated.

Which row gives the final charges on X and Y?

Charge on X / nCCharge on Y / nC
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Marking pointMarkNotes
Step 1Charge is conserved: the total is +6.0 + (−2.0) = +4.0 nC.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Identical conductors in contact share the charge equally: +2.0 nC each.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis adds the magnitudes, 6.0 + 2.0, ignoring the sign of the charge on Y.
  • BThe total is right, but identical conductors in contact share the charge equally.
  • CCorrect: the total +4.0 nC is conserved and shared equally between identical spheres.
  • DCharges are not exchanged between the spheres; the total charge is shared.
5D-1A-35Charged conducting sphere Paper 1AEasy1 mark

A hollow metal sphere carries a positive charge.

Which row describes the electric field inside the sphere and just outside it?

InsideJust outside
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Marking pointMarkNotes
Step 1The charge sits on the outer surface of the conductor and the field inside is zero; outside, the sphere behaves like a point charge at its centre, so the field is radial and points away from the positive charge.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: zero inside; outside, radial and outwards, as for a point charge at the centre.
  • BThese are the wrong way round — there is certainly a field outside a charged sphere.
  • CIn equilibrium there is no field inside a conductor; otherwise free charges would keep moving.
  • DOutside, the field is that of a point charge: radial, weakening with distance, not uniform.
6D-1A-49Charging by friction & induction Paper 1AEasy1 mark

A plastic comb is pulled through dry hair. Afterwards the comb is negatively charged.

Which statement explains this?

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Step 1Charging by friction transfers electrons; the protons stay in the nuclei. The comb gains electrons, so electrons move from the hair to the comb and, because charge is conserved, the hair is left positive by the same amount.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AProtons are bound in the nuclei of the atoms and are not transferred by rubbing; only electrons move.
  • BCorrect: the comb gains electrons from the hair, and the hair is left with an equal positive charge.
  • CThis reverses the transfer: a body that loses electrons becomes positive, so the comb could not then be negative.
  • DCharge is not created by friction; it is conserved, so the negative charge on the comb must be matched by an equal positive charge on the hair.
7D-1A-53Uniform fields between plates Paper 1AEasy1 mark

Two parallel metal plates are connected to a power supply that keeps the potential difference between them constant. P is a point between the plates, well away from the edges.

Which change increases the electric field strength at P?

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Marking pointMarkNotes
Step 1Between parallel plates the field is uniform with E = V/d; with V fixed, reducing d increases E. Plate area, position between the plates and the size of a test charge do not change E.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: E = V/d, so a smaller separation at the same potential difference gives a stronger field.
  • BAway from the edges E = V/d does not depend on the area of the plates.
  • CThe field between the plates is uniform, so it has the same strength at every point away from the edges.
  • DA larger test charge feels a larger force, but E = F/q is the force per unit charge and is unchanged.
8D-1A-72Electric field strength Paper 1AEasy1 mark

Inside a thundercloud, a raindrop carrying a charge of −3.2 pC is in an electric field of strength 5.0 × 104 N C−1 directed vertically upwards.

Which row gives the magnitude and the direction of the electric force on the raindrop?

Magnitude of force / NDirection of force
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Marking pointMarkNotes
Step 1F = qE = 3.2 × 10−12 × 5.0 × 104 = 1.6 × 10−7 N.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The charge is negative, so the force is opposite to the field: downwards.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: F = 1.6 × 10−7 N, and a negative charge is pushed opposite to the field.
  • BThe magnitude is right, but the field direction is the direction of the force on a positive charge; on a negative charge the force is opposite.
  • CThis converts picocoulombs as if they were nanocoulombs (10−9 C); 1 pC = 10−12 C.
  • DThis divides the field strength by the charge. Rearranging E = F/q gives F = qE.
9D-1A-76Quantization of charge Paper 1AEasy1 mark

In Millikan's experiment the electric charges on a large number of oil drops were determined.

Which statement describes what the results showed?

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Marking pointMarkNotes
Step 1The measured charges were different for different drops, but every value was an integer multiple of 1.6 × 10−19 C, the elementary charge: electric charge is quantized.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • ADifferent drops carried different charges; what they shared was that each charge was a whole number of elementary charges.
  • BThe experiment measured the charge on individual drops; it was evidence for quantization, not a test of the conservation of charge.
  • CDrops of the same mass were found with different charges; the charge depends on how many electrons were gained or lost, not on the mass.
  • DCorrect: the charges were always integer multiples of the elementary charge e.
10D-1A-80Magnetic field patterns Paper 1AEasy1 mark

A student sprinkles iron filings on a sheet of card placed over a bar magnet and uses small plotting compasses to find the direction of the magnetic field.

Which statement about the magnetic field lines of the bar magnet is correct?

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Marking pointMarkNotes
Step 1Magnetic field lines have no beginning or end: outside a bar magnet they run from the north pole to the south pole and they continue through the magnet from south to north, forming closed loops.✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AMagnetic field lines do not start or end anywhere; they continue inside the magnet back to the north pole.
  • BThe field is strongest where the lines are closest together, which is near the poles.
  • CThe field has a single direction at each point, so magnetic field lines can never cross.
  • DCorrect: the lines are closed loops, N → S outside and S → N inside the magnet.
11IBP2S-030-01-01Magnetic field patterns Paper 1AEasy1 mark

The main magnet of a hospital MRI scanner is a long solenoid carrying a large steady current. Which statement describes the magnetic field in the central region inside the solenoid?

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Marking pointMarkNotes
Inside a long solenoid the fields of all the turns combine to give straight, parallel, equally spaced field lines along the axis. Equal spacing means equal field strength, so the field is uniform. This is why a long solenoid is used where a uniform field is needed. Concentric circles describe the field around a single straight wire, not the field inside a solenoid. The fields of the turns add along the axis rather than cancel, so the field is not zero.✓ 1Answer A

Answer: A

Common error: Do not mix up the field patterns. Concentric circles belong to a straight wire. Parallel, evenly spaced lines belong to the inside of a solenoid.

12IBP2S-030-01-02Magnetic field of a straight wire Paper 1AEasy1 mark

A vertical copper earthing conductor on the side of a building carries a current directly upwards. Ignore the Earth's magnetic field. What is the direction of the magnetic field produced by the conductor at a point a short distance due east of it?

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Apply the right-hand grip rule with the thumb pointing up along the current. The fingers then curl anticlockwise when viewed from above. At a point due east of the wire, the anticlockwise tangent to the circle points north. South is the answer you get by curling the field the wrong way. East would be a radial field, and upwards would be a field parallel to the current. Neither radial nor parallel fields occur around a straight wire.✓ 1Answer A

Answer: A

Exam tip: Draw a quick top-down view: mark the wire as a dot, draw a circle round it, and add the arrow given by the grip rule. Then read off the tangent direction at the point you need.

13IBP2S-030-01-06Electric field lines Paper 1AEasy1 mark

In electrostatic spray painting, a charged spray nozzle is held near a car body panel. The metal panel is connected to earth. Which statement about the electric field lines between the nozzle and the panel is correct?

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Marking pointMarkNotes
In electrostatics, a conductor's surface is an equipotential, so field lines always meet it at 90°. The other options are wrong for these reasons:
• Field lines never cross, because the field has only one direction at each point.
• A field line shows the direction of the force, not the path. A droplet that already has velocity can leave a curved field line.
• Field strength is greatest where the lines are closest together.
✓ 1Answer A

Answer: A

Common error: A field line gives the direction of the force on a positive test charge. It is not the trajectory of a moving particle.

14IBP2S-030-01-10Quantization of charge Paper 1AEasy1 mark

A proton and an electron are each placed, one at a time, in the same uniform electric field inside a particle detector. Which statement is correct?

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The force is F = qE. The two charges have equal magnitude e, so the forces are equal in size, although opposite in direction. The acceleration is a = F/m. The electron's mass is about 1/1800 of the proton's, so its acceleration is about 1800 times larger. The electric force does not depend on mass, so options C and D are wrong.✓ 1Answer B

Answer: B

Common error: Mass affects acceleration, not the electric force. Find the force with F = qE first, then use Newton's second law.

15IBP2S-030-02-01Electric field strength Paper 1AEasy1 mark

A high-voltage switch has two flat parallel electrodes separated by an air gap. The electric field strength in the gap is E. The engineer doubles both the potential difference across the electrodes and their separation. What is the new electric field strength in the gap?

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For a uniform field between parallel plates, E = V/d. Doubling V doubles E, and doubling d halves it. The two changes cancel, so the field stays E. Option C ignores the change in d. Option D treats E as proportional to Vd. Option A treats E as depending on 1/d only.✓ 1Answer B

Answer: B

Exam tip: When two quantities change at once, write the formula as a ratio: Enew/E = (2V/2d)/(V/d).

16IBP2S-030-02-02Quantization of charge Paper 1AEasy1 mark

In a vacuum tube, an electron is released from rest between two horizontal parallel plates. The upper plate is positive and the lower plate is negative. Gravitational effects are negligible. Which describes the motion of the electron before it reaches a plate?

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The field between the plates is uniform and points from the positive plate to the negative plate, so downwards. The electron is negative, so the force on it is upwards, towards the positive plate. A uniform field gives a constant force and therefore a constant acceleration (A). Option B confuses a constant force with a constant velocity. Option C wrongly treats the field as stronger near a plate. Option D treats the electron as if it were a positive charge.✓ 1Answer A

Answer: A

Common error: Field direction is defined by the force on a positive test charge. An electron always accelerates against the field lines.

17IBP2S-030-02-07Uniform fields between plates Paper 1AEasy1 mark

Two parallel plates 4.0 cm apart have a potential difference of 500 V. An external agent moves a small charge of +2.0 μC slowly from the negative plate to the positive plate. What is the work done by the external agent on the charge?

positive platenegative plate4.0 cmp.d. = 500 V+2.0 μC charge
Diagram not to scale
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Marking pointMarkNotes
The work done moving a charge through a potential difference is W = qV.
Here W = 2.0 × 10−6 × 500 = 1.0 × 10−3 J, which is option B.
The work is positive because the positive charge is pushed towards the higher potential. The plate separation is not needed.
Option A wrongly multiplies by the separation as well (qVd).
Option C is qV/d. This is the size of the force in newtons, not the work.
Option D wrongly assumes that no work is needed to move a charge between the plates.
✓ 1Answer B

Answer: B

Exam tip: Check the units of your answer. qV/d has units of newtons, so it cannot be an energy.

18IBP2S-030-02-09Uniform fields between plates Paper 1AEasy1 mark

Two large parallel plates produce a uniform field. Point X is close to the positive plate, point Y is midway between the plates and point Z is close to the negative plate. All three points are far from the plate edges. Which relationship between the field strengths at X, Y and Z is correct?

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Marking pointMarkNotes
Away from the edges, the field between parallel plates is uniform. The field lines are parallel and equally spaced, so the field strength is the same everywhere, E = V/d (A). Option B confuses field strength with potential, which does fall steadily from the positive plate to the negative plate. Options C and D apply point-charge reasoning to plates.✓ 1Answer A

Answer: A

Common error: Potential changes linearly across the gap, but the field strength does not change at all.

19IBP2S-030-03-01Charging by friction & induction Paper 1AEasy1 mark

A child slides down an inflatable PVC play slide and, at the bottom, is found to be positively charged. The slide becomes negatively charged. Which statement explains this?

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Marking pointMarkNotes
In charging by friction only electrons move. Protons are bound in nuclei and cannot be transferred by rubbing. Charge is conserved, so friction cannot create it. The child loses electrons, leaving a net positive charge, and the slide gains those electrons and becomes negative. Option D would make the child negative.✓ 1Answer B

Answer: B

Common error: Saying that 'positive charges moved' in friction charging is wrong. Always describe the transfer as electrons gained or lost.

20IBP2S-030-03-02Conservation of charge Paper 1AEasy1 mark

An uncharged metal sphere stands on an insulating base. The following steps are carried out in order:
1. A negatively charged rod is held close to the sphere, without touching it.
2. The sphere is briefly connected to earth.
3. The earth connection is removed.
4. The rod is taken away.
What is the final charge on the sphere?

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The negative rod repels free electrons in the sphere. When the sphere is earthed, some of these electrons escape to earth. Breaking the earth connection before removing the rod traps the electron deficit, so the sphere stays positive after the rod is removed. Induction always gives a charge opposite in sign to the inducing charge.✓ 1Answer C

Answer: C

Exam tip: In induction, the earth connection must be removed before the inducing charge is taken away. Otherwise the charge flows back and the sphere ends up neutral.

21IBP2S-030-03-03Charged conducting sphere Paper 1AEasy1 mark

A sensitive sensor is mounted inside a hollow spherical metal shell of radius R in a high-voltage laboratory. The shell carries a total charge +Q, and there are no other charges inside the shell. What is the electric field strength produced by this charge at the sensor, a distance R/2 from the centre?

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In electrostatic equilibrium, all the charge on a conductor sits on its outer surface. A spherical surface of radius R/2 drawn inside the shell encloses no charge. By symmetry and Gauss's law, the field everywhere inside the hollow shell is therefore zero, so A is correct.
Option B, kQ/(2R2), wrongly uses the linearly increasing field found inside a uniformly charged solid insulating sphere.
Option C, kQ/R2, wrongly assumes the field inside equals the value at the surface.
Option D, 4kQ/R2, wrongly treats the charge as a point charge at the centre for an internal point.
✓ 1Answer A

Answer: A

Common error: The 'point charge at the centre' model applies only at points on or outside the surface of a charged conducting sphere. Inside a hollow conductor with no enclosed charge, the field is zero.

22IBP2S-030-00-01Electric field strength Paper 1AEasy1 mark

Two small charged spheres in a dust-filter test rig are a distance r apart and exert an electric force of magnitude F on each other. The charge on each sphere is doubled and the separation is increased to 3r. What is the new magnitude of the force?

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Coulomb's law gives F ∝ q1q2/r2. Doubling both charges multiplies the force by 2 × 2 = 4. Tripling the separation divides it by 32 = 9. The new force is 4F/9. Option B uses 1/r instead of 1/r2. Option C doubles only one charge. Option D makes both mistakes.✓ 1Answer A

Answer: A

Common error: Both charges appear in the product q₁q₂. Doubling each one multiplies the force by four, not by two.

23IBP2S-030-00-02Magnetic field of a straight wire Paper 1AEasy1 mark

A vertical cable inside a lamp post carries a steady current directed vertically upwards. The magnetic field lines around the cable are viewed from directly above. Which describes them?

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The field of a long straight current forms closed circles centred on the wire, in planes perpendicular to it. Use the right-hand grip rule: point the thumb along the current (upwards, towards the viewer above), and the fingers curl anticlockwise as seen from above. Radial lines describe the electric field of a charged wire. Lines parallel to the wire confuse the field with the current direction.✓ 1Answer A

Answer: A

Exam tip: When current flows towards you, the field is anticlockwise. When it flows away from you, the field is clockwise.

24IBP2S-030-00-06Electric field lines Paper 1AEasy1 mark

Which statement about electric field lines is correct?

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By definition, the direction of a field line is the direction of the force on a positive test charge. Lines can never cross, because the field has only one direction at each point. A particle's path follows a field line only in special cases, because its velocity need not be along the force. The field is strongest where the lines are closest together.✓ 1Answer D

Answer: D

Exam tip: Field lines give the direction of force, and hence of acceleration, not of velocity.

25IBP2S-030-05-01Quantization of charge Paper 1AEasy1 mark

A toner particle in a laser printer carries a net charge. Which of the following could be the magnitude of the charge on the particle?

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Charge is quantized. Any net charge on an object must be a whole-number multiple of the elementary charge e = 1.60×10−19 C. 4.8×10−19 C = 3e, so it is possible. The other values are 0.75e, 1.5e and 3.5e. These are not integer multiples of e, so they cannot occur.✓ 1Answer C

Answer: C

Exam tip: Divide each option by 1.60×10⁻¹⁹ C. Only a whole-number result is allowed.

26IBP2S-030-05-02Quantization of charge Paper 1AEasy1 mark

A volleyball is rubbed against a synthetic court surface. It ends up with a charge of −8.0×10−9 C. Which statement describes the ball?

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A negative charge means the ball has gained electrons. The number of electrons is n = Q/e = 8.0×10−9 / 1.60×10−19 = 5.0×1010. Option B has the sign reversed. Option C multiplies Q by e instead of dividing. Option D inverts the ratio. A number of electrons must also be a whole number, not a tiny fraction.✓ 1Answer A

Answer: A

Common error: A negative object has an excess of electrons. Electrons are never 'lost' to make something negative.

27IBP2S-030-05-06Quantization of charge Paper 1AEasy1 mark

A nanoparticle charge detector records the following charges on four particles:
3.2×10−19 C, 4.8×10−19 C, 8.0×10−19 C, 11.2×10−19 C.
What is the largest value of a basic unit of charge that is consistent with all four readings?

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The readings are 2, 3, 5 and 7 times 1.6×10−19 C. This is the largest common factor. 0.80×10−19 C also divides into every reading, but it is not the largest value. 3.2×10−19 C does not divide into 4.8×10−19 C a whole number of times. 4.8×10−19 C does not divide into 3.2×10−19 C.✓ 1Answer B

Answer: B

Common error: The smallest measured charge is not necessarily the basic unit. Look for the largest common factor of all the readings.

28IBP2S-030-05-07Quantization of charge Paper 1AEasy1 mark

A student claims that a charged speck of dust carries a charge of +1.0×10−19 C. Why is this claim impossible?

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Charge is quantized. The net charge on any object is ne, where n is an integer and e = 1.60×10−19 C. A value of 1.0×10−19 C is 0.625e, which is not allowed. Objects become positive by losing electrons, so B is wrong. Being hard to measure (A) does not make a value impossible. D has no physical basis.✓ 1Answer C

Answer: C

Exam tip: Quantization means the charge is a whole number of e, whatever the size of the object.

29IBP2S-030-04-01Combining two electric fields Paper 1AEasy1 mark

A student draws electric field lines to show the field around a charged dust particle inside an electrostatic air purifier. Which statement about electric field lines is correct?

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A field line is defined so that its direction at any point is the direction of the force on a small positive test charge. The field at a point has only one direction, so lines cannot cross. A released particle accelerates in the direction of the force, but once it has velocity its path can leave a curved line, so it only follows a straight line exactly. Lines start on positive charges and end on negative charges.✓ 1Answer B

Answer: B

Common error: Do not confuse a field line with a particle's trajectory. The two coincide only for straight field lines or for a particle that has negligible inertia.

30IBP2S-030-04-02Conservation of charge Paper 1AEasy1 mark

In a laser-printer demonstration, two identical small conducting spheres carry charges of +6.0 nC and −2.0 nC. They are touched together and then separated. What is the charge on each sphere afterwards?

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Charge is conserved, so the total is +6.0 + (−2.0) = +4.0 nC. The spheres are identical, so this total is shared equally and each sphere carries +2.0 nC. Option B adds the magnitudes and ignores the sign. Option C halves each charge separately. Option D assumes that opposite charges cancel completely.✓ 1Answer A

Answer: A

Exam tip: Add the charges with their signs first, then share the total equally between the identical conductors.

31IBP2S-030-04-07Conservation of charge Paper 1AEasy1 mark

A footballer pulls a synthetic training shirt off over her head. The shirt gains a charge of −4.8 nC. Which statement is correct? (e = 1.60×10−19 C)

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The shirt becomes negative, so it gained electrons. Only electrons move during friction charging, because protons are bound in nuclei. The number of electrons is N = 4.8×10−9 / 1.60×10−19 = 3.0×1010. The hair is left with +4.8 nC, so total charge is conserved. Option D counts the charge on both objects instead of the charge transferred.✓ 1Answer A

Answer: A

Common error: Positive charging is always a loss of electrons, never a gain of protons.

32D-1A-11Electric field strength Paper 1AMedium1 mark

A small charge +q placed at point P experiences an electric force of magnitude F directed due east.

The charge is removed and replaced by a charge −2q. Which row gives the electric field strength at P and the force on the charge −2q?

Electric field strength at PForce on −2q
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Marking pointMarkNotes
Step 1E = F/q is fixed by the charges producing the field, so it stays F/q towards the east whatever test charge is used.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Force on −2q = (−2q)E: magnitude 2F, directed west (opposite to the field).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the field does not depend on the test charge, and a negative charge is pushed opposite to the field.
  • BThis reverses the field with the sign of the test charge. The field direction is defined by the force on a positive charge.
  • CThis divides the new force, 2F, by the old charge q; the field is (2F)/(2q) = F/q.
  • DThis ignores the sign of the new charge: a negative charge experiences a force opposite to the field.
33D-1A-13Coulomb's law Paper 1AMedium1 mark

Two small charged spheres in air exert an electric force F on each other. The permittivity of air is ε0.

The spheres are immersed in an insulating liquid of permittivity 5.0ε0, with the same charges and the same separation. What is the force between them now?

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Marking pointMarkNotes
Step 1In Coulomb's law k = 1/4πε, so F ∝ 1/ε for fixed charges and separation.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Multiplying the permittivity by 5.0 divides the force by 5.0: F/5.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis multiplies by the permittivity ratio; k is inversely proportional to ε.
  • BThis treats k as a universal constant; k = 1/4πε depends on the medium between the charges.
  • CCorrect: F = q1q2/4πεr², so a five-fold permittivity gives F/5.
  • DThis divides by 5² — the permittivity appears to the first power, not squared like r.
34D-1A-16Quantization of charge Paper 1AMedium1 mark

In a Millikan-type experiment the charges measured on three oil drops are 3.2 × 10−19 C, 6.4 × 10−19 C and 8.0 × 10−19 C.

What is the largest value of the elementary charge that is consistent with all three results?

A charged oil drop held stationary between two charged horizontal plates, with the electric force balancing its weight.+−qEmgdthe drop hangs at rest when qE balances mg
Diagram NOT accurately drawn
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Marking pointMarkNotes
Step 1If charge is quantized, every measured charge is a whole-number multiple of the elementary charge.—All 2 steps must be completed — there is no mark for a part-answer.
Step 23.2, 6.4 and 8.0 are 2, 4 and 5 times 1.6, and no larger value divides all three: 1.6 × 10−19 C.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis is 3.2 × 10−19/3, but 8.0 × 10−19 is 7.5 times this value, not a whole number.
  • BCorrect: the charges are 2, 4 and 5 times 1.6 × 10−19 C.
  • C8.0 × 10−19 C is 2.5 times this value, not a whole-number multiple.
  • DThis is the largest charge measured; the smaller charges are not whole multiples of it.
35D-1A-17Combining two electric fields Paper 1AMedium1 mark

Two point charges of equal magnitude are fixed a short distance apart, and M is the midpoint between them. Either charge on its own would produce an electric field of magnitude E at M.

Which row gives the magnitude of the resultant electric field at M when the charges are both positive and when they have opposite signs?

Both positiveOpposite signs
Show mark scheme
Marking pointMarkNotes
Step 1Both positive: each charge pushes a positive test charge at M away from itself, so the two fields are opposite and cancel.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Opposite signs: both fields point from the positive charge towards the negative charge, so they add to 2E.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis adds the magnitudes without regard to direction; field strength is a vector.
  • BThese are the wrong way round: at the midpoint like charges cancel and unlike charges reinforce.
  • CSymmetry alone does not make the field zero: for opposite charges the two fields point the same way.
  • DCorrect: like charges give opposite fields that cancel; unlike charges give fields in the same direction.
36D-1A-36Magnetic field of a straight wire Paper 1AMedium1 mark

A long straight wire is perpendicular to the page and carries a current into the page (⊗).

Which diagram best shows the magnetic field lines around the wire in the plane of the page?

Four diagrams A to D of possible magnetic field-line patterns around a wire carrying a current into the pageA.B.C.D.
Diagrams NOT accurately drawn
Show mark scheme
Marking pointMarkNotes
Step 1Right-hand grip rule: with the thumb along the current (into the page), the fingers curl clockwise as seen from the front.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The field weakens with distance from the wire, so the circles get farther apart going outwards.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe direction is right, but the circles get closer together outwards, which would mean a field that strengthens away from the wire.
  • BThis is the pattern for a current out of the page; the grip rule has been applied the wrong way round.
  • CRadial lines are the electric field of a point charge. The magnetic field lines of a wire are circles round it.
  • DCorrect: clockwise circles, spaced farther apart as the field weakens with distance.
37D-1A-38Coulomb's law Paper 1AMedium1 mark

Two point charges exert a force of magnitude F on each other. Both charges are doubled and their separation is trebled.

What is the magnitude of the new force?

Show mark scheme
Marking pointMarkNotes
Step 1F ∝ q1q2/r²: doubling both charges multiplies the force by 4, trebling r divides it by 9.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2New force = 4F/9.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis doubles only one of the charges.
  • BCorrect: × 2 × 2 for the charges and ÷ 3² for the separation gives 4F/9.
  • CThis divides by 3 instead of 3²; the separation appears squared in Coulomb's law.
  • DThis multiplies by 3² instead of dividing; the force falls as the separation increases.
38D-1A-50Magnetic field patterns Paper 1AMedium1 mark

The north-seeking end of a compass needle points towards the Earth's geographic North Pole.

Which row describes the magnetic pole of the Earth that lies near its geographic North Pole and the direction of the Earth's magnetic field lines there?

Magnetic poleField lines near the geographic North Pole
Show mark scheme
Marking pointMarkNotes
Step 1Unlike poles attract, so the north-seeking pole of the needle is attracted towards a magnetic south pole: the pole near the geographic North Pole is a magnetic south pole.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Outside a magnet the field lines run from its north pole to its south pole, so near the geographic North Pole the Earth's field lines point into the Earth.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis names the magnetic pole after the geographic pole; a north-seeking needle is attracted to a magnetic south pole, not to a north pole.
  • BCorrect: the needle's north pole is attracted to a magnetic south pole, and field lines end on a south pole, so they enter the Earth there.
  • CThe pole is right, but outside a magnet the field lines run towards a south pole, so they point into the Earth there, not out of it.
  • DThe field lines do point into the Earth there, but that is because the pole is a south pole: field lines leave a north pole, they do not enter it.
39D-1A-51Quantization of charge Paper 1AMedium1 mark

In a Millikan-type experiment, two oil drops X and Y are each held stationary in turn between the same horizontal parallel plates by adjusting the potential difference V between the plates.

The drops are made of the same oil. Drop X has twice the radius of drop Y. X carries four electron charges and Y carries two. Buoyancy can be ignored.

What is VX/VY?

Show mark scheme
Marking pointMarkNotes
Step 1For a stationary drop the electric force balances the weight: qV/d = mg, so V = mgd/q.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Same oil, twice the radius: mX = 2³mY = 8mY; and qX = 2qY.—
Step 3VX/VY = 8/2 = 4.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the mass to be proportional to the radius squared instead of the radius cubed: 4/2 = 2.
  • BCorrect: V ∝ m/q, and the mass ratio is 8 while the charge ratio is 2.
  • CThis uses the mass ratio but ignores the different charges on the drops.
  • DThis multiplies by the charge ratio instead of dividing: a larger charge needs a smaller field, not a larger one.
40D-1A-52Conservation of charge Paper 1AMedium1 mark

Two uncharged metal spheres X and Y stand on insulating supports and touch each other. A negatively charged rod is held close to X, without touching it. With the rod still in place, Y is moved away from X. The rod is then removed.

Which row describes the final charges on X and Y?

Charge on XCharge on Y
Show mark scheme
Marking pointMarkNotes
Step 1The negative rod repels free electrons, which move from X into Y: X becomes positive and Y gains an equal negative charge.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The spheres are separated while the rod is still there, so the electrons cannot return. Charge is conserved, so the total stays zero: X is positive and Y is negative, with charges of equal size.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis reverses the transfer: the negative rod repels electrons away from X, so X, the sphere nearer the rod, becomes positive.
  • BCharge is conserved: the electrons that left X are now on Y, so Y carries a negative charge equal in size to the positive charge on X.
  • CThe spheres were separated before the rod was removed, so the displaced electrons cannot flow back and both spheres stay charged.
  • DCorrect: electrons pushed from X into Y are trapped on Y when the spheres are separated, so X is positive and Y equally negative.
41D-1A-54Magnetic field of a straight wire Paper 1AMedium1 mark

The diagram shows a cross-section, in the plane of the page, through two long, straight, parallel wires X and Y that are perpendicular to the page. The wires carry equal currents, out of the page in X and into the page in Y. M is midway between them.

Either current alone produces a magnetic field of magnitude B at M.

Which row gives the magnitude and the direction of the resultant magnetic field at M?

Cross-section of two long parallel wires X and Y with midpoint MXYMcurrent out of pagecurrent into pagedd
Diagram NOT accurately drawn
MagnitudeDirection
Show mark scheme
Marking pointMarkNotes
Step 1Right-hand grip rule: the field lines of X are anticlockwise circles, so at M, to the right of X, the field of X points towards the top of the page.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The field lines of Y are clockwise circles, so at M, to the left of Y, the field of Y also points towards the top of the page. The two fields add: 2B towards the top of the page.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe magnitude is right, but both directions have been reversed, as if the grip rule had been applied the wrong way.
  • BOpposite currents do not cancel at the midpoint: on the line between the wires their fields point the same way.
  • CCorrect: both fields point towards the top of the page at M, so they add to 2B.
  • DThe magnetic field of a straight wire is tangential to circles round the wire, not along the line from the wire.
42D-1A-70Charging by friction & induction Paper 1AMedium1 mark

A sensitive electronic instrument is placed inside a closed, uncharged metal box that stands on an insulating base. The box is then placed in a strong external electric field. The electric field inside the box is zero.

Which statement explains why the field inside the box is zero?

Show mark scheme
Marking pointMarkNotes
Step 1The external field pushes free electrons in the metal towards one side of the box, leaving the opposite side positive (induction); the net charge of the box stays zero.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The electrons stop moving only when the field of the separated charges cancels the external field in the metal and in the space it encloses, so the field inside the box is zero.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThe field is not absorbed: external field lines end on the induced negative charge on one side of the box and begin again on the induced positive charge on the other side.
  • BCorrect: the induced charges on the outer surface produce a field that cancels the external field everywhere inside.
  • CField lines pass through insulators such as plastic or glass; only a conductor, whose free electrons can move, shields the space inside it.
  • DCharge is conserved: an isolated box cannot gain a net charge by induction; its charges are only separated, and the field lines end and begin on them rather than being repelled.
43D-1A-71Combining two electric fields Paper 1AMedium1 mark

Two equal positive point charges are fixed at points P and R. X is a point on the straight line between P and R, at a distance x from P. E is the component of the resultant electric field at X in the direction from P towards R.

Which graph shows how E varies with x between P and R?

Four sketch graphs of the field component E against x between the two chargesA.Ex0B.Ex0C.Ex0D.Ex0
Sketch graphs NOT to scale
Show mark scheme
Marking pointMarkNotes
Step 1Between the charges, the field of P points away from P (towards R) and the field of R points away from R (towards P), so the two contributions have opposite signs.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Near P the field of P dominates (E positive and very large close to P); at the midpoint the two cancel (E = 0); near R the field of R dominates (E negative and very large in magnitude close to R).✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis adds the magnitudes of the two fields as if they were in the same direction. Between like charges the fields are opposite, so they cancel at the midpoint.
  • BThis has the directions reversed: the field of a positive charge points away from it, so close to P the component towards R is positive, not negative.
  • CCorrect: positive near P, zero at the midpoint, negative near R, and symmetrical about the midpoint.
  • DThis shows the field of P alone. The field of R must be added; it is opposite to that of P and grows large close to R.
44D-1A-73Charging by friction & induction Paper 1AMedium1 mark

After a journey, friction with the air has left the metal body of a road tanker, which rests on insulating rubber tyres, with a positive charge of 4.8 μC. Before unloading, the driver connects the body to earth with a metal cable, and the body becomes uncharged.

Which row gives the number of electrons that pass through the cable and the direction in which they move?

Number of electronsDirection of electron movement
Show mark scheme
Marking pointMarkNotes
Step 1The body is positive, so it has fewer electrons than protons; to become neutral it must gain electrons, which flow from the earth through the cable.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Charge is conserved: number of electrons = 4.8 × 10−6/1.60 × 10−19 = 3.0 × 1013.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis multiplies the charge by e instead of dividing; the number of electrons must be Q/e.
  • BThis assumes the tanker keeps half its charge, as if it shared the charge equally with the earth. The earth takes up as much charge as needed, so the tanker ends uncharged.
  • CThe number is right, but electrons leaving a positive body would make it more positive; neutralising a positive charge needs electrons to flow onto it.
  • DCorrect: 3.0 × 1013 electrons flow from the earth onto the positive tanker.
45D-1A-74Combining two electric fields Paper 1AMedium1 mark

Point charges of +2.0 nC and −2.0 nC are fixed in a vacuum at the corners A and B of an equilateral triangle ABC of side 0.30 m.

Which row gives the magnitude and the direction of the resultant electric field at C?

Equilateral triangle ABC with a positive charge at A and a negative charge at B+−A +2.0 nCB −2.0 nCC0.30 m
Diagram NOT to scale
Magnitude / N C−1Direction
Show mark scheme
Marking pointMarkNotes
Step 1Each charge alone gives E = kq/r² = 8.99 × 109 × 2.0 × 10−9/0.30² = 2.0 × 102 N C−1 at C: away from A along AC, and towards B along CB.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The two equal vectors are at 120° to each other. Their components at right angles to AB cancel and their components parallel to AB add: resultant = 2 × 2.0 × 102 × cos 60° = 2.0 × 102 N C−1, from A towards B.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the two fields have equal magnitudes and are at 120°, so the resultant has the same magnitude, directed parallel to AB from the positive charge towards the negative charge.
  • BThe magnitude is right, but both fields have been reversed: the field points away from a positive charge and towards a negative charge, so the resultant points from A towards B.
  • CThis takes the field of the negative charge at B as pointing away from B, as for a positive charge; it points towards B.
  • DThis adds the magnitudes as if the two fields were in the same direction. Field strength is a vector and the two fields are at 120° to each other.
46D-1A-77Coulomb's law Paper 1AMedium1 mark

In Coulomb's law the constant k = 1/4πε0, where ε0 is the permittivity of free space.

What are the SI base units of ε0?

Show mark scheme
Marking pointMarkNotes
Step 1From F = q1q2/4πε0r², ε0 = q1q2/4πFr², so its unit is C² N−1 m−2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With C = A s and N = kg m s−2: A² s² × kg−1 m−1 s² × m−2 = kg−1 m−3 s4 A2.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis is the unit of k. Because k = 1/4πε0, the unit of ε0 is the inverse of the unit of k.
  • BThis takes the coulomb as A s−1. Charge = current × time, so 1 C = 1 A s.
  • CCorrect: C² N−1 m−2 = kg−1 m−3 s4 A2.
  • DThis uses r instead of r² in Coulomb's law, leaving m−2 instead of m−3.
47D-1A-81Magnetic field patterns Paper 1AMedium1 mark

A flat circular coil of wire lies in the plane of the page. The current I in the coil is clockwise as seen in the diagram. Point X is at the centre of the coil. Point Y is in the plane of the page, a short distance outside the coil.

Which row gives the direction of the magnetic field due to the current at X and at Y?

Flat circular coil in the plane of the page with a clockwise current; X at the centre and Y just outside the coilIXYcoil
Diagram NOT to scale
Field at XField at Y
Show mark scheme
Marking pointMarkNotes
Step 1Right-hand grip rule for the coil: with the fingers curled in the direction of the clockwise current, the thumb points into the page, so the field at X is into the page.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Magnetic field lines are closed loops: the lines that pass into the page through the coil return outside it. Close to Y the nearby wire dominates, and its field there is out of the page.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis applies the grip rule with the wrong hand or against the current: both directions are reversed.
  • BThis takes the field to have the same direction everywhere in the plane of the coil. The field lines are closed loops, so outside the coil they return in the opposite direction.
  • CThis treats the coil like an ideal long solenoid, with no field outside. Just outside the wire of a flat coil the field is strong, like that close to a straight wire.
  • DCorrect: the field is into the page inside the coil and out of the page just outside it.
48D-1A-83Coulomb's law Paper 1AMedium1 mark

A proton and an alpha particle are released from rest a short distance apart in a vacuum. The charge of the alpha particle is 2e and its mass is 4 times the mass of the proton. Only the electric force acts.

Which row gives (magnitude of force on the proton)/(magnitude of force on the alpha particle) and (magnitude of acceleration of the proton)/(magnitude of acceleration of the alpha particle) when they are released?

Force ratioAcceleration ratio
Show mark scheme
Marking pointMarkNotes
Step 1The two forces are a Newton's third law pair, F = kq1q2/r² on each, so the force ratio is 1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2a = F/m: the same force on a quarter of the mass gives the proton 4 times the acceleration.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis takes the force on each particle as proportional to its own charge only. Coulomb's law involves the product of the charges, which is the same for both.
  • BThe forces are equal, but equal forces give equal accelerations only for equal masses.
  • CCorrect: equal and opposite forces, and a ∝ 1/m gives a ratio of 4.
  • DThis takes the force on each particle as proportional to the charge of the other particle alone; the force on each is kq1q2/r², the same for both.
49D-1A-84Uniform fields between plates Paper 1AMedium1 mark

Two large, parallel metal plates a small distance apart are connected to a power supply. Point X moves along the straight line through the centres of the plates, at right angles to them. It starts outside the plates, passes through the gap between them and ends outside the plates on the other side. The dashed lines in the sketch graphs show the positions of the positive and negative plates.

Which graph best shows how the magnitude E of the electric field strength at X varies with the position x of X?

Four sketch graphs of field strength E against position x along the axis through the platesA.Ex0B.Ex0C.Ex0D.Ex0+−+−+−+−
Sketch graphs NOT to scale
Show mark scheme
Marking pointMarkNotes
Step 1Between the plates, away from the edges, the field lines are parallel and equally spaced, so E is uniform (E = V/d).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The field lines start on the positive plate and end on the negative plate, so beyond the plates (away from the edges) there are almost no field lines and E ≈ 0.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis makes the field strongest next to each plate, as if it came from point charges on the plates. Between large parallel plates the field is uniform, so E is the same everywhere in the gap.
  • BCorrect: E is uniform between the plates and almost zero outside them.
  • CThis continues the uniform field beyond the plates. Field lines run only from the positive plate to the negative plate, so outside the plates there are almost no field lines.
  • DThis treats each plate like a point charge whose field weakens gradually with distance. Outside the plates the fields of the two oppositely charged plates are opposite and cancel, so E is almost zero.
50D-1A-85Electric field strength Paper 1AMedium1 mark

Small test charges q, some positive and some negative, are placed in turn at point P near a fixed charged sphere S. The graph shows how the electric force F on the test charge varies with q. F is positive when it is directed away from S. The test charges do not disturb the charge on S.

Which row gives the electric field strength at P and the magnitude of the force on a +2.0 nC test charge at P after the charge on S has been doubled?

Force on the test charge against test charge, a straight line through the origin−5−4−3−2−1012345q / nC−12−8−404812F / μN
Graph drawn to scale
Field strength at P / N C−1Force on the +2.0 nC charge / μN
Show mark scheme
Marking pointMarkNotes
Step 1E = F/q = gradient of the graph = 10 × 10−6/4.0 × 10−9 = 2.5 × 103 N C−1.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The field at P is produced by S, so doubling the charge on S doubles E to 5.0 × 103 N C−1: F = qE = 2.0 × 10−9 × 5.0 × 103 = 1.0 × 10−5 N = 10 μN.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe force is right, but the field strength has been found as q/F. Field strength is force per unit charge, F/q, the gradient of the graph.
  • BThis leaves the field at P unchanged. The field at P is produced by S, so doubling the charge on S doubles the field and the force.
  • CCorrect: E = 2.5 × 103 N C−1; doubling the charge on S doubles E, so F = 10 μN.
  • DThis multiplies the force by 4, as if both charges in kq1q2/r² had been doubled; only the charge on S changes, so the force doubles.
51IBP2S-030-01-03Charging by friction & induction Paper 1AMedium1 mark

In a continuous inkjet printer, charged ink droplets pass between two parallel deflecting plates. The plates are 2.0 mm apart and have a potential difference of 1.5 kV across them. What is the magnitude of the electric field strength between the plates?

Show mark scheme
Marking pointMarkNotes
For parallel plates, E = V/d = 1500 V ÷ 2.0×10−3 m = 7.5×105 V m−1. The distractors come from common slips:
• 3.0 V m−1 comes from multiplying V by d.
• 1.3×10−6 V m−1 comes from dividing d by V.
• 7.5×102 V m−1 comes from leaving the separation in millimetres.
✓ 1Answer D

Answer: D

Common error: Always convert mm to m before using E = V/d. A factor-of-1000 error is the most frequent slip with plate separations.

52IBP2S-030-01-04Gravity compared with electrostatics Paper 1AMedium1 mark

A dust grain above the lunar surface is modelled as a point charge. The electric field strength due to the grain is E at a distance r. The charge on the grain then doubles. What is the field strength due to the grain at a distance 3r?

Show mark scheme
Marking pointMarkNotes
For a point charge, E = kQ/r2. Doubling Q doubles E. Tripling r divides E by 32 = 9. The new field is therefore (2/9)E. The distractors come from these mistakes:
• 2E/3 treats E as proportional to 1/r.
• 6E multiplies both factors instead of dividing by the distance factor.
• 9E/2 inverts the ratio.
✓ 1Answer A

Answer: A

Exam tip: Handle each change as a separate factor: the charge factor goes on top, and the distance factor is squared and goes underneath.

53IBP2S-030-01-05Combining two electric fields Paper 1AMedium1 mark

In a test rig for an electrostatic air filter, two small spheres carry charges +4q and +q. Their centres are a distance d apart. At what distance from the +4q sphere, on the line between them, is the resultant electric field strength zero?

+4q+qd?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At the null point the two fields are equal in size: k(4q)/x2 = kq/(d − x)2. Taking square roots gives 2/x = 1/(d − x), so x = 2d/3. This point is closer to the smaller charge, as it must be. The distractors come from these mistakes:
• d/3 measures the distance from the wrong charge.
• d/2 ignores the different charges.
• 4d/5 shares the distance in the linear ratio of the charges and forgets the inverse-square law.
✓ 1Answer B

Answer: B

Common error: The null point between two like charges always lies nearer the smaller charge. Use this to check that your answer is sensible.

54IBP2S-030-01-07Magnetic field of a straight wire Paper 1AMedium1 mark

A domestic mains cable contains two long, parallel wires. At one instant they carry equal currents in opposite directions. At a distance x, one wire alone would produce a magnetic field of magnitude B. The point P is midway between the wires, a distance x from each. What is the magnitude of the resultant magnetic field at P?

Show mark scheme
Marking pointMarkNotes
Apply the grip rule to each wire. For currents in opposite directions, the two fields point the same way in the region between the wires, so they add to give 2B. They would cancel to zero only if the currents were in the same direction. This is the usual source of the wrong answer. Outside the cable, the fields from the two wires largely cancel, which is why paired conductors produce very little external field.✓ 1Answer B

Answer: B

Exam tip: Between two parallel wires, currents in the same direction give cancelling fields and currents in opposite directions give adding fields. Check this with the grip rule rather than memorising it.

55IBP2S-030-02-03Combining two electric fields Paper 1AMedium1 mark

In a test rig for electrostatic paint spraying, two small charged spheres are fixed a distance d apart. Sphere X carries charge +4Q and sphere Y carries charge +Q. At which point on the line between them is the resultant electric field zero?

+4Q+Qd?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The resultant field is zero where the two field magnitudes are equal: 4kQ/(d − x)2 = kQ/x2, where x is the distance from Y. Taking square roots gives d − x = 2x, so x = d/3 from Y. This point is closer to the smaller charge, as expected. Option B puts the point nearer the larger charge. Option C comes from forgetting the square in the inverse-square law (d − x = 4x). Option D assumes the charges are equal.✓ 1Answer A

Answer: A

Exam tip: The null point between two like charges always lies closer to the smaller charge. Use this to check your answer.

56IBP2S-030-02-04Combining two electric fields Paper 1AMedium1 mark

A simple model of a polar molecule uses a charge +q and a charge −q separated by a distance 2a. What is the electric field at the midpoint between the charges?

Show mark scheme
Marking pointMarkNotes
Each charge produces a field of magnitude kq/a2 at the midpoint. The field of +q points away from +q, which is towards −q. The field of −q points towards −q. Both point the same way, so they add to 2kq/a2 towards −q (C). Option A applies the result for like charges. Option B counts only one charge. Option D has the direction reversed.✓ 1Answer C

Answer: C

Common error: Do not assume the fields cancel at a midpoint. Draw each field vector separately before adding them.

57IBP2S-030-02-05Magnetic field patterns Paper 1AMedium1 mark

In an electron-beam welding machine, electrons travelling horizontally enter the uniform vertical field between two horizontal deflecting plates. Gravitational effects are negligible. Which quantity stays constant while an electron is between the plates?

Show mark scheme
Marking pointMarkNotes
The field is vertical, so the electric force on the electron is vertical only. There is no horizontal force, so the horizontal velocity component stays constant (A). The vertical velocity component increases, so the speed and kinetic energy both increase (B and C are wrong). The velocity direction turns steadily as the electron follows a parabolic path (D is wrong).✓ 1Answer A

Answer: A

Exam tip: Treat the motion like a horizontally launched projectile, with the electric force playing the role of the weight.

58IBP2S-030-02-06Quantization of charge Paper 1AMedium1 mark

In an inkjet printer, a charged ink droplet has acceleration a while it passes between two deflecting plates. The potential difference across the plates is doubled and the plate separation is halved. The charge and mass of the droplet are unchanged. What is the new acceleration of the droplet?

Show mark scheme
Marking pointMarkNotes
The acceleration is a = qE/m = qV/(md). Doubling V multiplies a by 2, and halving d multiplies it by another 2. The new acceleration is therefore 4a (D). Option C ignores the change in d. Option B wrongly assumes the two changes cancel. Option A inverts the dependence on both quantities.✓ 1Answer D

Answer: D

Common error: Halving the separation increases the field strength, because E is proportional to 1/d.

59IBP2S-030-03-05Charged conducting sphere Paper 1AMedium1 mark

A charged conducting sphere of radius R produces an electric field of strength E just outside its surface. What is the field strength at a point 2R from the surface of the sphere?

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Outside the sphere the field is that of a point charge at the centre, E ∝ 1/r2, where r is measured from the centre. A point 2R from the surface is 3R from the centre, so the field is E/32 = E/9. Option C measures distance from the surface instead of the centre. Options A and B use an inverse (not inverse-square) relationship.✓ 1Answer D

Answer: D

Common error: Distances in Coulomb's law are measured from the centre of the sphere, not from its surface.

60IBP2S-030-03-06Conservation of charge Paper 1AMedium1 mark

Two identical small conducting spheres carry charges of +6 nC and −2 nC. They attract each other with a force of magnitude F. The spheres are touched together and then returned to their original separation. What is the force between them now?

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The total charge is +6 − 2 = +4 nC, which is shared equally, so each sphere has +2 nC. Originally the product of charge magnitudes was 6 × 2 = 12. Now it is 2 × 2 = 4, so the force is F/3, and like charges repel. Option C wrongly adds the magnitudes (8 nC shared as 4 nC each). Option D assumes the charges cancel completely.✓ 1Answer A

Answer: A

Exam tip: When identical conductors touch, add the charges with their signs, then share the total equally.

61IBP2S-030-03-07Charging by friction & induction Paper 1AMedium1 mark

A hospital technician tests a plastic instrument tray for static charge. She uses a gold-leaf electroscope that has been given a positive charge. When the tray is brought near the cap, the leaves diverge further. What can be concluded about the tray?

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Further divergence means more positive charge has moved to the leaves. A positively charged object repels the remaining free electrons up towards the cap, so the leaves become more positive. A negative object, or an uncharged object (through induced or polarised charge), would attract positive charge towards the cap and make the leaves collapse. Only increased divergence identifies the sign with certainty.✓ 1Answer C

Answer: C

Exam tip: Only repulsion, here increased divergence, proves an object is charged. Attraction, here collapse, can also happen with neutral objects.

62IBP2S-030-00-03Magnetic field of a straight wire Paper 1AMedium1 mark

A magnetic sensor in an electric-vehicle battery pack is a distance r from a long straight busbar and measures a field strength B. The current in the busbar is tripled and the sensor is moved to a distance r/2 from it. What field strength does the sensor now measure?

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For a long straight wire, B = μ0I/(2πr), so B ∝ I/r. Tripling I gives a factor of 3. Halving r gives a factor of 2. The new field is 6B. Option C treats B as proportional to 1/r2, as in Coulomb's law. Option B divides by 2 instead of multiplying. Option D uses 1/r2 inverted.✓ 1Answer A

Answer: A

Common error: The field of a straight wire falls off as 1/r, not as 1/r². Do not carry over the inverse-square law from point charges.

63IBP2S-030-00-04Magnetic field of a straight wire Paper 1AMedium1 mark

A two-core cable for a medical scanner consists of two long parallel conductors a distance d apart. Each carries current I, and the currents are in opposite directions. What is the magnitude of the magnetic field at the point midway between the conductors?

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Each conductor is d/2 from the midpoint, so each produces B = μ0I/(2π·d/2) = μ0I/(πd). The currents are opposite, so by the right-hand grip rule the two fields point in the same direction between the wires. They therefore add, giving 2μ0I/(πd). Option A is the result for currents in the same direction. Option B is the field of one wire only. Option D uses d instead of d/2 for the distance.✓ 1Answer C

Answer: C

Exam tip: Between the wires, opposite currents give fields that add. Outside the pair, their fields largely cancel.

64IBP2S-030-00-05Combining two electric fields Paper 1AMedium1 mark

A sodium ion and a chloride ion are a fixed distance apart. When they are in vacuum, the electric force between them is F. The same pair is now placed at the same separation in water, which has a permittivity about 80 times that of vacuum. What is the force between them now?

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Coulomb's law is F = q1q2/(4πεr2), so F ∝ 1/ε. Increasing the permittivity by a factor of 80 reduces the force by the same factor, to F/80. This is one reason ionic solids dissolve readily in water. Option B inverts the dependence. Options C and D apply a square root or a square that is not in the law.✓ 1Answer A

Answer: A

Common error: Permittivity appears to the first power in the denominator. Only the separation is squared.

65IBP2S-030-00-07Combining two electric fields Paper 1AMedium1 mark

In an ion-trap design, a point charge +4q is fixed at x = 0 and a point charge +q is fixed at x = d. At what position between them is the resultant electric force on a small test charge zero?

+4q+qdtest charge
Diagram not to scale
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For zero resultant force, 4q/x2 = q/(d − x)2. Taking square roots gives 2(d − x) = x, so x = 2d/3. This point is closer to the smaller charge, as expected. Option C comes from using 1/r instead of 1/r2. Option B puts the point nearer the larger charge. Option D ignores the difference between the charges.✓ 1Answer A

Answer: A

Exam tip: The null point between two like charges always lies closer to the smaller charge.

66IBP2S-030-00-09Magnetic field of a straight wire Paper 1AMedium1 mark

A hiker's compass is placed near a long vertical cable carrying a current of 10 A. The horizontal component of the Earth's magnetic field is 2.0 × 10−5 T. At what distance from the cable does the cable's field have the same magnitude as this component? (μ0 = 4π × 10−7 T m A−1)

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Rearranging B = μ0I/(2πr) gives r = μ0I/(2πB) = (2 × 10−7 × 10)/(2.0 × 10−5) = 0.10 m. Option B omits the factor of 2 in 2π. Option C omits 2π altogether. Option D uses 4π instead of 2π.✓ 1Answer A

Answer: A

Exam tip: μ₀/(2π) = 2 × 10⁻⁷ T m A⁻¹ exactly. Using this value makes wire-field calculations quick.

67IBP2S-030-05-03Gravity compared with electrostatics Paper 1AMedium1 mark

Two protons in a particle-accelerator bunch are a distance d apart. The ratio of the electric force to the gravitational force between them is R. The separation is increased to 2d. What is the new ratio?

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Both forces follow an inverse-square law. FE = ke2/r2 and FG = Gm2/r2. The ratio is FE/FG = ke2/(Gm2), which does not depend on r. Each force falls to a quarter of its value, so the ratio stays at R.✓ 1Answer C

Answer: C

Exam tip: When two quantities depend on distance in the same way, their ratio does not depend on distance at all.

68IBP2S-030-05-04Gravity compared with electrostatics Paper 1AMedium1 mark

The electric force between two protons is far larger than the gravitational force between them. Yet gravity governs the motion of planets and moons. Which statement best explains this?

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Mass comes in only one sign, so gravitational effects of all particles add together. Charge comes in two signs, and planets are almost exactly neutral. The attractive and repulsive electric forces therefore nearly cancel, leaving gravity dominant. G (≈10−11) is much smaller than k (≈1010). Both forces are inverse-square. Electric fields do exist in a vacuum.✓ 1Answer B

Answer: B

Exam tip: A key difference to quote: gravity is always attractive, while electric forces can attract or repel.

69IBP2S-030-05-05Gravity compared with electrostatics Paper 1AMedium1 mark

Two identical dust grains in deep space each carry the same charge q. The electric repulsion between them exactly balances their gravitational attraction. Two other identical grains each have twice the mass. What charge must each of these grains carry for the forces to balance?

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For balance, kq2/r2 = Gm2/r2. This gives q = m√(G/k), so q is directly proportional to m. Doubling the mass means the charge must double. √2 q comes from wrongly taking q2 ∝ m. 4q comes from taking q ∝ m2.✓ 1Answer C

Answer: C

Common error: Both sides contain a product of two equal quantities (q·q and m·m). Take the square root of both sides before reasoning about proportionality.

70IBP2S-030-05-08Quantization of charge Paper 1AMedium1 mark

In an evacuated tube, a uniform vertical electric field is adjusted until the electric force on a single electron exactly balances its weight. What is the electric field strength?
(me = 9.1×10−31 kg, e = 1.6×10−19 C, g = 9.8 m s−2)

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For balance, eE = meg, so E = meg/e = (9.1×10−31 × 9.8)/(1.6×10−19) ≈ 5.6×10−11 N C−1. Option B uses the proton mass instead of the electron mass. Option C inverts the ratio, giving e/(meg). Option D multiplies by e instead of dividing by it. Such a tiny field shows how weak gravity is compared with electric forces on an electron.✓ 1Answer A

Answer: A

Common error: Use the mass of the particle named in the question. Electrons and protons have the same size of charge but very different masses.

71IBP2S-030-05-09Conservation of charge Paper 1AMedium1 mark

A sensor contains two identical tiny conducting beads. One bead carries a charge of +5e and the other carries −2e. The beads touch and are then separated. Which is a possible final distribution of charge?

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Charge is conserved, so the total charge is +5e − 2e = +3e. Charge is also quantized, so each bead must carry a whole number of e. An equal split of +1.5e each is therefore impossible. The beads must share the charge as +2e and +1e. Option C breaks conservation of charge, because it gives a total of 6e. Option D ignores the electron transfer that happens when conductors touch.✓ 1Answer B

Answer: B

Exam tip: Check two things for any charge-sharing problem: conservation (the total is unchanged) and quantization (each object has a whole number of e).

72IBP2S-030-04-03Electric field lines Paper 1AMedium1 mark

The metal skin of an aircraft becomes charged during flight. Once the charges have stopped moving, which describes the electric field lines just outside the aircraft's surface?

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In electrostatic equilibrium there is no field component along a conductor's surface, because such a component would move the free charges. The lines must therefore meet the surface at 90°. Charge density is greatest where the surface curvature is greatest, so the lines crowd together near sharp points. This is why static wicks are fitted at the wing tips. The field inside the metal is zero.✓ 1Answer C

Answer: C

Common error: A conductor in equilibrium has zero field inside it and no field component along its surface, so field lines are always perpendicular to the surface.

73IBP2S-030-04-04Combining two electric fields Paper 1AMedium1 mark

A field-line diagram shows two small charged spheres X and Y. Eighteen field lines leave X. Six of these lines end on Y, and the remaining twelve continue to very large distances. No lines leave Y. Which statement is correct?

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Lines leave positive charges and end on negative charges, so X is positive and Y is negative. The number of lines drawn is proportional to the magnitude of the charge. X has 18 lines and Y has 6, so |QX| = 3|QY|.✓ 1Answer D

Answer: D

Exam tip: In a correctly drawn diagram, the number of lines attached to each charge is proportional to the magnitude of that charge. Count them.

74IBP2S-030-04-05Conservation of charge Paper 1AMedium1 mark

Fluorine-18 (proton number 9) is used as a tracer in PET medical scanners. Which equation for its decay conserves electric charge?

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Compare the total charge on each side, in units of e. On the left the charge is +9. In B the right side gives +8 + 1 + 0 = +9, so charge is conserved. A gives +8 − 1 = +7. C gives +10 + 1 = +11. D gives +8. Only B conserves charge.✓ 1Answer B

Answer: B

Exam tip: Write the charge of every particle under the equation, using +1 for a positron, −1 for an electron and 0 for a neutrino, then check that both sides have the same total.

75IBP2S-030-04-06Charging by friction & induction Paper 1AMedium1 mark

In an industrial inkjet printer, a charged ink droplet passes through the central region between two parallel deflection plates held at a constant potential difference. Which statement about the electric force on the droplet in this region is correct?

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In the central region the field lines are parallel and equally spaced, so the field is uniform. The force F = qE therefore has the same magnitude and direction everywhere in that region, acting perpendicular to the plates. The field does not weaken towards the middle, and the two plates do not cancel each other: both contribute a field in the same direction.✓ 1Answer C

Answer: C

Common error: Treating each plate like a point charge, with a field that falls off with distance, is wrong for parallel plates. Between them the field is uniform, apart from the edges.

76IBP2S-030-04-09Electric field lines Paper 1AMedium1 mark

An isolated positive point charge is modelled by a field-line diagram. At distance r from the charge, n field lines pass through a small area held perpendicular to the lines. How many lines pass through the same area at distance 2r?

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The same total number of lines spreads over a sphere of area 4πr2. Doubling r multiplies this area by 4, so the number of lines per unit area falls to one quarter. The same small area therefore has n/4 lines, which matches E ∝ 1/r2. The total number of lines stays constant, but their density does not. Option n/2 wrongly assumes E ∝ 1/r. Option n/8 wrongly treats the spreading as depending on volume (r3).✓ 1Answer C

Answer: C

Exam tip: Field-line density represents field strength. For a point charge, the density follows the inverse-square law.

77D-1A-14Uniform fields between plates Paper 1AHard1 mark

Two parallel plates 0.050 m apart have a potential difference of 500 V between them. An electron is released from rest at the negative plate and accelerates to the positive plate.

Which row gives the kinetic energy of the electron when it reaches the positive plate?

positive platenegative plate0.050 mp.d. = 500 Velectron
Diagram not to scale
Kinetic energy / eVKinetic energy / J
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Step 1Work done on the electron = qV: a charge e moving through 500 V gains 500 eV, whatever the plate separation.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2In joules: 500 × 1.60 × 10−19 = 8.0 × 10−17 J.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis multiplies the p.d. by the separation (500 × 0.050 = 25). The energy gained is qV and does not involve d.
  • BCorrect: Ek = eV = 500 eV = 8.0 × 10−17 J.
  • CThe electronvolt value is right, but it has been divided by e instead of multiplied to convert to joules.
  • DThis uses the field strength V/d = 1.0 × 104 V m−1 in place of the p.d.
78D-1A-18Charging by friction & induction Paper 1AHard1 mark

A negatively charged rod is brought near one end of an isolated metal sphere on an insulating stand. The far side of the sphere is then briefly earthed, the earth connection is removed, and finally the rod is taken away.

What is the final charge on the sphere?

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Step 1The negative rod repels free electrons in the metal to the far side of the sphere.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Earthing gives those electrons a path to earth, so the sphere loses negative charge.—
Step 3The earth connection is broken before the rod is removed, so the electrons cannot return: the sphere is left positively charged.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: electrons repelled by the rod escape to earth, and removing the earth before the rod leaves a net positive charge.
  • BEarthing lets charge flow only while the rod holds the electrons away; once the earth is removed the sphere keeps a net charge.
  • CNo charge passes from the rod to the sphere — they never touch. This would be charging by contact.
  • DThat describes the sphere before it is earthed. Earthing lets electrons leave, so a net charge remains.
79D-1A-37Electric field lines Paper 1AHard1 mark

In a three-dimensional model of an electric field, neighbouring field lines in region P are twice as far apart as those in region Q, measured in every direction at right angles to the field.

What is (field strength in Q) : (field strength in P)?

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Step 1Field strength is shown by the number of lines crossing unit area at right angles to the field.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Halving the spacing in both directions across the lines divides the area per line by 2 × 2 = 4, so the field in Q is four times that in P.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • ACloser lines mean a stronger field, so Q must have the larger value.
  • BThis inverts the ratio; the closely spaced region is the stronger one.
  • CThis allows for the spacing in one direction only; the number of lines per unit area varies as 1/spacing².
  • DCorrect: lines per unit area in Q are 2² = 4 times those in P.
80D-1A-75Quantization of charge Paper 1AHard1 mark

A charged oil drop is held stationary between two horizontal parallel plates. The potential difference between the plates is then reversed, keeping the same magnitude. Ignore the buoyancy force of the air.

Which row gives the magnitude and the direction of the initial acceleration of the drop? g is the acceleration of free fall.

Magnitude of accelerationDirection
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Step 1While the drop is stationary, the electric force is upwards and equal in magnitude to the weight, mg.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Reversing the p.d. reverses the field, so the electric force on the same charge is now downwards with the same magnitude mg.—
Step 3The drop starts from rest, so there is no drag at first: resultant force = 2mg downwards and a = 2g downwards.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the weight and the reversed electric force, each of magnitude mg, both act downwards.
  • BThe magnitude is right, but the reversed electric force acts downwards, in the same direction as the weight.
  • CThis treats the electric force as switched off. Reversing the field reverses the electric force; it does not remove it.
  • DThe forces keep the same magnitudes, but the electric force has changed direction, so they no longer balance.
81D-1A-78Electric field lines Paper 1AHard1 mark

The diagram shows the electric field lines around two point charges X and Y. Twelve field lines leave X.

Which row gives the sign of the charge on X and the value of (magnitude of the charge on X)/(magnitude of the charge on Y)?

Electric field lines around two point charges X and Y, with arrowsXY
Diagram NOT to scale
Sign of charge on XCharge on X / charge on Y
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Step 1The arrows point away from X, so field lines start on X: X is positive (and Y, where lines end, is negative).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The number of lines at a charge is proportional to the magnitude of the charge: 12 lines leave X and 4 end on Y, so the ratio is 12/4 = 3.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: lines leave X, so X is positive, and X has three times as many lines as Y.
  • BThe ratio is right, but field lines start on positive charges and end on negative charges, so X, where the lines start, is positive.
  • CThis is (charge on Y)/(charge on X); X has more lines, so it has the larger charge.
  • DThis counts only the four lines that join X to Y. The eight lines that go off to a large distance also leave X and must be counted.
82D-1A-79Electric field lines Paper 1AHard1 mark

Which statement about electric field lines is correct?

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Step 1The tangent to a field line at a point gives the direction of the force on a positive test charge there. A force has only one direction at a point, so if two lines crossed, the force would have two directions.✓ 1Answer B

Answer: B  ·  1 stage of work, one mark

Every option, and why

  • AA field line gives the direction of the force, and so of the acceleration, not of the velocity. Once the charge is moving it cannot follow a curved line exactly; it follows a line only where the lines are straight.
  • BCorrect: the field has one direction at each point, so field lines never cross.
  • CThis reverses the convention: field lines start on positive charges and end on negative charges.
  • DField strength is shown by how close the lines are, and that changes along a line, for example a radial line from a point charge, along which the field weakens.
83D-1A-82Magnetic field patterns Paper 1AHard1 mark

The solenoid of an electric door lock is an air-core coil with ends P and Q. When the coil is viewed looking at end P, the current in the coil is clockwise. A small compass is placed inside the solenoid, on its axis.

Which row gives the magnetic pole at end P and the end towards which the north-seeking end of the compass needle points?

End view of the solenoid from end P showing a clockwise current, and a side view with ends P and Q and a compass insideend Pview looking at end PcompassPQside view
Diagram NOT to scale
Pole at end PCompass needle points towards
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Step 1Right-hand grip rule: fingers along the clockwise current give a thumb pointing away from the viewer, so inside the solenoid the field points from P towards Q.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Field lines enter the solenoid at P and leave at Q, so Q behaves as a north pole and P as a south pole.—
Step 3The compass needle lines up with the field, so its north-seeking end points along the field inside: towards Q.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis reverses the grip rule: a clockwise current seen from P gives a field pointing away from the viewer, so P is a south pole and the field inside points towards Q.
  • BCorrect: P is a south pole, and inside the solenoid the field (and so the needle) points from P towards Q.
  • CThe pole is right, but inside a magnet the field runs from south to north. A north-seeking pole points towards a south pole only outside the magnet.
  • DThe needle direction is right, but field lines leave a north pole; the field enters the solenoid at P, so P is a south pole.
84IBP2S-030-02-10Uniform fields between plates Paper 1AHard1 mark

In a particle-steering device, one pair of plates produces a uniform vertical field of 3.0 kV m−1. A second pair produces a uniform horizontal field of 4.0 kV m−1 in the same region. An electron is released from rest in this region. Gravitational effects are negligible. Which describes the subsequent motion of the electron?

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The two uniform fields add as vectors. The resultant is a single uniform field of magnitude √(3.02 + 4.02) = 5.0 kV m−1. The electron starts from rest under a constant force, so it accelerates in a straight line. Its charge is negative, so it moves opposite to the resultant field (C). A parabola (A) needs an initial velocity that is not parallel to the force. Option B ignores the negative charge. Option D adds the magnitudes as scalars.✓ 1Answer C

Answer: C

Exam tip: A parabolic path needs an initial velocity at an angle to a constant force. A particle starting from rest always moves in a straight line along the force.

85IBP2S-030-03-10Charging by friction & induction Paper 1AHard1 mark

A small charge +q is held at the centre of an uncharged hollow metal sphere without touching it. The sphere is connected to earth, and then the connection is removed. The charge +q remains at the centre. What are the charges on the inner and outer surfaces of the metal sphere?

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The field inside the metal must be zero, so the inner surface always carries −q, induced by +q. Before earthing, the outer surface carries +q. Earthing lets electrons flow up from earth and neutralise this outer charge, leaving 0 on the outer surface. The sphere's net charge is now −q. Option A describes the situation before earthing. Options C and D ignore the induced inner charge.✓ 1Answer B

Answer: B

Common error: Earthing removes charge from the outer surface only. The induced charge on the inner surface stays, because the enclosed charge holds it there.

86IBP2S-030-00-08Conservation of charge Paper 1AHard1 mark

Two identical small conducting spheres in a static-electricity sensor carry charges of +6 nC and −2 nC. They attract each other with a force of magnitude F. The spheres are briefly touched together and then returned to their original separation. What is the new force between them?

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Charge is conserved: the total is +6 − 2 = +4 nC. Identical spheres share this equally, so each carries +2 nC. The force is then repulsive, with magnitude proportional to 2 × 2 = 4 compared with 6 × 2 = 12 originally. The new force is therefore F/3. Option C shares the sum of the magnitudes (8 nC) instead of the net charge. Option B forgets that both charges are now positive. Option D assumes the charges cancel completely.✓ 1Answer A

Answer: A

Common error: When conductors touch, share the algebraic total charge, including signs, not the sum of the magnitudes.

87IBP2S-030-04-08Conservation of charge Paper 1AHard1 mark

Three identical conducting spheres are mounted on insulating stands. Sphere A carries +9.0 nC, sphere B carries −3.0 nC and sphere C is uncharged. A is touched to B and then separated. B is then touched to C and separated. What is the final charge on C?

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When A touches B, the total charge is +9.0 − 3.0 = +6.0 nC, so each sphere then carries +3.0 nC. When B (+3.0 nC) touches C (0 nC), each ends with +1.5 nC. Option A assumes all three spheres share the total at once. Option C comes from ignoring the sign of B: (9 + 3)/2 = 6, then 6/2 = 3. Option D assumes B is still negative.✓ 1Answer B

Answer: B

Exam tip: Work through the contacts one at a time, recording the charge on every sphere after each step.

88D-1B-04Charging by friction & induction Paper 1BEasy8 marks

A student investigates the charges produced when a polythene strip is rubbed with a woollen cloth. After rubbing, the strip and then the cloth are each lowered into a metal can (a Faraday pail) connected to a digital coulombmeter, which measures the charge on the object. The student holds the cloth in one hand and the strip by an insulating handle.

The procedure is repeated five times. Each charge reading has an uncertainty of ±0.4 nC.

TrialCharge on strip / nCCharge on cloth / nC
1−12.4+12.1
2−8.7+8.9
3−15.3+14.6
4−10.2+10.5
5−6.9+6.8
(a)

Explain, in terms of electrons, why the strip and the cloth gain charges of opposite sign.

(2)
(b)

Calculate the total charge on the strip and the cloth for trial 3, and its absolute uncertainty.

(2)
(c)

Deduce whether the results of all five trials are consistent with the conservation of charge.

(1)
(d)

Calculate the number of electrons transferred in trial 1.

(1)
(e)

In three of the trials the magnitude of the charge on the cloth is slightly smaller than that on the strip. Suggest a reason for this and one improvement to the method.

(2)
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Part (a)
Rubbing transfers electrons from the cloth to the polythene✓ 1Do not accept transfer of positive charge/protons.
The strip gains electrons and becomes negative; the cloth loses the same number and becomes (equally) positive✓ 1
Part (b)
total = −15.3 + 14.6 = −0.7 nC✓ 1
uncertainty = 0.4 + 0.4 = ±0.8 nC, so total = (−0.7 ± 0.8) nC✓ 1Absolute uncertainties add for a sum.
Part (c)
All five totals (−0.3, 0.2, −0.7, 0.3, −0.1 nC) are zero within ±0.8 nC, so the results are consistent with conservation of charge✓ 1Allow ECF from (b).
Part (d)
N = 12.4 × 10−9/1.60 × 10−19 = 7.8 × 1010✓ 1Accept 7.8 × 1010.
Part (e)
Some charge flows from the cloth through the student's hand/body to earth (the body is a conductor)✓ 1OWTTE
Hold the cloth with insulating tongs / an insulating handle or wear an insulating glove✓ 1

Answers: (b) (−0.7 ± 0.8) nC  ·  (d) 7.8 × 1010 electrons (the remaining parts are explanations — see the table above)

89D-1B-13Uniform fields between plates Paper 1BEasy8 marks

A small conducting ball of mass 0.250 g hangs from an insulating thread of length 0.300 m midway between two vertical parallel plates separated by (120 ± 1) mm. The ball carries a fixed charge q. When a potential difference V is applied across the plates, the ball is pushed sideways and its horizontal displacement x is read from a millimetre scale behind it, with an uncertainty of ±1 mm. The uncertainty in V is ±0.1 kV.

For small displacements, the horizontal electric force on the ball is F = mgx/L, where L is the length of the thread.

V / kVx / mm
1.05.5
2.010.0
3.015.5
4.020.0
5.026.0
6.030.5
Horizontal displacement of a charged ball against plate voltage, with error bars01234567V / kV05101520253035x / mm
Graph drawn to scale
(a)

Show that the electric force on the ball when V = 6.0 kV is about 2.5 × 10−4 N.

(1)
(b)

Draw the line of best fit for the data.

(1)
(c)

Determine the gradient of the line.

(2)
(d)

Determine the charge q on the ball.

(2)
(e)

Calculate the percentage uncertainty in the electric field strength between the plates when V = 6.0 kV.

(1)
(f)

Charge slowly leaks from the ball into the air during the experiment. Suggest one way to reduce the effect of this on the results.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = 0.250 × 10−3 × 9.81 × 0.0305/0.300 = 2.49 × 10−4 N✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (b)
Single straight line through all the error bars and (close to) the origin✓ 1
Part (c)
Gradient calculated from a large triangle✓ 1
gradient = 5.11 mm kV−1 (= 5.11 × 10−6 m V−1)✓ 1Accept 4.9 to 5.3 mm kV−1. Unit required.
Part (d)
qV/d = mgx/L, so gradient = qL/(mgd) and q = gradient × mgd/L✓ 1
q = 5.11 × 10−6 × 0.250 × 10−3 × 9.81 × 0.120/0.300 = 5.0 × 10−9 C✓ 1Allow ECF from (c). Accept 4.8–5.2 × 10−9 C.
Part (e)
(0.1/6.0 + 1/120) × 100 = 2.5 %✓ 1Accept 2 % to 3 %.
Part (f)
Recharge the ball to the same charge before each reading OR take all the readings quickly OR carry out the experiment in dry air✓ 1OWTTE

Answers: (a) 2.49 × 10−4 N  ·  (c) 5.11 mm kV−1  ·  (d) q = 5.0 × 10−9 C  ·  (e) 2.5 % (the remaining parts are explanations — see the table above)

90D-1B-20Uniform fields between plates Paper 1BEasy8 marks

Two square metal plates of side 0.10 m are held parallel and connected to a 2.00 kV supply. An electric field meter measures the field strength E at the point midway between the centres of the plates. The separation d of the plates is varied from 20 mm to 125 mm. The uncertainty in each value of E is ±0.5 kV m−1.

The graph shows E against 1/d.

d / mm1/d / m−1E / kV m−1
2050.0100.2
3033.366.1
4025.050.0
6016.733.1
8012.524.2
10010.018.3
1258.0013.7
Electric field strength midway between two plates against the reciprocal of their separation010203040501/d / m⁻¹020406080100E / kV m⁻¹
Graph drawn to scale
(a)

Draw the line of best fit for the four points with d ≤ 60 mm.

(1)
(b)

Determine the gradient of your line.

(2)
(c)

Compare your answer to (b) with the supply voltage.

(1)
(d)

For d = 125 mm, compare the measured value of E with the value predicted by E = V/d, and deduce whether the difference can be explained by the uncertainty in E.

(2)
(e)

Explain why E at large separations is smaller than V/d.

(1)
(f)

Suggest one change to the apparatus so that E = V/d holds over the whole range of separations.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Straight line through the origin passing within ±0.5 kV m−1 of the four points with d ≤ 60 mm✓ 1Line need not pass through the remaining three points.
Part (b)
Gradient calculated from a large triangle✓ 1
gradient = 2.00 kV (= 2000 V)✓ 1Accept 1.95–2.05 kV. Unit required.
Part (c)
The gradient equals V since E = V/d; it agrees with 2.00 kV (within about 1 %)✓ 1Allow ECF from (b).
Part (d)
V/d = 2000/0.125 = 16.0 kV m−1, but E = 13.7 kV m−1: a difference of 2.3 kV m−1✓ 1
The difference is much larger than ±0.5 kV m−1 (about 4 % of E), so it cannot be explained by the uncertainty: it is a real/systematic effect✓ 1OWTTE
Part (e)
Edge effects: when d is not small compared with the plate width the field lines bulge outwards beyond the edges, so the field between the plates is no longer uniform and is weaker at the centre✓ 1OWTTE
Part (f)
Use plates that are much larger (wider) compared with the largest separation✓ 1Accept: reduce the maximum separation.

Answers: (b) 2.00 kV  ·  (d) V/d = 16.0 kV m−1; difference 2.3 kV m−1 (the remaining parts are explanations — see the table above)

91D-1B-25Magnetic field of a straight wire Paper 1BEasy10 marks

A student uses a compass to determine the horizontal component BE of the Earth's magnetic field. A long straight horizontal wire runs from south to north and carries a steady current of (3.00 ± 0.02) A towards the north. A small compass rests on a horizontal card directly below the wire, at a distance r below the centre of the wire, and is deflected through an angle θ from north. The angle is read to ±1° and r is measured to ±1 mm.

At the compass, the magnetic field of the wire is horizontal and at right angles to BE, so tan θ = B/BE, where B = μ0I/(2πr) is the flux density due to the wire and μ0 = 4π × 10−7 T m A−1.

r / mmθ / °1/r / m−1tan θ
205850.01.60
3046
403825.00.781
503220.00.625
602816.70.532
8021
1001710.00.306
Tangent of compass deflection against reciprocal of distance from the wire, with error bars01020304050601/r / m⁻¹00.30.60.91.21.51.8tan θ
Graph drawn to scale
(a)

State the direction of the magnetic field due to the wire at the position of the compass.

(1)
(b)

Calculate the absolute uncertainty in 1/r for r = 20 mm.

(1)
(c)

Complete the table for r = 30 mm and r = 80 mm.

(1)
(d)

Plot the two missing points on the graph and draw the line of best fit for the data.

(2)
(e)

Determine the gradient of the line. State its unit.

(2)
(f)

The uncertainty in the gradient is ±4 %. Determine BE and its absolute uncertainty. Compare your answer with the value of 1.90 × 10−5 T given by a geomagnetic model for the laboratory.

(2)
(g)

The card rests on a laboratory bench with a steel frame. Suggest how the student could check whether the frame affects the results.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
West «horizontal, towards the west»✓ 1Do not accept “clockwise” or “anticlockwise” without a direction at the compass.
Part (b)
Δ(1/r) = (1/r) × (Δr/r) = 50 × 1/20 = ±2.5 m−1✓ 1Accept ±3 m−1.
Part (c)
r = 30 mm: 33.3 and 1.04; r = 80 mm: 12.5 and 0.384✓ 1All four values required.
Part (d)
Points plotted at (33.3, 1.04) and (12.5, 0.384) to within half a small square✓ 1Allow ECF from (c).
Single straight line through all the error bars, passing through or very close to the origin✓ 1
Part (e)
gradient = 31.6 × 10−3✓ 1Accept 30.3–33.6 × 10−3.
unit: m✓ 1
Part (f)
BE = μ0I/(2π × gradient) = 2 × 10−7 × 3.00/31.6 × 10−3 = (1.90 ± 0.09) × 10−5 T✓ 1Uncertainty from 4 % + 0.7 % ≈ 4.7 %. Allow ECF from (e).
1.90 × 10−5 T lies within the range of the student's value, so the two agree✓ 1The comparison must be consistent with the candidate's value and uncertainty.
Part (g)
Repeat a measurement with the apparatus moved well away from the bench (or on a wooden support) and check that the compass points north with no current and the deflection is unchanged✓ 1OWTTE. Accept reversing the current and checking that the deflection is equal and opposite.

Answers: (b) ±2.5 m−1  ·  (c) 33.3, 1.04; 12.5, 0.384  ·  (e) 31.6 × 10−3 m  ·  (f) BE = (1.90 ± 0.09) × 10−5 T (the remaining parts are explanations — see the table above)

92D-1B-33Charging by friction & induction Paper 1BEasy10 marks

A student investigates charging by friction. A motorised rig rubs a PTFE disc with a nylon brush, making n strokes. The disc is then lifted by an insulating handle and lowered into a metal cup connected to a digital coulombmeter, which measures the charge on the disc. The charge on the disc is negative; the table shows its magnitude Q. Before each trial the disc is wiped with a damp cloth and left to dry, to remove the charge from the previous trial.

The readings fluctuate slightly, and each value of Q has an uncertainty of ±0.3 nC. The graph shows the data with error bars.

n05101520304060
Q / nC0.86.710.914.516.820.422.124.0
Magnitude of the charge on the disc against number of strokes, measured points with error bars0102030405060n0481216202428Q / nC
Graph drawn to scale
(a)

State the resolution of the coulombmeter.

(1)
(b)

Calculate the percentage uncertainty in Q for n = 5.

(1)
(c)

Calculate the number of excess electrons on the disc for n = 40.

(1)
(d)

Draw the curve of best fit for the data.

(1)
(e)

The student's hypothesis is that Q is proportional to n. Discuss whether the data support this hypothesis.

(2)
(f)

Determine, using the graph, the charge transferred to the disc per stroke at the start of the rubbing.

(2)
(g)

The reading for n = 0 is not zero. Suggest a reason for this and state, with a reason, whether it affects your answer to (f).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.1 nC✓ 1Unit required.
Part (b)
0.3/6.7 × 100 = 4.5 %✓ 1Accept 5 %.
Part (c)
N = 22.1 × 10−9/1.60 × 10−19 = 1.4 × 1011✓ 1Accept 1.38 × 1011.
Part (d)
smooth curve through all the error bars, steep at first and levelling off at large n✓ 1Do not accept a straight line or point-to-point segments.
Part (e)
the graph is a curve that levels off «Q/n falls from 1.34 nC at n = 5 to 0.40 nC at n = 60», so Q is not proportional to n over the whole range✓ 1OWTTE
eg doubling n from 10 to 20 increases Q only from 10.9 nC to 16.8 nC «×1.54, not ×2», a difference far larger than the ±0.3 nC uncertainty, so the data do not support the hypothesis «Q rises roughly linearly only for the first few strokes»✓ 1Award [1] max for «not proportional» without a reason taken from the data. Do not credit the non-zero reading at n = 0 alone: it is a constant offset (see (g)).
Part (f)
tangent drawn to the curve at n = 0 and its gradient found from a large triangle✓ 1
1.3 nC per stroke✓ 1Accept 1.1–1.6 nC per stroke. Allow ECF from their curve.
Part (g)
the disc was not completely discharged before the trial / it kept some charge from the previous trial «PTFE is an insulator, so wiping does not remove all the charge»✓ 1OWTTE
no: the same residual charge is added to every reading, so the whole curve is shifted up and the gradient of the tangent is unchanged✓ 1Reason required.

Answers: (b) 4.5 %  ·  (c) 1.4 × 1011  ·  (f) 1.3 nC per stroke (the remaining parts are explanations — see the table above)

93D-1B-03Coulomb's law Paper 1BMedium11 marks

A student investigates how the electric force between two charged spheres depends on their separation. A small conducting sphere on an insulating rod stands on an electronic balance of resolution 0.001 g. An identical sphere carrying charge of the same sign is clamped vertically above it. The repulsion increases the balance reading by Δm, so the electric force is F = (Δm)g.

The centre-to-centre separation r is varied. The student tests the hypothesis F ∝ rn by plotting lg (Δm / g) against lg (r / m). Four of the points have been plotted.

r / mΔm / glg (r / m)lg (Δm / g)
0.0400.349−1.398−0.457
0.0500.224−1.301−0.650
0.0600.159
0.0750.101−1.125−0.996
0.0900.071
0.1200.040−0.921−1.398
Log of balance reading change against log of separation−1.4−1.3−1.2−1.1−1−0.9lg (r / m)−1.5−1.3−1.1−0.9−0.7−0.5−0.3lg (Δm / g)
Graph drawn to scale
(a)

Calculate the percentage uncertainty in Δm for r = 0.120 m.

(1)
(b)

Complete the table.

(2)
(c)

Plot the two missing points on the graph and draw the line of best fit for the data.

(2)
(d)

Determine the gradient of the line.

(2)
(e)

The uncertainty in the gradient is ±0.06. Discuss whether the data support Coulomb's law.

(1)
(f)

The readings were taken in order of increasing r. During the experiment charge slowly leaked from the spheres into the air. Explain the effect of this leakage on the gradient of the graph.

(2)
(g)

Suggest one change to the procedure that would reduce the effect described in (f).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.001/0.040 × 100 = 2.5 %✓ 1Accept 3 %. Accept ½ × resolution giving 1.3 %.
Part (b)
lg (r / m) = −1.222 and −1.046✓ 1Both needed. Accept 3 or 4 d.p.
lg (Δm / g) = −0.799 and −1.149✓ 1Both needed. Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (c)
Both points plotted correctly to within half a small square✓ 1Allow ECF from (b).
Single straight line of best fit with the points scattered on both sides✓ 1
Part (d)
Gradient calculated from a large triangle (at least half the length of the line)✓ 1
gradient = −1.97✓ 1Accept −1.92 to −2.02. Negative sign required. No unit.
Part (e)
Coulomb's law predicts a gradient of −2, which lies within the range −1.91 to −2.03, so the data support F ∝ 1/r²✓ 1Allow ECF from (d).
Part (f)
The charges, and so the forces, are smaller for the later readings at large r, so Δm at large r (the right-hand end of the graph) is too small✓ 1OWTTE
The line is steeper: the gradient becomes more negative (magnitude greater than it should be)✓ 1Award MP2 only if MP1 is scored or the reasoning is clear.
Part (g)
Recharge both spheres from the same supply before every reading OR take readings in a random order / in both increasing and decreasing r and average OR work in dry air so leakage is slower✓ 1OWTTE

Answers: (a) 2.5 %  ·  (b) −1.222, −0.799; −1.046, −1.149  ·  (d) −1.97 (the remaining parts are explanations — see the table above)

94D-1B-10Magnetic field patterns Paper 1BMedium11 marks

A student uses a long air-core solenoid to determine the magnetic constant μ0. The solenoid has 480 turns (counted) wound over a length of (0.240 ± 0.002) m. A magnetic field sensor placed at the centre of the solenoid, along its axis, measures the flux density B for different currents I. The uncertainty in each value of B is ±0.05 mT.

For a long solenoid, B = μ0nI, where n is the number of turns per unit length.

I / AB / mT
0.300.96
0.601.66
0.902.44
1.203.17
1.503.95
1.804.64
Magnetic flux density at the centre of a solenoid against current, with error bars00.40.81.21.62I / A012345B / mT
Graph drawn to scale
(a)

Calculate n and its absolute uncertainty.

(2)
(b)

Draw the line of best fit for the data.

(1)
(c)

The line does not pass through the origin. Suggest a reason for this and state its effect, if any, on a value of μ0 found from the gradient.

(2)
(d)

Determine the gradient of the line and its absolute uncertainty. Draw lines of maximum and minimum gradient on the graph.

(3)
(e)

Determine μ0 and its absolute uncertainty.

(2)
(f)

The accepted value of μ0 is 1.26 × 10−6 T m A−1. Comment on the student's result.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
n = 480/0.240 = 2000 m−1✓ 1
Δn = 2000 × (0.002/0.240) = ±17 m−1, so n = (2000 ± 17) m−1✓ 1Accept ±20 m−1. The number of turns has no uncertainty.
Part (b)
Single straight line through all the error bars, with a small positive intercept✓ 1Do not accept a line forced through the origin.
Part (c)
A systematic (zero) error: the sensor was not zeroed with the current switched off / a constant background field is added to every reading✓ 1OWTTE. Accept a nearby magnet or magnetised object as the source of the background field. The Earth's field (about 0.05 mT) is too small on its own to give the intercept of about 0.2 mT, but credit it if the idea of a constant added field is clear.
A constant offset shifts the whole line but does not change its gradient, so μ0 is unaffected✓ 1
Part (d)
Best gradient = 2.48 mT A−1 (2.48 × 10−3 T A−1)✓ 1Accept 2.44 to 2.52 mT A−1.
Max and min gradient lines through all the error bars, with gradients ≈ 2.52 and 2.43 mT A−1✓ 1
Uncertainty = ½(2.52 − 2.43) ≈ ±0.04 mT A−1✓ 1Accept ±0.04 to ±0.10 mT A−1.
Part (e)
μ0 = gradient/n = 2.48 × 10−3/2000 = 1.24 × 10−6 T m A−1✓ 1Allow ECF from (a) and (d).
percentage uncertainty = 1.6 % + 0.8 % ≈ 2.4 %, so μ0 = (1.24 ± 0.03) × 10−6 T m A−1✓ 1Fractional uncertainties of gradient and n must be added. MP2 is for matching precision of value and uncertainty.
Part (f)
The accepted value lies within the range 1.21–1.27 × 10−6 T m A−1, so the result is consistent with it✓ 1Allow ECF from (e).

Answers: (a) n = (2000 ± 17) m−1  ·  (d) (2.48 ± 0.04) mT A−1  ·  (e) μ0 = (1.24 ± 0.03) × 10−6 T m A−1 (the remaining parts are explanations — see the table above)

95D-1B-16Magnetic field of a straight wire Paper 1BMedium10 marks

A student investigates how the magnetic flux density B around a long straight vertical wire varies with the distance r from the centre of the wire. The wire carries a steady current of 8.00 A. A magnetic field sensor, aligned tangentially to a circle around the wire, is placed at different distances r. At each position the student records the sensor reading Bon with the current switched on and Boff with it switched off; each reading has an uncertainty of ±0.5 μT. The flux density due to the wire is B = Bon − Boff.

Theory predicts B = μ0I/(2πr). The graph shows lg (B / μT) against lg (r / mm) with the line of best fit.

r / mmBon / μTBoff / μTB / μTlg (r / mm)lg (B / μT)
10179.018.6160.41.0002.205
15124.818.4106.41.1762.027
2098.918.780.21.3011.904
3071.518.5
4554.518.635.91.6531.555
6044.918.426.51.7781.423
Log of magnetic flux density against log of distance from a wire, with line of best fit0.91.11.31.51.71.9lg (r / mm)1.31.51.71.92.12.3lg (B / μT)
Graph drawn to scale
(a)

Explain why the student records a reading with the current switched off at each position.

(1)
(b)

Complete the table for r = 30 mm.

(2)
(c)

Determine the absolute uncertainty in B for r = 60 mm.

(1)
(d)

Determine the gradient of the line.

(2)
(e)

State what the gradient shows about the relationship between B and r.

(1)
(f)

Determine, using the line, the current in the wire. Compare your answer with the ammeter reading.

(1)
(g)

Another student measures each distance from the surface of the wire instead of from its centre. Explain the effect on this student's graph.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
To measure the background field (e.g. the Earth's magnetic field) so that it can be subtracted, leaving the field due to the current alone✓ 1OWTTE
Part (b)
B = 71.5 − 18.5 = 53.0 μT✓ 1
lg (r / mm) = 1.477 AND lg (B / μT) = 1.724✓ 1Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (c)
0.5 + 0.5 = ±1.0 μT✓ 1Absolute uncertainties add for a difference. Do not accept ±0.5 μT.
Part (d)
Gradient calculated from a large triangle on the line✓ 1
gradient = −1.00✓ 1Accept −0.96 to −1.04. No unit.
Part (e)
gradient ≈ −1, so B ∝ r−1: B is inversely proportional to r, as predicted✓ 1Allow ECF from (d).
Part (f)
e.g. at lg (r / mm) = 1.00, lg (B / μT) = 2.205 so B = 160 μT and I = 2π × 0.010 × 160 × 10−6/(4π × 10−7) = 8.0 A, in agreement with 8.00 A✓ 1Accept 7.8–8.2 A with a comparison. Any point on the line may be used.
Part (g)
Every distance is too small by the same amount (the radius of the wire)✓ 1
The fractional error is largest at small r, so the left-hand points move left: the graph is no longer straight (it curves/flattens at small r) and its gradient magnitude is less than 1✓ 1OWTTE. Do not accept “the line shifts” without reference to the change in shape or gradient.

Answers: (b) B = 53.0 μT; 1.477; 1.724  ·  (c) ±1.0 μT  ·  (d) −1.00  ·  (f) I = 8.0 A (the remaining parts are explanations — see the table above)

96D-1B-24Uniform fields between plates Paper 1BMedium11 marks

In a recycling plant, a mixture of PET and PVC plastic flakes is tumbled in a drum, which charges the flakes by friction. The flakes then fall from rest between two large vertical parallel plates (250 ± 2) mm apart, starting midway between them. The length of fall between the plates is (1.200 ± 0.005) m. A potential difference V between the plates deflects the PET flakes towards the negative plate.

An engineer measures the mean horizontal displacement x of the PET flakes at the bottom of the plates for different values of V. Each value of x has an uncertainty of ±3 mm. If air resistance is negligible and the field is uniform, x = (q/m) × VH/(dg), where q/m is the charge-to-mass ratio of a flake, H is the length of fall and d is the separation of the plates.

V / kVx / mm
514
1026
1541
2053
2568
3080
Horizontal displacement of plastic flakes against potential difference between the plates05101520253035V / kV020406080100x / mm
Graph drawn to scale
(a)

Show that the time taken by a flake to fall between the plates is about 0.49 s.

(1)
(b)

Deduce the sign of the charge on the PET flakes and explain, in terms of electrons, how the PVC flakes become charged in the drum.

(2)
(c)

Draw the line of best fit for the data.

(1)
(d)

Determine the gradient of the line, in mm kV−1.

(2)
(e)

The uncertainty in the gradient is ±4 %. Determine q/m for the PET flakes and its absolute uncertainty.

(3)
(f)

Near the top and bottom edges of the plates the electric field strength is smaller than V/d. Explain how this affects the value of q/m obtained in (e).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
t = √(2H/g) = √(2 × 1.200/9.81) = 0.495 s✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
PET flakes are positive (attracted to the negative plate)✓ 1
Electrons are transferred from the PET flakes to the PVC flakes by friction, so the PVC flakes gain electrons and become negative✓ 1Do not accept transfer of positive charge/protons.
Part (c)
Single straight line through all the error bars, passing through or very close to the origin✓ 1
Part (d)
Gradient calculated from a triangle using at least half the length of the line✓ 1
gradient = 2.68 mm kV−1✓ 1Accept 2.60–2.76 mm kV−1.
Part (e)
q/m = gradient × dg/H = 2.68 × 10−6 × 0.250 × 9.81/1.200 = 5.48 × 10−6 C kg−1✓ 1Gradient must be converted to m V−1 (× 10−6). Allow ECF from (d).
fractional uncertainty = 4 % + 2/250 + 0.005/1.200 = 5.2 %✓ 1
q/m = (5.5 ± 0.3) × 10−6 C kg−1✓ 1MP3 is for matching the precision of value and uncertainty.
Part (f)
Near the edges the horizontal force qE on a flake is smaller than qV/d, and the flakes move slowly near the top, so they spend a large part of the fall there; x is smaller than the model predicts for each V✓ 1OWTTE
so the gradient is too small and q/m is underestimated✓ 1MP2 only if the direction of the effect on x is consistent.

Answers: (a) 0.495 s  ·  (d) 2.68 mm kV−1  ·  (e) q/m = (5.5 ± 0.3) × 10−6 C kg−1 (the remaining parts are explanations — see the table above)

97D-1B-34Coulomb's law Paper 1BMedium12 marks

A student uses the electric force between two charged spheres to determine the permittivity of free space ε0. Two small conducting spheres A and B, each of radius 12 mm, are charged positively. A is fixed to a digital force sensor (resolution 0.01 mN) and B is mounted on an insulating rod on a sliding track. The charges, measured with a coulombmeter, are qA = (30 ± 1) nC and qB = (27 ± 1) nC. The centre-to-centre separation r is measured to ±1 mm and the repulsive force F is recorded. Assume the air between the spheres has the permittivity of free space.

For point charges, F = qAqB/(4πε0r²). The graph shows F against 1/r². Two of the points have not been plotted.

r / mm1/r² / m−2F / mN
406254.18
504002.80
602781.96
801.14
1001000.72
1200.49
15044.40.33
Electric force against 1/r squared, five measured points; two not plotted01002003004005006007001/r² / m⁻²012345F / mN
Graph drawn to scale
(a)

Calculate the percentage uncertainty in 1/r² for r = 50 mm.

(1)
(b)

Complete the table and plot the two missing points on the graph.

(2)
(c)

Draw, on the graph, the straight line of best fit through the origin for the five points with r ≥ 60 mm.

(1)
(d)

Determine the gradient of your line. State its unit.

(2)
(e)

The uncertainty in the gradient is ±3 %. Determine ε0 and its absolute uncertainty.

(3)
(f)

The accepted value of ε0 is 8.85 × 10−12 C2 N−1 m−2. Comment on the student's result.

(1)
(g)

The point for r = 40 mm lies below the line. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
2 × (1/50) × 100 = 4.0 %✓ 1Accept 4 %.
Part (b)
156 AND 69.4 «m−2»✓ 1Both needed.
both points plotted to within half a small square✓ 1Allow ECF from MP1.
Part (c)
single straight line from the origin passing through or very close to the five points with r ≥ 60 mm✓ 1The line passes above the point for 40 mm.
Part (d)
gradient from a large triangle = 7.12 × 10−3 «mN m²» = 7.12 × 10−6✓ 1Accept 6.9–7.3 × 10−6 N m².
unit N m² «or mN m²»✓ 1Unit must match the value.
Part (e)
ε0 = qAqB/(4π × gradient) = 30 × 10−9 × 27 × 10−9/(4π × 7.12 × 10−6) = 9.05 × 10−12 «C2 N−1 m−2»✓ 1Allow ECF from (d). Accept 8.83–9.34 × 10−12.
fractional uncertainty = 3 % + 1/30 + 1/27 = 10.0 %✓ 1
ε0 = (9.0 ± 0.9) × 10−12 C2 N−1 m−2✓ 1MP3 is for matching the precision of value and uncertainty.
Part (f)
8.85 × 10−12 C2 N−1 m−2 lies within the range of the student's value, so the result is consistent with the accepted value✓ 1Allow ECF from (e). The comparison must use the uncertainty.
Part (g)
at small separations the spheres cannot be treated as point charges «the separation is not large compared with the radius»✓ 1
the like charges repel each other to the far sides of the spheres, so the charges are on average farther apart than r and the force is less than qAqB/(4πε0r²)✓ 1OWTTE

Answers: (a) 4.0 %  ·  (d) 7.12 × 10−6 N m²  ·  (e) (9.0 ± 0.9) × 10−12 C2 N−1 m−2 (the remaining parts are explanations — see the table above)

98D-1B-35Uniform fields between plates Paper 1BMedium12 marks

A student investigates the electrical breakdown of air in a uniform electric field, using a current-limited high-voltage supply inside a safety enclosure. Two large, flat, parallel brass electrodes with rounded edges are separated by a gap d, set with a micrometer to ±0.01 mm. The potential difference across the gap is increased slowly until a spark crosses it, and the breakdown potential difference Vb is read from a high-voltage meter. Five sparks are recorded for each gap and the table gives the mean. The graph shows the means with error bars of ±0.3 kV, except for d = 4.00 mm.

The five readings for d = 4.00 mm were 13.0 kV, 13.6 kV, 13.1 kV, 13.5 kV and 13.3 kV.

Over this range the student models the results with Vb = Ebd + V0, where Eb is the breakdown field strength of air and V0 is a constant.

d / mmVb / kV
1.004.5
2.007.3
3.0010.4
4.00
5.0016.2
6.0019.3
Breakdown potential difference against gap width, measured points with error bars01234567d / mm04812162024breakdown p.d. V / kV
Graph drawn to scale
(a)

Calculate the mean value of Vb for d = 4.00 mm and its uncertainty.

(2)
(b)

Plot the point for d = 4.00 mm on the graph and draw the line of best fit for the data.

(2)
(c)

Determine Eb from your line, in V m−1.

(2)
(d)

Draw lines of maximum and minimum gradient on the graph and determine the absolute uncertainty in Eb.

(2)
(e)

The breakdown field strength of dry air is often quoted as 3.0 × 106 V m−1. Discuss whether the student's result is consistent with this value.

(1)
(f)

Calculate Vb/d for d = 1.00 mm and for d = 6.00 mm, and outline what these values show about using Vb/d alone as the breakdown field strength.

(2)
(g)

Explain why the electrodes are large and flat with rounded edges.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
mean = 13.3 kV✓ 1
uncertainty = ½ × range = ½ × (13.6 − 13.0) = ±0.3 kV✓ 1Award [1] max for ±0.6 kV.
Part (b)
point plotted at (4.00, 13.3) to within half a small square✓ 1Allow ECF from (a).
single straight line through all the error bars, with a positive intercept on the Vb axis✓ 1Do not accept a line forced through the origin.
Part (c)
gradient from a large triangle = 2.96 kV mm−1✓ 1Accept 2.85–3.10 kV mm−1.
Eb = 2.96 × 106 V m−1✓ 1Conversion: 1 kV mm−1 = 106 V m−1. Allow ECF from the gradient.
Part (d)
steepest and shallowest lines through all the error bars, with gradients ≈ 3.08 and 2.85 kV mm−1✓ 1
uncertainty = ½(3.08 − 2.85) ≈ ±0.11 × 106 V m−1✓ 1Accept ±0.10 to ±0.20 × 106 V m−1.
Part (e)
3.0 × 106 V m−1 lies within the range 2.85–3.07 × 106 V m−1, so the result is consistent with it✓ 1Allow ECF from (c) and (d).
Part (f)
4.50 × 106 V m−1 AND 3.22 × 106 V m−1 «4.50 and 3.22 kV mm−1»✓ 1Both needed.
Vb/d is not constant: it is larger for small gaps «because of the intercept V0», so the gradient, not Vb/d, should be used✓ 1OWTTE
Part (g)
so that the field in the gap is uniform, E = V/d: there are no edge effects or sharp points where the field lines crowd together and the field is stronger✓ 1OWTTE

Answers: (a) (13.3 ± 0.3) kV  ·  (c) 2.96 × 106 V m−1  ·  (d) ±0.11 × 106 V m−1  ·  (f) 4.50 × 106 V m−1 and 3.22 × 106 V m−1 (the remaining parts are explanations — see the table above)

99D-1B-37Electric field lines Paper 1BMedium11 marks

In fine weather there is a vertical electric field in the atmosphere, directed downwards towards the negatively charged ground. A student uses secondary data: values of the electric field strength E at different heights h above flat ground, averaged over many fine-weather balloon soundings. Each value of E has an uncertainty of ±5 %.

The student tests whether E decreases exponentially with height by plotting lg(E/V m−1) against h. Six of the points have been plotted.

h / kmE / V m−1lg(E/V m−1)
0.599.21.997
1.079.61.901
1.567.2
2.052.81.723
3.036.51.562
4.024.2
5.015.91.201
6.011.11.045
lg E against height, points with error bars01234567h / km0.81.01.21.41.61.82.02.2lg(E / V m⁻¹)
Graph drawn to scale
(a)

Calculate the absolute uncertainty in lg(E/V m−1) that corresponds to an uncertainty of ±5 % in E.

(1)
(b)

Complete the table and plot the two missing points on the graph.

(2)
(c)

Draw the line of best fit for the data.

(1)
(d)

Determine the gradient of the line. State its unit.

(2)
(e)

Show that E halves for every increase in height of about 1.7 km.

(1)
(f)

Determine, using your line, the electric field strength at ground level predicted by the exponential model.

(2)
(g)

Explain, in terms of electric field lines, how the decrease of E with height shows that the air contains electric charge, and deduce the sign of this charge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
lg 1.05 ≈ ±0.02✓ 1Accept 0.434 × 0.05 = 0.022. Accept ±0.02.
Part (b)
1.827 AND 1.384✓ 1Both needed. Accept 2 or more d.p.
both points plotted to within half a small square✓ 1Allow ECF from MP1.
Part (c)
single straight line through all the error bars✓ 1Do not accept a curve.
Part (d)
gradient from a large triangle = −0.174✓ 1Accept −0.186 to −0.162. Negative sign required.
unit km−1✓ 1
Part (e)
Δh = lg 2/|gradient| = 0.301/0.174 = 1.73 «km»✓ 1Must see full substitution OR answer to at least 3 s.f. Allow ECF from (d). ALTERNATIVE: read two values of h where E differs by a factor of 2.
Part (f)
line extrapolated to h = 0: intercept lg(E/V m−1) ≈ 2.08✓ 1Accept 2.06–2.10. Allow ECF from (c).
E = 102.08 ≈ 120 V m−1✓ 1Accept 115–126 V m−1. Unit required.
Part (g)
field strength is shown by the density of field lines, so fewer field lines cross each square metre at greater heights: some field lines must begin in the air below✓ 1OWTTE
the lines point downwards and field lines begin on positive charges, so the air contains «net» positive charge✓ 1MP2 only scores with a correct reference to where lines begin.

Answers: (a) ±0.02  ·  (d) −0.174 km−1  ·  (e) 1.73 km  ·  (f) 120 V m−1 (the remaining parts are explanations — see the table above)

100D-1B-39Magnetic field patterns Paper 1BMedium10 marks

A student investigates the magnetic field along the axis of an air-core solenoid of length (150 ± 1) mm and radius 20 mm, wound with 270 turns (counted). The current in the solenoid is (2.00 ± 0.02) A. A magnetic field sensor, zeroed with the current switched off, measures the flux density B on the axis at a distance x from the centre of the solenoid. The ends of the solenoid are at x = ±75 mm. Each reading of B has an uncertainty of ±0.02 mT.

For a long solenoid, B = μ0nI, where n is the number of turns per unit length and μ0 = 4π × 10−7 T m A−1.

x / mm020405060708090100120
B / mT4.374.334.203.993.602.781.700.880.490.18
Magnetic flux density on the axis against distance from the centre of the solenoid, measured points0102030405060708090100110120130x / mm012345B / mT
Graph drawn to scale
(a)

Show that μ0nI is about 4.5 mT.

(1)
(b)

Determine the percentage uncertainty in μ0nI.

(2)
(c)

Draw the curve of best fit for the data.

(1)
(d)

Deduce whether the difference between the measured value of B at the centre and μ0nI can be explained by the uncertainties in the measurements.

(2)
(e)

Suggest a reason for the difference in (d).

(1)
(f)

Determine, using your curve, (B at an end of the solenoid)/(B at the centre).

(2)
(g)

Outline, in terms of magnetic field lines, why B is smaller at the ends of the solenoid than at the centre.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B = 4π × 10−7 × (270/0.150) × 2.00 = 4.52 × 10−3 «T»✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (b)
percentage uncertainties added: 1/150 «length» + 0.02/2.00 «current» «the number of turns has no uncertainty»✓ 1
= 1.7 %✓ 1Accept 1.7 % or 2 %.
Part (c)
smooth curve through the points: almost flat near x = 0, falling most steeply near x = 75 mm and levelling off towards zero beyond the end✓ 1Do not accept point-to-point segments.
Part (d)
μ0nI = (4.52 ± 0.08) mT, ie 4.45 to 4.60 mT, but the measured value is (4.37 ± 0.02) mT✓ 1Allow ECF from (b).
the ranges do not overlap, so the difference cannot be explained by the uncertainties✓ 1
Part (e)
the solenoid is not long compared with its radius/diameter, so B = μ0nI «for a long solenoid» does not apply exactly «field lines spread out near the ends even affect the centre»✓ 1OWTTE. Do not accept «the Earth's field», because the sensor was zeroed.
Part (f)
B at x = 75 mm read from the curve ≈ 2.24 mT✓ 1Accept 2.15–2.35 mT.
ratio = 2.24/4.37 ≈ 0.51✓ 1Accept 0.49–0.54. Allow ECF from MP1.
Part (g)
the field lines spread out as they leave «or enter» the end of the solenoid, so they are further apart there and the field is weaker✓ 1OWTTE

Answers: (a) 4.52 mT  ·  (b) 1.7 %  ·  (f) 0.51 (the remaining parts are explanations — see the table above)

101D-1B-18Quantization of charge Paper 1BHard11 marks

In a Millikan-type experiment, charged oil drops are observed between two horizontal parallel plates separated by (6.00 ± 0.05) mm. For each drop the potential difference V across the plates is adjusted until the drop is held stationary; the uncertainty in V is ±1 V. The mass m of each drop has been found separately (method not required) with an uncertainty of ±3 %.

When a drop is stationary the electric force balances its weight, so qV/d = mg.

Dropm / 10−15 kgV / Vq / 10−19 C
A1.541864.87
B0.86320
C2.392166.51
D1.16217
E3.09225
(a)

Drop B.

(i)

Show that the charge on drop B is about 1.6 × 10−19 C.

(1)
(ii)

Determine the absolute uncertainty in the charge on drop B.

(3)
(b)

Calculate the charges on drops D and E.

(1)
(c)

Deduce the number of elementary charges on each drop and hence determine a value for the elementary charge.

(2)
(d)

Compare your answer to (c) with the accepted value of the elementary charge.

(1)
(e)

The student later finds that the voltmeter reads 2.0 % higher than the true potential difference. Explain the effect of this on the value obtained for the elementary charge.

(2)
(f)

Suggest why the student chose to study only drops carrying small charges.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
q = mgd/V = 0.86 × 10−15 × 9.81 × 6.00 × 10−3/320 = 1.58 × 10−19 C✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
Percentage uncertainties added: 3 % (m) + 0.83 % (d) + 0.31 % (V)✓ 1
= 4.1 %, so Δq = 4.1 % × 1.58 × 10−19 ≈ 0.07 × 10−19 C✓ 1Accept 0.06–0.07 × 10−19 C.
q = (1.58 ± 0.07) × 10−19 C✓ 1MP3 is for matching the precision of value and uncertainty. Accept (1.6 ± 0.1) × 10−19 C.
Part (b)
D: 3.15 × 10−19 C AND E: 8.08 × 10−19 C✓ 1Both needed.
Part (c)
Numbers of elementary charges: A 3, B 1, C 4, D 2, E 5✓ 1Allow ECF from (b).
e = total charge/total number = 24.2 × 10−19/15 = 1.61 × 10−19 C✓ 1Accept the mean of q/n for each drop (1.60–1.62 × 10−19 C).
Part (d)
The accepted value 1.60 × 10−19 C differs by less than 1 %, well within the ≈ 4 % uncertainty of each charge, so they agree✓ 1Allow ECF.
Part (e)
The recorded V is larger than the true p.d., so q = mgd/V is too small for every drop✓ 1
The value of e is too small, by about 2 % (a systematic error)✓ 1Award MP2 only if the direction is correct.
Part (f)
The ≈ 4 % uncertainty in q is a larger fraction of e for large n; beyond about 12 elementary charges it exceeds ±½e, so the whole number of charges on the drop could not be identified✓ 1OWTTE

Answers: (a)(i) 1.58 × 10−19 C  ·  (a)(ii) (1.58 ± 0.07) × 10−19 C  ·  (b) 3.15 and 8.08 × 10−19 C  ·  (c) e = 1.61 × 10−19 C (the remaining parts are explanations — see the table above)

102D-1B-23Coulomb's law Paper 1BHard11 marks

A student tests whether the electric force between two charged spheres is proportional to the charge on one of them. Two identical small conducting spheres A and B, each of radius 20 mm, are given charges of the same sign. A stands on an insulating rod on a top-pan balance; B is clamped directly above A with their centres 100 mm apart. The repulsion increases the balance reading by Δm.

After each reading, B is touched with a third identical sphere C, which is uncharged and held by an insulating handle. C is then earthed before it is used again. The table shows Δm after n such touches. The readings fluctuate, and each value of Δm has an uncertainty of ±0.001 g. Two of the points have not been plotted.

nΔm / glg (Δm / g)
00.330−0.48
10.153−0.82
20.072−1.14
30.034−1.47
40.016−1.80
50.007−2.15
Logarithm of balance reading against number of halvings of the charge, with error bars0123456n−2.4−2−1.6−1.2−0.8−0.40lg (Δm / g)
Graph drawn to scale
(a)

Explain why each touch with C halves the charge on B.

(1)
(b)

Calculate the percentage uncertainty in Δm for n = 5.

(1)
(c)

Plot the points for n = 2 and n = 4 on the graph and draw the line of best fit for the data.

(2)
(d)

Show that, if the force is proportional to the charge on B, the gradient of the line should be about −0.30.

(1)
(e)

Determine the gradient of the line.

(2)
(f)

The uncertainty in the gradient is ±0.01. Discuss whether the data support the hypothesis.

(1)
(g)

The air in the laboratory is humid, so charge leaks slowly from both A and B during the experiment. Explain how this leakage affects the gradient of the line.

(2)
(h)

Suggest one change to the method that would reduce the effect described in (g).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B and C are identical (and C starts uncharged), so the charge on B is shared equally between them; charge is conserved, so B keeps half✓ 1OWTTE
Part (b)
0.001/0.007 × 100 = 14 %✓ 1Accept 14 % or 15 %.
Part (c)
Points plotted at (2, −1.14) and (4, −1.80) to within half a small square✓ 1
Single straight line passing through (or very close to) all six points and within the error bar for n = 5✓ 1Do not accept a line joining the dots.
Part (d)
Each touch multiplies the force (and Δm) by ½, so lg Δm falls by lg 2 = 0.301 for each increase of 1 in n✓ 1Must see lg 2 or log of 0.5 OR the change in lg Δm per step = −0.30.
Part (e)
Gradient calculated from a triangle using at least half the length of the line✓ 1
gradient = −0.33✓ 1Accept −0.32 to −0.35. The negative sign is required.
Part (f)
−0.30 lies outside the range −0.32 to −0.34, so the data do not support the hypothesis as they stand / there is a systematic error✓ 1Allow ECF from (e).
Part (g)
Between readings the charge on both spheres falls, so Δm falls by more than a factor of 2 for each touch✓ 1
The line is steeper (gradient more negative than −0.30), and the effect grows with n because more time has passed✓ 1Consistent with (f). OWTTE
Part (h)
Take the readings quickly / at equal short time intervals and correct for the measured loss, OR work in dry air (e.g. with a dehumidifier), OR measure the remaining charge directly each time✓ 1Do not accept “repeat the experiment” alone.

Answers: (b) 14 %  ·  (e) −0.33 (the remaining parts are explanations — see the table above)

103D-1B-36Magnetic field patterns Paper 1BHard11 marks

A student uses the magnetometer app of a smartphone to investigate the magnetic field on the axis of a small cylindrical neodymium magnet. The phone lies flat on a bench and the magnet is placed on a ruler with its axis along the ruler, pointing at the magnetometer, with its north pole facing the phone. The distance x from the top edge of the phone to the centre of the magnet is varied. The magnetometer is inside the phone, an unknown distance x0 from the top edge. The app displays the component of the magnetic field along the length of the phone.

With the magnet removed, the app reads 21.6 μT, due to the Earth's magnetic field. Each reading of the app has an uncertainty of ±0.5 μT. B is the magnitude of the field due to the magnet alone. The point for x = 50 mm has not been plotted.

For distances large compared with the size of the magnet, B = C/(x + x0)³, where C is a constant.

x / mmapp reading / μTB / μT(B/μT)−1/3
30331.5309.90.148
40175.1153.50.187
50108.8
6075.954.30.264
7057.335.70.304
8046.625.00.342
10034.913.30.422
12029.88.20.496
B to the power minus one third against x, measured points with error bars−20020406080100120140x / mm00.10.20.30.40.5(B / μT) to the power −1/3
Graph drawn to scale
(a)

Show that a graph of (B/μT)−1/3 against x is a straight line that meets the x axis at x = −x0.

(2)
(b)

Complete the table for x = 50 mm and plot the point on the graph.

(2)
(c)

Determine the absolute uncertainty in (B/μT)−1/3 for x = 120 mm.

(2)
(d)

Draw the line of best fit for the data.

(1)
(e)

Determine x0.

(2)
(f)

Explain why the phone must not be rotated on the bench between the reading with the magnet removed and the other readings.

(1)
(g)

State the direction of the magnetic field of the magnet at the magnetometer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B−1/3 = C−1/3(x + x0) = C−1/3x + C−1/3x0, which has the form y = mx + c with constant gradient C−1/3✓ 1
B−1/3 = 0 when x + x0 = 0, so the line meets the x axis at x = −x0✓ 1
Part (b)
B = 108.8 − 21.6 = 87.2 «μT» AND (B/μT)−1/3 = 0.226✓ 1Both needed. Accept 3 or 4 s.f. for the second value.
point plotted to within half a small square✓ 1Allow ECF from MP1.
Part (c)
ΔB = 0.5 + 0.5 = 1.0 μT, so ΔB/B = 1.0/8.2 = 12 %✓ 1Uncertainties of the two readings add for a difference.
fractional uncertainty in B−1/3 = ⅓ × 12 % ≈ 4.1 %, so absolute uncertainty = ±0.020✓ 1Accept ±0.02. Award [1] max for ±0.06 «⅓ omitted».
Part (d)
single straight line through all the error bars, extended to cross the x axis✓ 1
Part (e)
line extrapolated to (B/μT)−1/3 = 0 OR x0 = intercept on the y axis ÷ gradient✓ 1
x0 = 8 mm✓ 1Accept 5–11 mm. A positive value is required.
Part (f)
the component of the Earth's field measured by the magnetometer depends on the orientation of the phone, so the background subtracted would no longer be correct «the error is largest at large x, where B is small»✓ 1OWTTE
Part (g)
along the axis, away from the magnet «towards the phone»✓ 1Field lines leave a north pole.

Answers: (b) 87.2 μT; 0.226  ·  (c) ±0.020  ·  (e) 8 mm (the remaining parts are explanations — see the table above)

104D-1B-38Charged conducting sphere Paper 1BHard11 marks

A student investigates the electric field outside charged conducting spheres. Two hollow metal spheres on insulating stands, A of radius 40 mm and B of radius 80 mm, are each charged to (25 ± 1) nC, measured with a coulombmeter. An electric field meter measures the field strength E at different distances r from the centre of each sphere in turn, well away from other objects. Each value of E has an uncertainty of ±0.2 kV m−1.

The graph shows lg(E/kV m−1) against lg(r/m) for both spheres. Two points have not been plotted.

spherer / mmE / kV m−1lg(r/m)lg(E/kV m−1)
A6063.1−1.2221.800
A8034.8−1.097
A10022.5−1.0001.352
A15010.1−0.8241.004
A2005.5−0.6990.740
B10022.3−1.0001.348
B12015.8−0.9211.199
B1509.89−0.8240.995
B2005.62−0.6990.750
B2503.67−0.602
B3002.47−0.5230.393
lg E against lg r for both spheres, measured points only−1.3−1.2−1.1−1.0−0.9−0.8−0.7−0.6−0.5lg(r / m)0.20.40.60.81.01.21.41.61.82.0lg(E / kV m⁻¹)
Graph drawn to scale
(a)

Calculate the percentage uncertainty in E for r = 300 mm.

(1)
(b)

Complete the table and plot the two missing points on the graph.

(2)
(c)

Draw a single line of best fit for the data from both spheres.

(1)
(d)

Determine the gradient of the line and state what it shows about the variation of E with r.

(2)
(e)

Determine, using the line, the charge on each sphere.

(2)
(f)

Comment on your answer to (e).

(1)
(g)

Explain why the data for both spheres lie on the same line, even though the spheres have different radii.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.2/2.47 × 100 = 8 %✓ 1
Part (b)
lg(E/kV m−1) = 1.542 «A, 80 mm» AND 0.565 «B, 250 mm»✓ 1Accept 2 or more d.p.
both points plotted to within half a small square✓ 1Allow ECF from MP1.
Part (c)
single straight line passing through or very close to all the points of both spheres✓ 1
Part (d)
gradient = −2.00✓ 1Accept −1.95 to −2.05. Negative sign required.
gradient ≈ −2, so E ∝ 1/r² «E is inversely proportional to the square of the distance»✓ 1Allow ECF from MP1.
Part (e)
a point read from the line, eg at lg(r/m) = −1.000, lg(E/kV m−1) = 1.352 so E = 22.5 kV m−1 at r = 0.100 m✓ 1
Q = Er²/k = 22500 × 0.100²/8.99 × 109 = 25.0 × 10−9 C✓ 1Accept 23.5–26.5 nC. Allow ECF.
Part (f)
the value agrees with the coulombmeter reading, (25 ± 1) nC «it lies within the uncertainty»✓ 1Allow ECF from (e). The comparison must use the value and the uncertainty.
Part (g)
the charge on a conducting sphere is spread evenly over its outer surface, and outside the sphere the field is the same as that of a point charge at the centre✓ 1OWTTE
so outside each sphere E = kQ/r² depends only on Q and r, and Q is the same for both✓ 1MP2 only scores if MP1 scores.

Answers: (a) 8 %  ·  (d) −2.00  ·  (e) 25.0 nC (the remaining parts are explanations — see the table above)

105D-2-05Coulomb's law Paper 2Easy10 marks

In a linear ion trap, three identical ions X, Y and Z are held at rest on a straight line in a vacuum. Each ion has a charge of +1.60 × 10−19 C. Neighbouring ions are 6.0 μm apart, as shown.

Three positive ions X, Y and Z on a straight horizontal line, with 6.0 micrometres between neighbouring ions. There is empty space to the right of Z.+X+Y+Z6.0 μm6.0 μm
Diagram NOT to scale
(a)
(i)

Calculate the magnitude of the force on Z due to Y.

(2)
(ii)

Calculate the magnitude of the force on Z due to X.

(1)
(iii)

Draw, on the diagram, arrows to represent the force on Z due to X and the force on Z due to Y.

(2)
(b)

Electrodes in the trap produce an additional electric field that holds Z at rest. Determine the magnitude and direction of this additional field at the position of Z.

(3)
(c)

State and explain the resultant electric force exerted on Y by X and Z.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
F = 8.99 × 109 × (1.60 × 10−19)²/(6.0 × 10−6)²✓ 1
F = 6.4 × 10−18 N✓ 1Award [2] for CNA.
Part (a)(ii)
the distance is doubled so the force is a quarter: 1.6 × 10−18 N✓ 1Allow ECF from (a)(i).
Part (a)(iii)
two arrows starting at Z, both pointing to the right «away from X and Y, because like charges repel»✓ 1
arrow for the force due to Y about four times as long as the arrow for the force due to X✓ 1Allow ECF from (a)(ii).
Part (b)
resultant force on Z from X and Y = 6.4 × 10−18 + 1.6 × 10−18 = 8.0 × 10−18 N «to the right»✓ 1Allow ECF from (a).
E = F/q = 8.0 × 10−18/1.60 × 10−19 = 50 N C−1✓ 1
directed to the left «towards Y»✓ 1The field must give a force opposite to the resultant repulsion because the ion is positive.
Part (c)
zero✓ 1Award MP1 only with a valid reason.
X and Z have equal charges at equal distances on opposite sides of Y, so their forces on Y are equal in magnitude and opposite in direction✓ 1

Answers: (a)(i) 6.4 × 10−18 N  ·  (a)(ii) 1.6 × 10−18 N  ·  (b) 50 N C−1, to the left (the remaining parts are explanations — see the table above)

106D-2-06Uniform fields between plates Paper 2Easy12 marks

In an electrostatic precipitator, smoke rises between two large, vertical, parallel metal plates X and Y that are 0.25 m apart. The potential difference between the plates is 50 kV, with X positive. Charged dust particles in the smoke are pulled sideways onto a plate and removed from the smoke.

Two vertical parallel plates 0.25 m apart seen from the side: plate X on the left is positive and plate Y on the right is negative. Smoke enters the gap from below.+−+−+−+−plate Xplate Y0.25 msmoke
Diagram NOT to scale
(a)
(i)

Draw, on the diagram, the electric field lines between the plates, including the regions near the top and bottom edges of the plates.

(2)
(ii)

Calculate the electric field strength between the plates, away from the edges.

(1)
(b)

A dust particle of mass 2.0 × 10−13 kg carries a charge of −4.8 × 10−16 C.

(i)

Show that the particle carries 3000 excess electrons.

(1)
(ii)

Calculate the magnitude of the electric force on the particle and state its direction.

(2)
(iii)

Determine whether the weight of the particle can be neglected when its sideways motion is considered.

(3)
(c)

The particle starts midway between the plates. Calculate, in electronvolts, the work done on the particle by the electric field as it moves to the plate.

(2)
(d)

Suggest why the precipitator removes dust less effectively near the top and bottom edges of the plates.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
parallel, equally spaced lines perpendicular to the plates, with arrows from X to Y, in the central region✓ 1At least three lines.
lines near the edges curve «bulge» outwards and are further apart✓ 1Do not award if lines cross or do not start and end on the plates.
Part (a)(ii)
E = V/d = 5.0 × 104/0.25 = 2.0 × 105 V m−1✓ 1Accept N C−1.
Part (b)(i)
n = 4.8 × 10−16/1.60 × 10−19 = 3000✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
F = qE = 4.8 × 10−16 × 2.0 × 105 = 9.6 × 10−11 N✓ 1Allow ECF from (a)(ii).
towards X «the positive plate», horizontally✓ 1
Part (b)(iii)
weight = 2.0 × 10−13 × 9.81 = 2.0 × 10−12 N✓ 1
this is only about 2 % of the electric force✓ 1Allow ECF from (b)(ii).
and it acts vertically, at right angles to the sideways electric force, so it has no effect on the sideways motion: it can be neglected✓ 1Award MP3 for a valid conclusion supported by either the size comparison or the direction argument.
Part (c)
W = Fs = 9.6 × 10−11 × 0.125 = 1.2 × 10−11 J✓ 1ALTERNATIVE: the particle moves through a potential difference of 25 kV, so W = 3000 × 25 × 103 eV.
W = 1.2 × 10−11/1.60 × 10−19 = 7.5 × 107 eV✓ 1Award [2] for CNA.
Part (d)
the field is weaker «and non-uniform» near the edges «field lines further apart», so the force on the particles is smaller✓ 1OWTTE

Answers: (a)(ii) 2.0 × 105 V m−1  ·  (b)(ii) 9.6 × 10−11 N towards X  ·  (b)(iii) weight 2.0 × 10−12 N  ·  (c) 7.5 × 107 eV (the remaining parts are explanations — see the table above)

107D-2-14Magnetic field patterns Paper 2Easy10 marks

A flat circular coil is held with its plane perpendicular to the page. The diagram shows a cross-section through the centre of the coil. The wire of the coil passes through the page at A, where the current is out of the page, and at B, where the current is into the page.

Cross-section through a flat circular coil whose plane is perpendicular to the page. The coil passes through the page at A (current out of the page) and at B (current into the page). The axis of the coil is horizontal.axis of coilABleft-hand faceright-hand face
Diagram NOT to scale
(a)
(i)

Draw, on the diagram, the magnetic field lines of the coil in the plane of the page. Show the direction of the field.

(3)
(ii)

State which face of the coil, left-hand or right-hand, behaves like the north pole of a bar magnet.

(1)
(iii)

Explain why the field lines very close to A are almost circles centred on A.

(1)
(b)

A small bar magnet is placed on the axis of the coil, a short distance to the right of the coil, with its north pole facing the coil. The current in the coil is unchanged.

(i)

State and explain whether the magnet is attracted towards the coil or repelled from it.

(2)
(ii)

State the direction of the magnetic force that the magnet exerts on the coil.

(1)
(iii)

Suggest one change to the coil, other than increasing the current, that would increase the force on the magnet.

(1)
(c)

State the effect on the field pattern of reversing the current in the coil.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
closed loops around A and around B, nearly circular close to each wire✓ 1
lines through the middle of the coil roughly parallel to the axis, the central line straight along the axis, pattern symmetrical about the axis✓ 1
anticlockwise around A and clockwise around B, so the field at the centre points from left to right✓ 1Arrows needed on at least one loop around each wire or on the central line.
Part (a)(ii)
the right-hand face✓ 1Field lines leave the coil on this side. Allow ECF from (a)(i).
Part (a)(iii)
very close to A, the field due to the current at A is much stronger than the field due to the rest of the coil, so the pattern is that of a long straight wire✓ 1OWTTE
Part (b)(i)
the right-hand face of the coil behaves like a north pole «field lines leave the coil there»✓ 1Allow ECF from (a)(ii).
like poles repel, so the magnet is repelled «pushed to the right, away from the coil»✓ 1MP2 only scores if MP1 scores.
Part (b)(ii)
to the left «away from the magnet»✓ 1Newton's third law: equal and opposite to the force on the magnet. Allow ECF from (b)(i).
Part (b)(iii)
use more turns «carrying the same current», so the field of the coil is stronger✓ 1Accept: insert a soft-iron core. Do not accept moving the magnet.
Part (c)
the shape of the pattern is unchanged but every field line reverses direction «the north and south faces swap»✓ 1
108D-2-25Conservation of charge Paper 2Easy12 marks

Three identical small metal spheres X, Y and Z stand on insulating supports. Initially X carries a charge of +12.0 nC, Y carries a charge of −4.0 nC and Z is uncharged.

k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C.

+−X +12.0 nCY −4.0 nC0.060 mZuncharged
Diagram not to scale. Distance is between the centres.
(a)

X and Y are placed with their centres 0.060 m apart.

Calculate the magnitude of the electric force between them and state whether the force is attractive or repulsive.

(2)
(b)

X and Y are touched together, separated, and placed with their centres 0.060 m apart again.

(i)

State the charge on each sphere now.

(1)
(ii)

Calculate the magnitude of the force between X and Y now and state whether it is attractive or repulsive.

(2)
(iii)

Determine the number of electrons that moved when X and Y touched, and the direction in which they moved.

(2)
(c)

Y is then touched with Z and separated from it. Finally Z is touched with X and separated from it.

(i)

Determine the final charge on each of the three spheres.

(2)
(ii)

Outline how your answer to (c)(i) shows that electric charge is conserved.

(1)
(d)

Explain why none of the spheres can ever carry a charge of exactly 2.4 × 10−19 C.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = 8.99 × 109 × 12.0 × 10−9 × 4.0 × 10−9 / 0.060² = 1.2 × 10−4 N✓ 1Accept 1.20 × 10−4 N.
attractive, because the charges have opposite signs✓ 1
Part (b)(i)
+4.0 nC on each «the total of +8.0 nC is shared equally»✓ 1
Part (b)(ii)
F = 8.99 × 109 × (4.0 × 10−9)² / 0.060² = 4.0 × 10−5 N✓ 1Allow ECF from (b)(i).
repulsive «both charges are now positive»✓ 1
Part (b)(iii)
n = 8.0 × 10−9 / 1.60 × 10−19 = 5.0 × 1010✓ 1Award MP1 for use of a charge transfer of 8.0 nC.
from Y to X✓ 1Do not accept “positive charge moved from X to Y”.
Part (c)(i)
after Y and Z touch: +2.0 nC on each✓ 1
after Z and X touch: +3.0 nC on each; so X = +3.0 nC, Y = +2.0 nC, Z = +3.0 nC✓ 1Allow ECF from (b)(i).
Part (c)(ii)
total = 3.0 + 2.0 + 3.0 = 8.0 nC, which equals the initial total «12.0 − 4.0 + 0» = 8.0 nC✓ 1
Part (d)
2.4 × 10−19 / 1.60 × 10−19 = 1.5, which is not a whole number✓ 1
charge is quantized: it is transferred as whole electrons, so any charge is a whole-number multiple of e✓ 1OWTTE. Accept reference to Millikan's experiment as the evidence.

Answers: (a) 1.2 × 10−4 N, attractive  ·  (b)(ii) 4.0 × 10−5 N, repulsive  ·  (b)(iii) 5.0 × 1010 electrons from Y to X  ·  (c)(i) X +3.0 nC, Y +2.0 nC, Z +3.0 nC (the remaining parts are explanations — see the table above)

109D-2-26Charged conducting sphere Paper 2Easy11 marks

The dome of a Van de Graaff generator is a hollow metal sphere of radius R = 0.15 m on an insulating column. The dome carries a positive charge. The electric field strength at the surface of the dome is Es.

Air breaks down and a spark occurs when the electric field strength reaches 3.0 × 106 N C−1. k = 8.99 × 109 N m² C−2.

Blank axes: electric field strength E against distance r from the centre of the dome, from 0 to 3R0R2R3Rdistance from the centre of the dome r0EₛE
Figure 1 — axes for (a)
(a)

Sketch, on Figure 1, a graph to show how the electric field strength E varies with the distance r from the centre of the dome, from r = 0 to r = 3R.

(3)
(b)

Show that the largest charge the dome can hold without a spark is about 7.5 × 10−6 C.

(1)
(c)

Calculate the electric field strength at a distance of 0.40 m from the centre of the dome when it carries this charge.

(2)
(d)

Determine whether a dome of radius 0.25 m could hold a charge of 25 μC without a spark.

(2)
(e)

A student standing on an insulating mat places a hand on the uncharged dome. The generator is then switched on and the student's hair stands on end.

Explain this observation.

(2)
(f)

Outline why the student must stand on an insulating mat for this to happen.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = 0 for r < R✓ 1
E rises to its maximum value Es at r = R✓ 1Accept a vertical step at R.
for r > R, a decreasing curve passing through about Es/4 at 2R «and Es/9 at 3R»✓ 1Do not award MP3 for a straight line.
Part (b)
Q = ER²/k = 3.0 × 106 × 0.15² / 8.99 × 109 = 7.51 × 10−6 «C»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)
E = kQ/r² = 8.99 × 109 × 7.5 × 10−6 / 0.40² OR E = 3.0 × 106 × (0.15/0.40)²✓ 1
E = 4.2 × 105 N C−1✓ 1Accept 4.2–4.3 × 105 N C−1. Award [2] for CNA.
Part (d)
E at the surface = 8.99 × 109 × 25 × 10−6 / 0.25² = 3.6 × 106 N C−1✓ 1ALTERNATIVE: largest charge = 3.0 × 106 × 0.25² / 8.99 × 109 = 2.1 × 10−5 C.
this is greater than 3.0 × 106 N C−1 «25 μC is more than the largest charge», so it could not; a spark would occur✓ 1Conclusion must follow from a calculation.
Part (e)
charge flows onto the student by contact «the student is a conductor connected to the dome»✓ 1
each hair gains charge of the same sign, so the hairs repel each other «and the head» and spread out✓ 1OWTTE
Part (f)
otherwise the charge would flow through the student to the ground «the student would be earthed»✓ 1OWTTE

Answers: (b) 7.51 × 10−6 C  ·  (c) 4.2 × 105 N C−1  ·  (d) No: 3.6 × 106 N C−1 at the surface (the remaining parts are explanations — see the table above)

110D-2-27Electric field lines Paper 2Easy11 marks

The base of a thundercloud is 1.2 km above flat, level ground and carries a large negative charge. The potential difference between the base of the cloud and the ground is 3.0 × 107 V. A building has a pointed metal lightning conductor on its roof, connected to the earth by a thick copper strip.

Air breaks down, allowing a discharge, when the electric field strength reaches 3.0 × 106 V m−1.

Base of a thundercloud above flat ground; a building with a pointed lightning conductor stands on the ground (not to scale)−−−−−−−−−−−lightning conductorground
Diagram NOT to scale
(a)

Explain why the surface of the ground below the cloud becomes positively charged.

(2)
(b)

Draw, on the diagram, the electric field lines between the base of the cloud and the ground, including the region close to the lightning conductor.

(3)
(c)

Determine whether the air breaks down in the region well away from the building. Treat the field there as uniform.

(2)
(d)

Explain, with reference to your diagram, why a discharge is most likely to begin at the tip of the lightning conductor.

(2)
(e)

Suggest why the lightning conductor is connected to the earth by a thick copper strip.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
free electrons in the ground are repelled by the negative charge on the cloud✓ 1
they move away «deeper into the earth», leaving the surface below the cloud with a positive charge «charging by induction»✓ 1OWTTE
Part (b)
away from the building: straight, parallel, equally spaced vertical lines✓ 1
arrows pointing upwards, from the ground to the cloud✓ 1Do not award MP2 for arrows pointing down.
lines meet the ground and the conductor at right angles and are crowded together at the tip of the conductor✓ 1
Part (c)
E = V/d = 3.0 × 107 / 1200 = 2.5 × 104 V m−1✓ 1
this is about 120 times smaller than 3.0 × 106 V m−1, so the air does not break down there✓ 1Conclusion must follow from a calculation.
Part (d)
the field lines are closest together at the sharp tip✓ 1
so the field strength there is far greater than in the uniform region and can reach the breakdown value✓ 1MP2 only scores if MP1 scores.
Part (e)
in a strike a very large charge «current» flows to the earth through the conductor✓ 1
thick copper has a very small resistance, so little energy is dissipated in it «P = I²R» and the current flows through the strip rather than through the building✓ 1OWTTE

Answers: (c) 2.5 × 104 V m−1; no (the remaining parts are explanations — see the table above)

111D-2-43Magnetic field patterns Paper 2Easy12 marks

A long air-core solenoid has 600 turns wound uniformly over a length of 0.30 m and carries a current of 2.0 A. The diagram shows a cross-section through its axis. In the top row of turns the current is out of the page; in the bottom row it is into the page.

The magnetic flux density inside a long solenoid is B = μ0nI, where n is the number of turns per unit length. μ0 = 4π × 10−7 T m A−1; e = 1.60 × 10−19 C.

Cross-section of an air-core solenoid with ends A and B; current out of the page in the top row of turns and into the page in the bottom row; an electron gun fires along the axisABaxiselectron gun
Diagram NOT accurately drawn
(a)
(i)

Draw, on the diagram, the magnetic field lines inside and outside the solenoid. Show the direction of the field.

(3)
(ii)

Deduce which end of the solenoid, A or B, behaves like the north pole of a bar magnet.

(1)
(iii)

Outline one similarity and one difference between the magnetic field of the solenoid and the magnetic field of a bar magnet of the same size.

(2)
(b)

Show that the magnetic flux density inside the solenoid is about 5.0 × 10−3 T.

(1)
(c)

Electrons leave the electron gun with a speed of 1.87 × 107 m s−1.

(i)

Explain why an electron moving along the axis of the solenoid is not deflected.

(2)
(ii)

Calculate the magnitude of the magnetic force on an electron that moves inside the solenoid at 10° to the axis.

(2)
(iii)

State, with a reason, whether the speed of the electron in (c)(ii) changes while it is inside the solenoid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
inside: straight, parallel, equally spaced lines along the axis✓ 1At least three lines.
outside: closed loops leaving one end and returning to the other end, more widely spaced than inside✓ 1Lines must not cross or touch the turns.
direction: from A to B inside the solenoid and from B round to A outside✓ 1Use of the right-hand grip rule. Award MP3 only if the arrows are consistent on all lines.
Part (a)(ii)
B «field lines leave the solenoid at B»✓ 1Allow ECF from the direction drawn in (a)(i).
Part (a)(iii)
similarity: the field outside has the same shape, with lines leaving the north end and returning to the south end✓ 1
difference: inside the solenoid the field is uniform and in air «so a beam can pass through it» OR the field of the solenoid can be switched off, varied or reversed by changing the current✓ 1Any one difference.
Part (b)
B = 4π × 10−7 × (600/0.30) × 2.0 = 5.03 × 10−3 «T»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(i)
the velocity is parallel to the magnetic field, so θ = 0 «and sin θ = 0»✓ 1
F = qvB sin θ = 0, so there is no magnetic force and the velocity does not change✓ 1OWTTE
Part (c)(ii)
F = 1.60 × 10−19 × 1.87 × 107 × 5.03 × 10−3 × sin 10°✓ 1Accept B = 5.0 × 10−3 T.
F = 2.6 × 10−15 N✓ 1Award [1] max for 1.5 × 10−14 N «sin θ omitted».
Part (c)(iii)
no, because the magnetic force is always perpendicular to the velocity, so it does no work «and the kinetic energy stays constant»✓ 1Reason required.

Answers: (b) 5.03 × 10−3 T  ·  (c)(ii) 2.6 × 10−15 N (the remaining parts are explanations — see the table above)

112D-2-45Coulomb's law Paper 2Easy11 marks

A bumblebee flying through dry air gains a positive charge of 32 pC. The bee hovers above a flower, which is earthed through the plant and the soil. A pollen grain of mass 2.0 × 10−11 kg carrying a charge of −5.0 fC lies on a petal directly below the bee. Treat the bee and the pollen grain as point charges and ignore any other forces holding the grain to the petal.

k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C; 1 fC = 10−15 C.

A positively charged bee hovering above a flower petal; a pollen grain lies on the petal directly below the bee; the flower is earthed through its stembeepetalpollen grainstem
Diagram NOT to scale
(a)

Explain how the bee becomes positively charged as it flies.

(2)
(b)

Calculate the number of electrons that the bee has lost.

(1)
(c)

Draw, on the diagram, the electric field lines between the bee and the petal.

(2)
(d)

Determine the greatest height of the bee above the pollen grain for which the grain can jump up to the bee.

(3)
(e)

Suggest one reason why the real greatest height may be different from your answer to (d).

(1)
(f)

When the bee lands on the flower it loses most of its charge.

Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
friction «rubbing» between the bee and the air molecules transfers electrons✓ 1
electrons move from the bee to the air, so the bee is left with more protons than electrons✓ 1Do not accept «the bee gains positive charges/protons».
Part (b)
N = 32 × 10−12 / 1.60 × 10−19 = 2.0 × 108✓ 1
Part (c)
lines start on the bee and end on the petal, with arrows pointing from the bee towards the petal✓ 1
lines meet the petal at right angles AND are closest together near the bee «directly below it»✓ 1Do not award MP2 if lines cross or end in mid-air.
Part (d)
weight of grain = 2.0 × 10−11 × 9.81 = 1.96 × 10−10 N✓ 1
kQq/r² = mg ⇒ r² = 8.99 × 109 × 32 × 10−12 × 5.0 × 10−15 / 1.96 × 10−10✓ 1The attraction must at least equal the weight.
r = 2.7 × 10−3 m «2.7 mm»✓ 1Award [3] for CNA. Award [2] max for 7.3 × 10−6 m «square root not taken».
Part (e)
the earthed petal has charges induced on it that also exert forces on the grain OR the grain is held by adhesive forces OR the bee is not a point charge «its charge is spread over its body»✓ 1Any one.
Part (f)
the bee touches an earthed conductor «charge transfer by contact»✓ 1
electrons flow from the earth through the plant to the bee, neutralising its positive charge✓ 1Do not accept «positive charge flows into the bee».

Answers: (b) 2.0 × 108  ·  (d) 2.7 mm (the remaining parts are explanations — see the table above)

113D-2-58Electric field strength Paper 2Easy13 marks

In gel electrophoresis, fragments of DNA are separated according to their length. A slab of gel soaked in a conducting solution lies between two flat electrodes X and Y that are 0.15 m apart. The potential difference between the electrodes is 120 V, with Y positive. Samples of DNA are placed in wells near X. Treat the electric field in the gel as uniform.

In a simple model, each base pair of a DNA fragment in the solution carries a charge of −2e. e = 1.60 × 10−19 C.

Top view of an electrophoresis tank: electrode X (negative) on the left and electrode Y (positive) on the right, 0.15 m apart, with a gel slab between them, wells near X and a DNA fragment F in the middle of the gelX (−)Y (+)gelwellsF0.15 m
Figure 1 — Diagram NOT to scale
(a)
(i)

Calculate the electric field strength in the gel.

(1)
(ii)

State the direction of the electric field in the gel and explain why the DNA fragments move towards Y.

(2)
(b)

Fragment F has 1500 base pairs.

(i)

Show that the magnitude of the electric force on F is about 4 × 10−13 N.

(1)
(ii)

Outline why the charge on any fragment must be a whole-number multiple of e.

(1)
(c)

Fragment F moves 60 mm through the gel in 40 minutes at constant speed.

(i)

Draw, on Figure 1, labelled arrows to represent the horizontal forces acting on F.

(2)
(ii)

Explain why F moves at a constant speed rather than accelerating.

(2)
(iii)

Calculate the speed of F.

(1)
(d)

The resistive force of the gel on a fragment is proportional to the speed of the fragment. Predict the distance that F moves in 40 minutes when the potential difference is increased to 180 V.

(2)
(e)

A fragment with 4500 base pairs carries three times the charge of F but moves a shorter distance in the same time. Suggest what this shows about the resistive force on the longer fragment.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
E = V/d = 120/0.15 = 800 V m−1✓ 1Accept N C−1.
Part (a)(ii)
from Y to X «from the positive to the negative electrode»✓ 1
the fragments are negatively charged, so the force on them is opposite to the field, towards Y «unlike charges attract»✓ 1Do not accept «DNA is attracted to the positive field».
Part (b)(i)
q = 1500 × 2 × 1.60 × 10−19 = 4.8 × 10−16 C AND F = qE = 4.8 × 10−16 × 800 = 3.84 × 10−13 «N»✓ 1Must see full substitution OR answer to 3 s.f. Allow ECF from (a)(i).
Part (b)(ii)
charge is quantized: a fragment's charge comes from a whole number of charged groups / electrons, each of charge «±»e✓ 1OWTTE. Accept reference to Millikan's experiment as evidence.
Part (c)(i)
arrow towards Y labelled electric force AND arrow towards X labelled resistive force / drag✓ 1Both arrows must start on F.
the two arrows are of equal length✓ 1Ignore vertical forces if they are also drawn.
Part (c)(ii)
the resistive force «of the gel» increases with speed until it is equal in magnitude to the electric force✓ 1
the resultant force is then zero, so «by Newton's first law» the velocity is constant✓ 1OWTTE.
Part (c)(iii)
v = 0.060/(40 × 60) = 2.5 × 10−5 m s−1✓ 1
Part (d)
the field and the electric force increase by a factor 180/120 = 1.5, so the speed at which the resistive force balances it is 1.5 times larger✓ 1
distance = 1.5 × 60 = 90 mm✓ 1Award [2] for CNA.
Part (e)
at the same speed, the resistive force on the longer fragment is more than three times that on F «the gel obstructs a longer molecule more»✓ 1OWTTE. Do not accept «it is heavier» alone.

Answers: (a)(i) 800 V m−1  ·  (b)(i) 3.8 × 10−13 N  ·  (c)(iii) 2.5 × 10−5 m s−1  ·  (d) 90 mm (the remaining parts are explanations — see the table above)

114D-2-65Charging by friction & induction Paper 2Easy12 marks

In a laser printer, the image of a page is formed on a rotating drum (Figure 1). The drum is an earthed metal cylinder covered with a thin coating of a photoconductor, a material that is an insulator in the dark but conducts when light falls on it. The outer surface of the coating is first given a uniform negative charge. A laser beam then scans the drum, and wherever the light falls the coating conducts.

e = 1.60 × 10−19 C.

End view of a laser-printer drum: an earthed metal cylinder with a photoconductor coating, a developer roller beside it and a laser beam striking the drumearthed metalcylinderphotoconductor coatingdeveloperdeveloper rollerlaser
Figure 1 — Diagram NOT to scale
(a)

Explain why the regions of the coating that are lit by the laser lose their charge.

(2)
(b)

Toner particles are tiny plastic grains. In the developer they are mixed and rubbed with carrier beads, and the toner particles become negatively charged. Explain, in terms of electrons, how the toner particles become charged, and deduce the sign of the charge on the carrier beads.

(2)
(c)

The developer roller carries toner particles past the drum, 0.20 mm from its surface. Opposite a lit region of the drum, the roller is 300 V negative relative to the surface of the coating. Treat the field in the gap as uniform.

(i)

Calculate the electric field strength in the gap.

(1)
(ii)

State the direction of the electric field in the gap and deduce the direction of the electric force on a toner particle.

(2)
(iii)

A toner particle carries a charge of −6.4 × 10−15 C. Calculate the number of excess electrons on the particle.

(1)
(iv)

Calculate the magnitude of the electric force on this particle in the gap.

(1)
(d)

Opposite an unlit region, the surface of the coating is 400 V negative relative to the roller. Explain why toner does not move onto the unlit regions.

(2)
(e)

Suggest why a laser printer may produce faint prints when the air is very humid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
where the light falls the coating becomes a conductor, so electrons can move through it✓ 1
the electrons flow to the earthed metal cylinder «to earth», so the charge on those regions is removed✓ 1Do not accept «positive charge flows in from earth» without the idea of electrons.
Part (b)
electrons are transferred by friction from the carrier beads to the toner particles✓ 1Do not accept transfer of positive charge/protons.
the beads become positive «with a total charge equal and opposite to that on the toner», because charge is conserved✓ 1
Part (c)(i)
E = V/d = 300/0.20 × 10−3 = 1.5 × 106 V m−1✓ 1Accept N C−1.
Part (c)(ii)
from the drum surface towards the roller «from the more positive to the more negative surface»✓ 1
the force on a negative toner particle is opposite to the field, so towards the drum✓ 1Allow ECF for a force opposite to the field direction given.
Part (c)(iii)
n = 6.4 × 10−15/1.60 × 10−19 = 4.0 × 104✓ 1
Part (c)(iv)
F = qE = 6.4 × 10−15 × 1.5 × 106 = 9.6 × 10−9 N✓ 1Allow ECF from (c)(i).
Part (d)
the field in the gap is now reversed, pointing from the roller towards the drum✓ 1
so the force on the negative toner is towards the roller / away from the drum «the negative coating repels the negative toner»✓ 1OWTTE.
Part (e)
water on surfaces / in the air conducts, so charge leaks away from the drum «and toner», making the fields and forces smaller✓ 1OWTTE.

Answers: (c)(i) 1.5 × 106 V m−1  ·  (c)(iii) 4.0 × 104  ·  (c)(iv) 9.6 × 10−9 N (the remaining parts are explanations — see the table above)

115D-2-68Magnetic field patterns Paper 2Easy12 marks

In transcranial magnetic stimulation (TMS), a short pulse of current in a coil held against the scalp produces a strong magnetic field inside the head, which is used to stimulate nerve cells. A “figure-of-eight” coil consists of two identical flat circular coils, C1 and C2, lying side by side in the same plane and touching at J. Each coil has 10 turns of radius 40 mm.

Figure 1 is a vertical cross-section through the centres of both coils. At one instant during a pulse the current in each coil is 4.0 kA, in the directions shown. X is a point in the head 20 mm below J.

The magnetic flux density at the centre of a flat circular coil of N turns and radius R is B = μ0NI/2R. The magnetic flux density at a distance r from a long straight wire is B = μ0I/2πr. μ0 = 4π × 10−7 T m A−1.

Vertical cross-section through the centres of two flat circular coils C1 and C2 lying side by side on the scalp. From left to right the wire bundles carry current into, out of, out of and into the page. J is the junction of the coils; X is 20 mm below J inside the head.scalpheadC1C2JX20 mm
Figure 1 — Diagram NOT to scale
(a)

Draw, on Figure 1, the magnetic field lines due to the two coils together in the region below the coils. Show the direction of the field.

(3)
(b)
(i)

Calculate the magnetic flux density at the centre of C1 due to C1 alone.

(1)
(ii)

State the direction of the magnetic field at the centre of C2 due to C2 alone.

(1)
(c)

Explain why the magnetic field just below J is stronger than the field due to either coil alone.

(2)
(d)

Estimate the magnetic flux density at X. Treat the wires of both coils at J as a single long straight wire.

(2)
(e)

Suggest one advantage of a figure-of-eight coil over a single circular coil for stimulating a small region of the brain.

(1)
(f)

State the effect on the magnetic field at X of reversing the current in both coils.

(1)
(g)

State and explain whether the wires of C1 and C2 at J attract or repel each other.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
lines pass downwards «into the head» through the middle of C1 and upwards through the middle of C2✓ 1Right-hand grip rule; both directions needed.
lines curve from below C1 to below C2, running horizontally from left to right «C1 towards C2» directly below J✓ 1Lines must not cross. At least three lines.
lines most closely spaced just below J and spreading further apart deeper in the head✓ 1OWTTE.
Part (b)(i)
B = 4π × 10−7 × 10 × 4.0 × 103/(2 × 0.040) = 0.63 T✓ 1
Part (b)(ii)
upwards «out of the head, perpendicular to the plane of the coil»✓ 1Accept «up the page».
Part (c)
the wires of C1 and C2 at J carry currents in the same direction «both out of the page»✓ 1
so below J the fields of the two coils are in the same direction and add «and these are the wires closest to points below J»✓ 1OWTTE.
Part (d)
current = 2 × 10 × 4.0 × 103 = 8.0 × 104 A✓ 1
B = 4π × 10−7 × 8.0 × 104/(2π × 0.020) = 0.80 T✓ 1Award [2] for CNA. Award [1] max for 0.40 T «one coil only».
Part (e)
the strongest field is concentrated in a small region below J, so only the nerve cells there are stimulated✓ 1OWTTE.
Part (f)
the field at X reverses direction «points from C2 towards C1» with the same magnitude✓ 1Both direction and unchanged magnitude are needed.
Part (g)
attract, because the currents in the two wires are in the same direction✓ 1Reason required.

Answers: (b)(i) 0.63 T  ·  (d) 0.80 T (the remaining parts are explanations — see the table above)

116D-2-70Magnetic field patterns Paper 2Easy12 marks

Magnetotactic bacteria contain a chain of tiny magnetic crystals that behaves like a small bar magnet. The Earth's magnetic field turns each bacterium so that the chain lies along a field line with its north pole pointing in the direction of the field. The bacterium then swims in the direction in which the north pole of its chain points.

The Earth drawn as a circle with its geographic North Pole at the top, the equator dashed, and point P in the northern hemispheregeographic North Polegeographic South PoleequatorP
Figure 1 — Diagram NOT to scale
Vertical cross-section through a pair of coaxial coils seen from the east, axis horizontal pointing north to the right; the wire of each coil crosses the page at the top and the bottom, with a dish of bacteria at the centredishnorthcoil 1coil 2
Figure 2 — Diagram NOT to scale
(a)

Sketch, on Figure 1, the magnetic field lines of the Earth outside the Earth. Show the direction of the field.

(3)
(b)

Explain why the bacteria in a pond at P swim downwards, towards the mud at the bottom.

(2)
(c)

At P the Earth's magnetic flux density is 49 μT, directed at 66° below the horizontal. Calculate the horizontal and vertical components of the flux density.

(2)
(d)

Researchers place a dish of bacteria at the centre of a pair of identical coaxial coils whose axis is horizontal and points north–south. Figure 2 is a vertical cross-section seen from the east. The coils carry equal currents in the same sense. At the centre, the coils produce a horizontal field of flux density B = 0.716μ0NI/R, where N = 40 is the number of turns in each coil and R = 0.20 m is the radius of each coil. μ0 = 4π × 10−7 T m A−1.

(i)

The researchers want the field of the coils at the centre to point south. State whether the current in the wires at the top of the coils in Figure 2 should be into or out of the page. Explain your answer.

(2)
(ii)

Determine the current needed so that the resultant horizontal field at the centre has the same magnitude as before but points south.

(2)
(iii)

Predict how the bacteria in the dish move once this current is switched on.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
pattern of a bar magnet: lines leave the surface in the southern hemisphere and re-enter in the northern hemisphere, curving round outside the Earth✓ 1At least four lines, symmetrical about the axis.
arrows pointing from the southern to the northern hemisphere outside the Earth «into the Earth near the geographic North Pole»✓ 1Do not award MP2 for arrows leaving near the North Pole.
lines closest together near the poles and parallel to the surface «horizontal» at the equator✓ 1
Part (b)
in the northern hemisphere the field lines point downwards into the Earth «the field has a downward component»✓ 1Allow ECF from (a).
the chain's north pole points along the field, north and down, so the bacteria swim downwards✓ 1
Part (c)
horizontal: 49 cos 66° = 20 μT «19.9 μT»✓ 1Accept 19.9–20 μT.
vertical: 49 sin 66° = 45 μT «44.8 μT»✓ 1Award [1] max if the components are interchanged.
Part (d)(i)
into the page✓ 1
by the right-hand grip rule, a current into the page at the top «and out of the page at the bottom» gives a field pointing left, towards the south, at the centre✓ 1MP2 can be awarded for a correct grip-rule argument even if the answer is stated as a field direction. OWTTE.
Part (d)(ii)
the coils must cancel the northward 20 μT and add 20 μT southwards: B = 2 × 19.9 = 39.8 μT✓ 1Allow ECF from (c).
I = BR/(0.716μ0N) = 39.8 × 10−6 × 0.20/(0.716 × 4π × 10−7 × 40) = 0.22 A✓ 1Award [1] max for 0.11 A «field cancelled only».
Part (d)(iii)
they turn round and swim southwards «and still downwards, because the vertical component is unchanged»✓ 1OWTTE.

Answers: (c) 20 μT horizontal, 45 μT vertical  ·  (d)(ii) 0.22 A (the remaining parts are explanations — see the table above)

117IBP2S-030-02-11Uniform fields between plates Paper 2Easy5 marks

An electrostatic precipitator removes smoke particles from the exhaust gases of a power-station chimney. Gas flows between two large vertical parallel metal plates 0.25 m apart. The potential difference between the plates is 40 kV. A smoke particle in the gas carries a negative charge of magnitude 3.2 × 10−18 C. Assume the field between the plates is uniform.

0.25 mparticle (−)gas flowp.d. = 40 kV
Diagram not to scale
(a)

Calculate the electric field strength between the plates.

(2)
(b)

Calculate the magnitude of the electric force on the smoke particle.

(2)
(c)

State, with a reason, the plate towards which the smoke particle moves.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses E = V/d = 40 × 103/0.25✓ 1
E = 1.6 × 105 V m−1 (or N C−1)✓ 1
Part (b)
Uses F = qE = 3.2 × 10−18 × 1.6 × 105✓ 1
F = 5.1 × 10−13 N✓ 1
Allow error carried forward from the field strength calculated in the first part.—
Part (c)
It moves towards the positive plate, because opposite charges attract. Also accept: the force on a negative charge is opposite to the field direction.✓ 1

Exam tip: Convert kV to V before substituting: 40 kV is 40 × 10³ V.

118IBP2S-030-03-11Charging by friction & induction Paper 2Easy6 marks

After a long drive, a driver slides across a polyester car seat and becomes negatively charged. She is wearing rubber-soled shoes. When she then touches the earthed metal casing of a fuel pump, she feels a small spark.

(a)

State which charged particles are transferred as she slides across the seat, and in which direction.

(2)
(b)

Outline why she stays charged while walking in rubber-soled shoes, but discharges when she touches the fuel pump.

(2)
(c)

Her charge just before touching the pump is −15 nC. Calculate the number of excess electrons on her.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Electrons ; transferred from the seat to the driver .✓ 2
Part (b)
Rubber is an insulator, so the charge cannot flow from her to the ground ; the metal casing is a conductor connected to earth, providing a path for the excess electrons to flow to earth .✓ 2
Part (c)
n = Q/e = 15×10−9 / 1.60×10−19 ; n = 9.4×1010 .✓ 2

Common error: Writing 'charge flows to earth' without saying that it is electrons, or giving their direction, often loses a mark.

119IBP2S-030-03-12Charged conducting sphere Paper 2Easy7 marks

In a science museum, the spherical metal (conducting) dome of an electrostatic generator has a radius of 0.12 m. It carries a charge of +2.0 μC, which is distributed uniformly over its surface. The dome is far from other objects.

(a)

Describe the pattern of the electric field lines outside the dome.

(2)
(b)

State the electric field strength inside the dome.

(1)
(c)

Calculate the electric field strength just outside the surface of the dome.

(2)
(d)

Calculate the electric field strength at a point 0.36 m from the centre of the dome.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Lines are radial, leaving the surface at right angles and evenly spaced around the sphere ; the direction is away from the dome, outwards .✓ 2
Part (b)
Zero .✓ 1
Part (c)
E = kQ/R2 = 8.99×109 × 2.0×10−6 / 0.122 ; E = 1.25×106 N C−1 .✓ 2
Part (d)
Uses E = kQ/r2 with r = 0.36 m measured from the centre, or divides the previous answer by 9 because the distance is three times larger ; E = 1.39×105 N C−1 . Allow error carried forward from the previous part.✓ 2

Exam tip: Field lines always meet the surface of a conductor at right angles. Outside a charged sphere, measure r from the centre, not from the surface.

120IBP2S-030-00-11Conservation of charge Paper 2Easy5 marks

In a packaging machine, two small charged polystyrene beads can be treated as point charges. One bead carries +1.5 nC and the other carries −2.4 nC. Their centres are 4.0 cm apart. (k = 8.99 × 109 N m2 C−2)

(a)

State whether the force between the beads is attractive or repulsive.

(1)
(b)

Calculate the magnitude of the electric force between the beads.

(2)
(c)

Determine the separation at which the magnitude of the force would be 5.0 × 10−6 N.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Attractive, because the charges have opposite signs✓ 1
Part (b)
Substitutes into F = kq1q2/r2 = 8.99 × 109 × 1.5 × 10−9 × 2.4 × 10−9/(0.040)2, with r converted to metres✓ 1
F = 2.0 × 10−5 N✓ 1
Part (c)
Rearranges to r = √(kq1q2/F) = √(3.24 × 10−8/5.0 × 10−6)✓ 1
r = 0.080 m, i.e. 8.0 cm✓ 1
Alternative: the force falls to a quarter of the previous value, so by the inverse-square law the separation doubles to 8.0 cm. For this method, allow error carried forward from the previous part's force.—

Common error: Convert nC to C and cm to m before substituting. Leaving the separation in cm makes the force 10⁴ times too small.

121IBP2S-030-05-11Quantization of charge Paper 2Easy5 marks

A medical linear accelerator used in radiotherapy produces pulses of electrons. Each pulse carries a total charge of magnitude 0.40 μC. The machine delivers 200 pulses per second.

(a)

State what is meant by the statement that electric charge is quantized.

(1)
(b)

Calculate the number of electrons in one pulse.

(2)
(c)

Calculate the number of electrons delivered per second.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any charge exists only as a whole-number (integer) multiple of the elementary charge e = 1.60×10−19 C. Accept equivalent wording, e.g. "charge comes in discrete units of e"✓ 1
Part (b)
n = Q/e = 0.40×10−6 / 1.60×10−19✓ 1
n = 2.5×1012✓ 1
Part (c)
Multiplies the number of electrons per pulse by 200, i.e. 2.5×1012 × 200. Allow error carried forward from the previous part. Alternative: charge per second = 0.40×10−6 × 200 = 8.0×10−5 C, then divides by 1.60×10−19 C✓ 1
Answer: 5.0×1014 s−1✓ 1

Common error: Convert μC to C (×10⁻⁶) before dividing by e. Otherwise the answer is wrong by a factor of a million.

122IBP2S-030-04-11Charging by friction & induction Paper 2Easy7 marks

In a science museum, a moving belt transfers charge to the dome of a Van de Graaff generator. In 8.0 s the dome gains a charge of +2.4 μC. Assume this charge results entirely from electrons leaving the dome and being carried away by the belt. The charge on an electron is −1.60×10−19 C.

(a)

State the law of conservation of charge.

(1)
(b)

Calculate the average rate at which charge is delivered to the dome.

(2)
(c)

Calculate the number of electrons removed from the dome in this time.

(2)
(d)

Outline how the charging of the dome is consistent with the law of conservation of charge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The total electric charge of an isolated system remains constant; charge is neither created nor destroyed, only transferred .✓ 1
Part (b)
I = Q/t = 2.4×10−6 / 8.0✓ 1
= 3.0×10−7 A (or C s−1)✓ 1
Part (c)
N = Q/e = 2.4×10−6 / 1.60×10−19✓ 1
= 1.5×1013 electrons✓ 1
Part (d)
The electrons that leave the dome are not destroyed. They are transferred to the belt and lower parts of the machine (comb, roller or earth). An equal negative charge of −2.4 μC therefore appears elsewhere .✓ 1
The total charge of the whole isolated system, for example the generator plus its surroundings or earth, is unchanged. Charge has been separated or transferred, not created .✓ 1

Exam tip: A positively charged object has lost electrons. To find the number of electrons, divide the magnitude of the charge by e.

123D-2-03Gravity compared with electrostatics Paper 2Medium11 marks

Two protons are a distance r apart in a vacuum. The mass of a proton is 1.67 × 10−27 kg.

(a)
(i)

Show that (electric force between the protons)/(gravitational force between the protons) is about 1036.

(1)
(ii)

Explain why this ratio does not depend on r.

(1)
(iii)

State one difference, other than their sizes, between the two forces in (a)(i).

(1)
(b)

Two identical dust grains drift in space, far from any other body. Each grain has mass m and charge q.

(i)

Show that the electric force and the gravitational force between the grains have equal magnitudes when q = m√(G/k).

(2)
(ii)

Each grain has a mass of 3.0 × 10−15 kg. Determine whether the electric force and the gravitational force between the two grains can cancel exactly.

(3)
(c)

Suggest why the motion of the planets around the Sun is determined by gravitational forces and not by electric forces, even though the electric force is the stronger interaction.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
ratio = ke²/(Gmp²) = 8.99 × 109 × (1.60 × 10−19)²/(6.67 × 10−11 × (1.67 × 10−27)²) = 1.24 × 1036✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
both forces are inversely proportional to r², so r cancels in the ratio✓ 1
Part (a)(iii)
the gravitational force is attractive whereas the electric force between two protons is repulsive✓ 1Accept «gravitational force acts between masses, electric force between charges». OWTTE
Part (b)(i)
kq²/r² = Gm²/r²✓ 1
q² = Gm²/k, so q = m√(G/k)✓ 1
Part (b)(ii)
q = 3.0 × 10−15 × √(6.67 × 10−11/8.99 × 109) = 2.6 × 10−25 C✓ 1
this is about 10−6 of the elementary charge «1.60 × 10−19 C»✓ 1Allow ECF from MP1.
charge is quantized, so a grain can only carry a whole-number multiple of e; with any charge at all the electric force is far larger than the gravitational force, so the forces cannot cancel✓ 1MP3 only scores if a comparison with e is made.
Part (c)
the Sun and the planets are electrically neutral overall «equal amounts of positive and negative charge»✓ 1
so the attractive and repulsive electric forces between their particles cancel and the resultant electric force is negligible✓ 1
gravitational forces are always attractive, so the forces on all the particles add, and the masses involved are very large✓ 1OWTTE

Answers: (a)(i) 1.24 × 1036  ·  (b)(ii) q = 2.6 × 10−25 C (the remaining parts are explanations — see the table above)

124D-2-24Charging by friction & induction Paper 2Medium11 marks

An uncharged metal sphere stands on an insulating support. A negatively charged plastic rod is brought close to the sphere but does not touch it, as shown in Figure 1.

k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C.

A negatively charged plastic rod held close to an uncharged metal sphere on an insulating support−−−−−−−−−charged plastic rodmetal sphereinsulating support
Figure 1
(a)

Annotate Figure 1 to show the distribution of charge on the sphere.

(2)
(b)

Explain why the sphere is attracted towards the rod even though the total charge on the sphere is zero.

(2)
(c)

In a simple model, the charge on the rod is a point charge of −36 nC. The charges induced on the sphere are point charges of +4.0 nC and −4.0 nC, at distances of 0.050 m and 0.090 m from the charge on the rod, on the line through the centre of the sphere.

Determine the resultant electric force on the sphere.

(3)
(d)

With the rod held in place, the far side of the sphere is touched briefly with a wire connected to earth. The wire is then removed, and finally the rod is taken away.

(i)

State and explain the direction in which electrons flow through the wire.

(2)
(ii)

Calculate the number of electrons that flowed through the wire, given that the sphere carries a charge of +4.0 nC once the rod has been removed.

(1)
(e)

Suggest one limitation of the model used in (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
positive charges shown on the side of the sphere nearest the rod✓ 1
an equal number of negative charges shown on the far side of the sphere✓ 1Do not award MP2 if charges are drawn inside the metal.
Part (b)
the induced positive charge is closer to the rod than the induced negative charge✓ 1
the force varies as 1/r², so the attraction of the near charges is larger than the repulsion of the far charges, giving a resultant attraction✓ 1OWTTE
Part (c)
attraction = 8.99 × 109 × 36 × 10−9 × 4.0 × 10−9 / 0.050² = 5.18 × 10−4 N✓ 1
repulsion = 8.99 × 109 × 36 × 10−9 × 4.0 × 10−9 / 0.090² = 1.60 × 10−4 N✓ 1
resultant = 3.6 × 10−4 N towards the rod✓ 1Direction needed for MP3. Award [3] for CNA.
Part (d)(i)
from the sphere to the earth✓ 1
the «free» electrons in the sphere are repelled by the negative rod, and the wire lets them move further away from it✓ 1OWTTE
Part (d)(ii)
n = 4.0 × 10−9 / 1.60 × 10−19 = 2.5 × 1010✓ 1
Part (e)
any one of: the induced charge is spread over the surface of the sphere, not at two points / the charge on the rod is spread along its length / the induced charges change as the separation changes✓ 1OWTTE

Answers: (c) 3.6 × 10−4 N towards the rod  ·  (d)(ii) 2.5 × 1010 electrons (the remaining parts are explanations — see the table above)

125D-2-28Coulomb's law Paper 2Medium13 marks

In a simple model of the hydrogen atom, an electron moves at constant speed in a circular orbit of radius 5.3 × 10−11 m around a stationary proton.

k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.

protonelectronr = 5.3 × 10⁻¹¹ m
Diagram not to scale
(a)
(i)

Calculate the magnitude of the electric force on the electron.

(2)
(ii)

State the direction of this force and outline why it does not change the speed of the electron.

(2)
(b)

Calculate the electric field strength due to the proton at the position of the electron.

(1)
(c)

Determine the number of orbits the electron completes in one second.

(4)
(d)

Outline one similarity and one difference between this model and a planet moving in a circular orbit around the Sun.

(2)
(e)

In this model the electron could orbit at any radius, so the energy of the atom could take any value.

Explain how the emission spectrum of hydrogen shows that the model is incomplete.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
F = 8.99 × 109 × (1.60 × 10−19)² / (5.3 × 10−11)²✓ 1
F = 8.2 × 10−8 N✓ 1Award [2] for CNA.
Part (a)(ii)
towards the proton «the centre of the orbit»✓ 1
the force is perpendicular to the velocity, so it does no work on the electron «and its kinetic energy is constant»✓ 1OWTTE
Part (b)
E = F/e OR ke/r² = 5.1 × 1011 N C−1✓ 1Allow ECF from (a)(i).
Part (c)
the electric force provides the centripetal force: F = mev²/r✓ 1
v = √(8.19 × 10−8 × 5.3 × 10−11 / 9.11 × 10−31) = 2.18 × 106 m s−1✓ 1Allow ECF from (a)(i).
T = 2πr/v = 1.53 × 10−16 s✓ 1
number per second = 1/T = 6.6 × 1015✓ 1Accept 6.5–6.7 × 1015. Award [4] for CNA.
Part (d)
similarity: in both, an inverse-square force directed towards the centre provides the centripetal force✓ 1OWTTE
difference: the electric force depends on the charges whereas gravity depends on the masses / the electric force is attractive here only because the charges are opposite, whereas gravity is always attractive✓ 1Accept any one valid difference.
Part (e)
the spectrum consists of discrete lines, so only photons of certain frequencies, and hence certain energies «E = hf», are emitted✓ 1
each photon energy equals the difference between two energy levels, so the atom has discrete energy levels, whereas the model allows a continuous range «and so predicts a continuous spectrum»✓ 1OWTTE

Answers: (a)(i) 8.2 × 10−8 N  ·  (b) 5.1 × 1011 N C−1  ·  (c) 6.6 × 1015 orbits («v = 2.18 × 106 m s−1») (the remaining parts are explanations — see the table above)

126D-2-29Magnetic field patterns Paper 2Medium10 marks

A long straight vertical wire W passes through a horizontal card. The diagram shows the card from above; the current in W is upwards, out of the page. A small compass C rests on the card 0.050 m due south of W.

The horizontal component of the Earth's magnetic field is 2.0 × 10−5 T, directed north. The magnetic flux density at a distance r from a long straight wire carrying current I is B = μ0I/2πr. μ0 = 4π × 10−7 T m A−1.

Top view: vertical wire W carrying current upwards (out of the page); compass C 0.050 m due south of W; north is up the page and east to the rightWC0.050 mNE
Diagram NOT accurately drawn
(a)

Draw, on the diagram, three magnetic field lines due to the current in W. Show the direction of the field.

(2)
(b)

State the direction of the magnetic field due to W at the position of C.

(1)
(c)

Calculate the current in W for which its magnetic field at C is equal in magnitude to the horizontal component of the Earth's field.

(1)
(d)

The current in W is set to the value calculated in (c).

(i)

Determine the magnitude of the resultant horizontal magnetic field at C and the direction in which the compass needle points.

(3)
(ii)

State the direction in which the needle points when the current is reversed.

(1)
(iii)

Determine the angle between the needle and north when, with the current in its original direction, the compass is moved to a point 0.10 m due south of W.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
three concentric circles centred on W, with the spacing increasing with distance from W✓ 1
arrows anticlockwise «as seen from above»✓ 1Right-hand grip rule with the current out of the page.
Part (b)
east «to the right»✓ 1Allow ECF from (a).
Part (c)
I = 2πrB/μ0 = 2π × 0.050 × 2.0 × 10−5 / 4π × 10−7 = 5.0 A✓ 1
Part (d)(i)
two equal fields at right angles, one north and one east, combine as vectors✓ 1
resultant = √2 × 2.0 × 10−5 = 2.8 × 10−5 T✓ 1
the needle points north-east «45° east of north»✓ 1Allow ECF from (b).
Part (d)(ii)
north-west «45° west of north»✓ 1
Part (d)(iii)
the field due to W halves to 1.0 × 10−5 T, so tan θ = 1.0 × 10−5 / 2.0 × 10−5✓ 1
θ = 27° east of north✓ 1Accept 26°–27°. Direction not required.

Answers: (c) 5.0 A  ·  (d)(i) 2.8 × 10−5 T; north-east  ·  (d)(iii) 27° (the remaining parts are explanations — see the table above)

127D-2-42Electric field strength Paper 2Medium13 marks

In fine weather there is an electric field of strength 120 N C−1 directed vertically downwards just above flat ground. Some small spiders travel through the air by "ballooning": a spider releases fine silk threads that carry a negative charge, and the electric field helps to lift it.

Model the Earth as a charged conducting sphere of radius 6.37 × 106 m. k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C.

A spider hanging below fine silk threads above flat ground, with blades of grass to the leftflat groundgrasssilk threadsspider
Diagram NOT to scale
(a)
(i)

State what is meant by electric field strength.

(1)
(ii)

Deduce the sign of the charge on the surface of the Earth.

(1)
(iii)

Show that the magnitude of the charge on the Earth is about 5.4 × 105 C.

(1)
(iv)

Calculate the number of excess electrons on each square metre of the Earth's surface.

(2)
(b)

A very small spider floats at rest above the flat ground, well away from the grass. Ignore the field due to the spider's own charge and any air resistance.

Draw, on the diagram, three electric field lines in the region around the spider, and arrows to represent the forces acting on the spider.

(2)
(c)

A spider of mass 0.50 mg and its threads carry a total charge of −2.0 nC.

(i)

Determine whether the fine-weather field above flat ground can lift this spider.

(2)
(ii)

Calculate the initial acceleration of the spider when it lets go of the tip of a blade of grass, where the electric field strength is 4.0 kV m−1.

(2)
(d)

Suggest, with reference to field-line density, why the electric field alone cannot carry the spider high above the grass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
the «electric» force per unit charge acting on a small positive «test» charge placed at the point✓ 1Accept E = F/q with symbols defined. Do not accept «force on a charge».
Part (a)(ii)
negative, because field lines point towards «end on» negative charge✓ 1Reason required.
Part (a)(iii)
Q = ER²/k = 120 × (6.37 × 106)² / 8.99 × 109 = 5.42 × 105 «C»✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(iv)
charge per square metre = 5.42 × 105 / 4π(6.37 × 106)² = 1.06 × 10−9 «C m−2»✓ 1
number = 1.06 × 10−9 / 1.60 × 10−19 = 6.6 × 109 «m−2»✓ 1Allow ECF from MP1. Award [2] for CNA.
Part (b)
at least three straight, parallel, equally spaced vertical lines with arrows pointing down✓ 1Lines must not start or end on the spider.
downward arrow labelled weight AND upward arrow labelled electric force, of equal length, both starting on the spider✓ 1
Part (c)(i)
electric force = 2.0 × 10−9 × 120 = 2.4 × 10−7 N AND weight = 0.50 × 10−6 × 9.81 = 4.9 × 10−6 N✓ 1Accept comparison of the field needed, mg/q = 2.5 × 103 N C−1, with 120 N C−1, OR of the charge needed, mg/E = 4.1 × 10−8 C, with 2.0 nC.
the electric force is much less than the weight «about 5 %», so no✓ 1MP2 only scores if MP1 scores.
Part (c)(ii)
F = 2.0 × 10−9 × 4.0 × 103 = 8.0 × 10−6 N, so resultant upward force = 8.0 × 10−6 − 4.9 × 10−6 «N»✓ 1
a = 3.1 × 10−6 / 5.0 × 10−7 = 6.2 m s−2 upwards✓ 1Award [1] max for 16 m s−2 «weight omitted».
Part (d)
the field lines are crowded near the tips «the field is strong only there»; above the grass they spread out to the uniform spacing of the fine-weather field✓ 1OWTTE
so the field strength falls towards 120 N C−1, where the electric force is less than the weight «from (c)(i)», so the spider then needs moving air to keep rising✓ 1Allow ECF from (c)(i).

Answers: (a)(iii) 5.42 × 105 C  ·  (a)(iv) 6.6 × 109 m−2  ·  (c)(i) 2.4 × 10−7 N < 4.9 × 10−6 N — no  ·  (c)(ii) 6.2 m s−2 (the remaining parts are explanations — see the table above)

128D-2-44Charging by friction & induction Paper 2Medium15 marks

In a continuous inkjet printer, a jet of electrically conducting ink leaves a nozzle connected to an earthed ink supply (Figure 1). The jet breaks up into droplets inside a charging electrode that is held at a positive potential difference relative to the earth. Each droplet keeps the charge it has when it breaks away from the jet. The droplets then pass between two deflection plates 2.0 mm apart, with a potential difference of 1.5 kV between them, before they reach the paper.

k = 8.99 × 109 N m² C−2; e = 1.60 × 10−19 C.

Continuous inkjet: an earthed ink supply feeds a jet that breaks into droplets inside a positive charging electrode; the droplets then pass between deflection plates towards the paperink supplycharging electrode (+)+1.5 kV0 Vdeflection platespaper
Figure 1 — Diagram NOT to scale
(a)

Annotate Figure 1 to show the charge on the part of the jet that is inside the charging electrode.

(2)
(b)

Explain, with reference to the earthed ink supply, why each droplet is negatively charged when it breaks away from the jet.

(3)
(c)

In a test, every droplet carries a charge of −0.50 pC and 8.0 × 104 droplets are formed each second.

(i)

Calculate the number of excess electrons on one droplet.

(1)
(ii)

Calculate the current in the earth connection of the ink supply.

(2)
(d)

Calculate the magnitude of the electric force on a droplet between the deflection plates.

(2)
(e)

Neighbouring droplets in the stream are 0.20 mm apart.

(i)

Determine whether the electric force between two neighbouring droplets can be neglected compared with the force in (d).

(3)
(ii)

Suggest, with a calculation, why printers that charge only every second droplet, leaving the droplets between them uncharged, print more accurately.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
negative charges «− signs» drawn on the part of the jet inside the electrode✓ 1
no positive charges drawn on the jet, nozzle or ink supply✓ 1Positive charges may be shown on the electrode.
Part (b)
the positive electrode attracts electrons in the conducting ink «towards the end of the jet»✓ 1
because the ink supply is earthed, electrons flow from the earth into the jet «so the jet does not become positive anywhere»✓ 1Do not accept «positive charge flows to earth» unless electrons are also mentioned.
a droplet that breaks away is insulated from the earth, so it keeps its excess electrons✓ 1OWTTE
Part (c)(i)
N = 0.50 × 10−12 / 1.60 × 10−19 = 3.1 × 106✓ 1
Part (c)(ii)
charge is conserved, so the charge carried away by the droplets each second flows in through the earth connection: I = Δq/Δt = 8.0 × 104 × 0.50 × 10−12✓ 1
I = 4.0 × 10−8 A✓ 1Award [2] for CNA.
Part (d)
E = V/d = 1.5 × 103 / 2.0 × 10−3 = 7.5 × 105 V m−1✓ 1
F = qE = 0.50 × 10−12 × 7.5 × 105 = 3.8 × 10−7 N✓ 1Award [2] for CNA.
Part (e)(i)
F = kq²/r² = 8.99 × 109 × (0.50 × 10−12)² / (0.20 × 10−3)²✓ 1
F = 5.6 × 10−8 N✓ 1
this is about 15 % of the deflecting force, so it cannot be neglected✓ 1Allow ECF from (d). MP3 only scores if MP2 scores.
Part (e)(ii)
the charged droplets are now 0.40 mm apart, so the force between them is ¼ as large «1.4 × 10−8 N»✓ 1Inverse-square reasoning required.
this is only about 4 % of the deflecting force, so the droplets land closer to their intended positions «the uncharged droplets exert no Coulomb force»✓ 1OWTTE

Answers: (c)(i) 3.1 × 106  ·  (c)(ii) 4.0 × 10−8 A  ·  (d) 3.8 × 10−7 N  ·  (e)(i) 5.6 × 10−8 N — about 15 %, not negligible (the remaining parts are explanations — see the table above)

129D-2-59Charging by friction & induction Paper 2Medium15 marks

In electrostatic flocking, short nylon fibres are made to stand upright on a surface to give it a velvet finish. The fibres have a coating that lets them conduct slightly. They rest on a horizontal metal mesh M that is held at a potential difference of 40 kV relative to earth, M being negative. The object to be flocked is an earthed metal tube coated with wet adhesive, held horizontally above M with its lowest point 0.10 m above M,

as shown in Figure 1.

Cross-section: a horizontal metal mesh M at the bottom with fibres lying on it, and an earthed metal tube coated with adhesive held horizontally 0.10 m above itM — horizontal metal mesh, 40 kV negative relative to earthtubeearthearthed metal tube coated with wet adhesive0.10 m
Figure 1 — Diagram NOT to scale
(a)

Explain how a fibre resting on M becomes negatively charged and why it then leaves M.

(2)
(b)

Draw, on Figure 1, at least five electric field lines in the region between M and the tube.

(3)
(c)

A charged fibre lines up along the field line on which it travels. Suggest, with reference to your answer to (b), why the fibres stick into the adhesive at right angles to the surface of the tube.

(1)
(d)

A fibre has a mass of 2.0 × 10−10 kg and a charge of −1.5 × 10−14 C. Close to M, treat the field as uniform with the strength of a uniform field across the 0.10 m gap.

(i)

Determine whether the fibre can rise from M towards the tube.

(3)
(ii)

Explain, with reference to your answer to (b), how the electric force on the fibre changes as it rises towards the tube.

(2)
(e)

Explain why the tube must be earthed for flocking to continue.

(2)
(f)

Some fibres land on a small area of the tube where there is no adhesive. These fibres fall back to M and then rise again, many times. Explain this behaviour.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
by contact: electrons flow from M onto the conducting fibre «the fibre gets charge of the same sign as M»✓ 1Do not accept «positive charge leaves the fibre» without electrons.
like charges repel, so the fibre is pushed away from M✓ 1
Part (b)
lines start on the tube and end on M, with arrows pointing towards M «downwards»✓ 1Lines must not cross or stop in mid-air.
lines meet the surface of the tube and the mesh at right angles✓ 1
lines closest together near the lowest part of the tube, with outer lines curving round to reach the sides of the tube✓ 1OWTTE.
Part (c)
the field lines meet the conducting surface at right angles, and each fibre arrives lying along a field line, so it enters the surface perpendicular to it✓ 1Allow ECF from (b).
Part (d)(i)
E = V/d = 4.0 × 104/0.10 = 4.0 × 105 V m−1✓ 1
electric force = 1.5 × 10−14 × 4.0 × 105 = 6.0 × 10−9 N AND weight = 2.0 × 10−10 × 9.81 = 2.0 × 10−9 N✓ 1
the «upward» electric force is greater than the weight «about 3 times», so the fibre rises✓ 1MP3 only scores if MP2 scores. Direction: the force on a negative charge is opposite to the downward field.
Part (d)(ii)
the field lines become closer together near «the lowest part of» the tube, so the electric field strength increases as the fibre rises✓ 1Allow ECF from (b).
the charge on the fibre does not change, so the «upward» electric force F = qE increases✓ 1OWTTE. Do not accept «the force increases because the fibre is closer to the tube» without reference to field strength or field lines.
Part (e)
each fibre that arrives carries negative charge «electrons» to the tube✓ 1
if the tube were not earthed it would become negative and repel / weaken the force on the arriving fibres; earthing lets the electrons flow away to earth✓ 1OWTTE.
Part (f)
on touching the bare earthed metal, the fibre loses its electrons / becomes positively charged «takes the induced positive charge of the tube surface»✓ 1
the electric force «and its weight» then push it down to M, where it is charged negatively by contact again and is repelled upwards✓ 1OWTTE.

Answers: (d)(i) E = 4.0 × 105 V m−1; electric force 6.0 × 10−9 N > weight 2.0 × 10−9 N (the remaining parts are explanations — see the table above)

130D-2-61Magnetic field of a straight wire Paper 2Medium12 marks

A DC current clamp measures the current in a cable without breaking the circuit. Its jaws close around the cable. A magnetic field sensor S in the jaws, 15 mm from the centre of the jaws, measures the component of the magnetic flux density along the jaws. The clamp calculates the current from this reading, assuming that the cable passes through the centre of the jaws.

The magnetic flux density at a distance r from a long straight wire carrying a current I is B = μ0I/2πr. μ0 = 4π × 10−7 T m A−1.

Cross-section of the clamp closed round a single wire carrying a current into the page; the sensor S is at the top of the jaws, 15 mm from the centre of the wireSjaws of clampwire
Figure 1 — Diagram NOT to scale
Cross-section of the clamp closed round a two-core cable: core P carries current into the page and core Q current out of the page, 3.0 mm apartSPQjaws of clamp
Figure 2 — Diagram NOT to scale
(a)

The clamp is closed around a single straight wire that carries a current of 12.0 A into the page, as shown in Figure 1.

(i)

State the direction of the magnetic field at S.

(1)
(ii)

Calculate the magnetic flux density at S.

(1)
(iii)

The wire is moved inside the jaws so that its centre is 5.0 mm from the centre of the jaws, on the line towards S. Determine the percentage error in the current displayed by the clamp.

(2)
(iv)

A second sensor S′ is added on the opposite side of the jaws, also 15 mm from the centre. With the wire in the position described in (a)(iii), the clamp now calculates the current from the mean of the readings of S and S′. Determine whether this reduces the error found in (a)(iii).

(2)
(b)

The clamp is then closed around a two-core cable that connects a solar panel to a battery (Figure 2). Core P carries a current of 12.0 A into the page and core Q carries the return current of 12.0 A out of the page. The centres of the cores are 3.0 mm apart.

(i)

Explain why the reading of the clamp is now much smaller than in (a).

(2)
(ii)

Suggest how the clamp could be used to measure the current in this cable.

(1)
(c)

The clamp is again closed around the single wire of (a), with the wire at the centre of the jaws. A second long straight wire W, outside the jaws and parallel to the first wire, carries a current of 20.0 A out of the page. W lies on the straight line through the centre of the jaws and S, 30 mm beyond S. The clamp calculates the current from the reading of sensor S only, as in (a)(iii). Determine the current displayed by the clamp.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
to the right «clockwise round the wire, along the jaws»✓ 1Right-hand grip rule with the current into the page.
Part (a)(ii)
B = 4π × 10−7 × 12.0/(2π × 0.015) = 1.6 × 10−4 T✓ 1
Part (a)(iii)
B at S = 4π × 10−7 × 12.0/(2π × 0.010) = 2.4 × 10−4 T «1.5 times the value in (a)(ii)»✓ 1Allow ECF from (a)(ii).
displayed current = 18 A, so the error is «+»50 % «too large»✓ 1Award [2] for CNA. Accept «the reading is 50 % too high».
Part (a)(iv)
B at S′ = 4π × 10−7 × 12.0/(2π × 0.020) = 1.2 × 10−4 T AND mean = 1.8 × 10−4 T✓ 1Allow ECF from (a)(iii).
displayed current = 13.5 A «error ≈ 13 %», so yes, the error is reduced «S′ reads too low and partly compensates for S»✓ 1MP2 only scores if MP1 scores. Accept 12–13 %.
Part (b)(i)
the currents are in opposite directions, so at S the fields of P and Q are in «almost» opposite directions✓ 1
S is nearly the same distance from both cores «15 mm compared with a 3.0 mm spacing», so the fields have similar magnitudes and largely cancel✓ 1OWTTE.
Part (b)(ii)
separate the two cores and close the clamp round one core only✓ 1
Part (c)
B due to W at S = 4π × 10−7 × 20.0/(2π × 0.030) = 1.3 × 10−4 T✓ 1
S is below W and the current in W is out of the page, so at S the field of W is to the right «along the jaws», the same direction as the field of the wire in the jaws: the fields add, 1.6 × 10−4 + 1.3 × 10−4 = 2.9 × 10−4 T✓ 1Right-hand grip rule. Do not accept subtracting the fields because the currents are in opposite directions.
displayed current = 12.0 × (2.93 × 10−4)/(1.6 × 10−4) = 22 A✓ 1Accept 21–22 A. Allow ECF from MP2. Award [1] max for 2.0 A «fields subtracted».

Answers: (a)(ii) 1.6 × 10−4 T  ·  (a)(iii) 18 A displayed; error 50 %  ·  (a)(iv) mean 1.8 × 10−4 T; 13.5 A displayed (error ≈ 13 %) — reduced  ·  (c) 22 A displayed (the remaining parts are explanations — see the table above)

131D-2-62Coulomb's law Paper 2Medium13 marks

In a nuclear reactor, a nucleus of uranium-235 (proton number 92) absorbs a slow neutron and splits into a nucleus of barium-141 (proton number 56), a nucleus of krypton-92 and three neutrons.

Immediately after the split, the barium and krypton nuclei are at rest with their centres 1.7 × 10−14 m apart. Treat each nucleus as a point charge.

Mass of krypton-92 nucleus = 92 u; 1 u = 1.66 × 10−27 kg; k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C.

BaKrZ = 56Kr-921.7 × 10⁻¹⁴ m
Diagram not to scale. Distance is between the centres.
(a)
(i)

State the proton number of krypton-92 and the conservation law that determines it.

(2)
(ii)

Show that the electric force between the barium and krypton nuclei is about 1.6 × 103 N.

(1)
(b)

Explain why the two nuclei fly apart, even though the protons and neutrons inside each nucleus are held together by the strong nuclear force.

(2)
(c)
(i)

Calculate the initial acceleration of the krypton nucleus.

(2)
(ii)

Calculate the electric field strength due to the barium nucleus at the centre of the krypton nucleus.

(1)
(d)

As the nuclei move apart, the electric force between them decreases.

(i)

Calculate the separation of the centres of the nuclei when the electric force between them has fallen to 1.0 % of its initial value.

(1)
(ii)

Explain, without further calculation, why the nuclei gain most of their final kinetic energy before their separation reaches the value in (d)(i).

(2)
(e)

In a different fission event, the uranium-235 nucleus splits into two nuclei that each contain 46 protons, with their centres 1.7 × 10−14 m apart. Deduce how the initial electric force between these two nuclei compares with the force in (a)(ii).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
36✓ 1
conservation of «electric» charge «92 = 56 + 36; neutrons are uncharged»✓ 1Do not accept «conservation of nucleon number» alone.
Part (a)(ii)
F = 8.99 × 109 × 56 × 36 × (1.60 × 10−19)2/(1.7 × 10−14)2 = 1.61 × 103 «N»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)
the strong nuclear force has a very short range «about 10−15 m», so it is negligible between nuclei whose centres are 1.7 × 10−14 m apart✓ 1
both nuclei are positive, so the only significant force between them is the «repulsive» electric force, which pushes them apart✓ 1OWTTE.
Part (c)(i)
a = F/m = 1.61 × 103/(92 × 1.66 × 10−27)✓ 1Allow ECF from (a)(ii).
a = 1.1 × 1028 m s−2✓ 1Award [2] for CNA. Accept 1.0–1.1 × 1028 m s−2.
Part (c)(ii)
E = F/q = 1.61 × 103/(36 × 1.60 × 10−19) = 2.8 × 1020 N C−1✓ 1ALTERNATIVE: E = kQ/r2 with Q = 56e. Accept V m−1. Allow ECF from (a)(ii).
Part (d)(i)
F ∝ 1/r2, so the separation is 10 times larger: 1.7 × 10−13 m✓ 1
Part (d)(ii)
the kinetic energy gained equals the work done by the electric force «force × distance moved / area under the force–separation graph»✓ 1
the force falls rapidly «as 1/r2», so beyond 1.7 × 10−13 m it is less than 1 % of its initial value and does very little further work✓ 1OWTTE. Allow ECF from (d)(i).
Part (e)
at the same separation F ∝ q1q2: 46 × 46 = 2116 compared with 56 × 36 = 2016✓ 1
so the force is about 5 % larger «1.05 times, 1.7 × 103 N»✓ 1Award [2] for a correct calculation of 1.69 × 103 N with a comparison.

Answers: (a)(i) 36  ·  (a)(ii) 1.6 × 103 N  ·  (c)(i) 1.1 × 1028 m s−2  ·  (c)(ii) 2.8 × 1020 N C−1  ·  (d)(i) 1.7 × 10−13 m  ·  (e) about 5 % larger (1.7 × 103 N) (the remaining parts are explanations — see the table above)

132D-2-64Combining two electric fields Paper 2Medium14 marks

A weakly electric fish senses its surroundings using an electric field that it produces itself. In a simple electrostatic model, at one instant the head H of the fish carries a charge of +0.50 nC and the tail T a charge of −0.50 nC. H and T are 0.20 m apart and are treated as point charges (Figure 1).

The fish is in fresh water, which has a permittivity of 80ε0.

k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C.

Top view of a fish 0.20 m long: positive charge at the head H, negative charge at the tail T, and point P on the axis 0.10 m in front of the head+−HTP0.10 m0.20 m
Figure 1 — Diagram NOT to scale
(a)

Sketch, on Figure 1, the electric field pattern around the fish in the plane of the page.

(3)
(b)

Outline why the electric field at a given point near the fish is much weaker than it would be if the same charges were in air.

(1)
(c)

Point P lies on the line through T and H, 0.10 m in front of H.

(i)

Show that the electric field strength at P due to the charge at H alone is about 5.6 V m−1.

(1)
(ii)

Determine the magnitude and direction of the resultant electric field at P.

(3)
(iii)

Calculate the magnitude of the force on a sodium ion, of charge +e, at P.

(1)
(d)

Point S lies on the same line, 0.40 m in front of H. Determine whether the resultant field strength in front of the fish decreases with distance from H according to an inverse-square law.

(3)
(e)

A small metal object lies near the side of the fish. Suggest how the electric field lines close to the object differ from those in (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
field lines leave H and end on T, with arrows pointing away from H and towards T✓ 1
lines curve round from H to T on both sides, pattern symmetrical about the line TH, no lines crossing✓ 1
lines most closely spaced near H and near T, with a straight line along the axis beyond H pointing away from the fish✓ 1Do not award MP3 if equal spacing is shown everywhere.
Part (b)
the permittivity of water is about 80 times that of air «≈ ε0», and k = 1/4πε, so the field «and force» is about 80 times smaller✓ 1
Part (c)(i)
E = kQ/(80r2) = 8.99 × 109 × 0.50 × 10−9/(80 × 0.102) = 5.62 «V m−1»✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
field due to T = 8.99 × 109 × 0.50 × 10−9/(80 × 0.302) = 0.62 V m−1✓ 1
the field due to H points away from H, the field due to T points towards T, so they are opposite: E = 5.62 − 0.62 = 5.0 V m−1✓ 1Award [1] max for 6.2 V m−1 «fields added».
directed away from the fish «along TH, away from H»✓ 1
Part (c)(iii)
F = eE = 1.60 × 10−19 × 5.0 = 8.0 × 10−19 N✓ 1Allow ECF from (c)(ii).
Part (d)
ES = (8.99 × 109 × 0.50 × 10−9/80) × (1/0.402 − 1/0.602) = 0.20 V m−1✓ 1
an inverse-square law would give 5.0 × (0.10/0.40)2 = 0.31 V m−1 OR the measured ratio EP/ES ≈ 26 compared with (0.40/0.10)2 = 16✓ 1Allow ECF from (c)(ii).
the field falls faster than an inverse-square law, so no «the opposite field of T cancels a growing fraction of the field of H»✓ 1MP3 only scores if MP2 scores.
Part (e)
the lines bend so that they meet the surface of the object at right angles✓ 1
there are no field lines inside the metal «charges are induced on its surface, where the lines start and end»✓ 1OWTTE.

Answers: (c)(i) 5.6 V m−1  ·  (c)(ii) 5.0 V m−1, away from the fish  ·  (c)(iii) 8.0 × 10−19 N  ·  (d) ES = 0.20 V m−1; inverse-square value 0.31 V m−1 (the remaining parts are explanations — see the table above)

133D-2-67Charged conducting sphere Paper 2Medium13 marks

In an electrospray ion source for a mass spectrometer, a conducting liquid flows out of the tip of a thin metal capillary that is held at +3.0 kV relative to an earthed plate (Figure 1). Small droplets break away from the tip, each carrying a positive charge. As a droplet drifts towards the plate, liquid evaporates from it, but the droplet does not lose any charge. Treat each droplet as an isolated conducting sphere.

k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C.

A metal capillary at +3.0 kV on the left releasing positively charged droplets that drift towards an earthed plate on the rightmetal capillary, +3.0 kV+++earthed platecharged droplets
Figure 1 — Diagram NOT to scale
(a)

Explain how the droplets become positively charged.

(2)
(b)

A droplet has a radius of 2.0 μm and a charge of +8.0 × 10−15 C.

(i)

Calculate the number of elementary charges on the droplet.

(1)
(ii)

Calculate the electric field strength at the surface of the droplet.

(2)
(iii)

State the electric field strength at the centre of the droplet. Explain your answer.

(2)
(c)
(i)

Explain why the electric field strength at the surface of the droplet increases as the liquid evaporates.

(1)
(ii)

The droplet breaks apart when the electric field strength at its surface reaches 2.0 × 108 V m−1. Determine the radius of the droplet when this happens.

(2)
(d)

When it breaks apart, the droplet throws off 12 small droplets, each carrying 250 elementary charges. Deduce the number of elementary charges left on the main droplet.

(1)
(e)

Two of the small droplets are 1.5 μm apart. Calculate the magnitude of the electric force between them.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the liquid is in contact with the positive metal capillary, so electrons are drawn out of the liquid into the metal «towards the positive supply»✓ 1Accept: positive ions in the liquid are repelled to the tip.
a droplet that breaks away has fewer electrons than protons, so it carries a positive charge✓ 1Do not accept «positive charge flows into the liquid» without the idea of electron loss.
Part (b)(i)
n = 8.0 × 10−15/1.60 × 10−19 = 5.0 × 104✓ 1
Part (b)(ii)
E = kq/r2 = 8.99 × 109 × 8.0 × 10−15/(2.0 × 10−6)2✓ 1
E = 1.8 × 107 V m−1✓ 1Award [2] for CNA.
Part (b)(iii)
zero✓ 1
the charge on a conductor resides on its outer surface; if there were a field inside, free charges would move until it was zero✓ 1MP2 can score without MP1 only if zero is clearly implied.
Part (c)(i)
the charge q is unchanged «conserved» while r decreases, so E = kq/r2 increases✓ 1
Part (c)(ii)
r = √(kq/E) = √(8.99 × 109 × 8.0 × 10−15/2.0 × 108)✓ 1
r = 6.0 × 10−7 m «0.60 μm»✓ 1Award [2] for CNA. ALTERNATIVE: r = 2.0 μm × √(1.8 × 107/2.0 × 108).
Part (d)
5.0 × 104 − 12 × 250 = 4.7 × 104 «charge is conserved»✓ 1Allow ECF from (b)(i).
Part (e)
q = 250 × 1.60 × 10−19 = 4.0 × 10−17 C AND F = 8.99 × 109 × (4.0 × 10−17)2/(1.5 × 10−6)2✓ 1
F = 6.4 × 10−12 N✓ 1Award [2] for CNA.

Answers: (b)(i) 5.0 × 104  ·  (b)(ii) 1.8 × 107 V m−1  ·  (c)(ii) 0.60 μm  ·  (d) 4.7 × 104  ·  (e) 6.4 × 10−12 N (the remaining parts are explanations — see the table above)

134D-2-69Electric field lines Paper 2Medium14 marks

A small model aircraft with no moving parts is driven by ion-wind thrusters. In each thruster unit, a thin wire W is held at a potential difference of +20 kV relative to a broad, aerofoil-shaped collector C that is 50 mm behind it (Figure 1). Close to W the electric field is strong enough to ionize the air. Positive ions drift from W to C, colliding with neutral air molecules and pushing them backwards. The current from W to C is 0.60 mA.

Assume that the ions cross the gap at a constant mean speed of 150 m s−1. e = 1.60 × 10−19 C.

Cross-section of one ion-wind thruster unit: a thin wire W on the left and a broad aerofoil-shaped collector C 50 mm behind it; the aircraft flies to the leftWCthin wire, +20 kVcollector (earthed)direction of flight50 mm
Figure 1 — Diagram NOT to scale
(a)

Draw, on Figure 1, the electric field lines between W and C.

(2)
(b)

Explain why the air is ionized only close to W.

(2)
(c)

Calculate the mean electric field strength between W and C.

(1)
(d)
(i)

Show that the number of ions in the gap between W and C at any instant is about 1.3 × 1012.

(2)
(ii)

Determine the total electric force on the ions in the gap. Use the mean field strength.

(2)
(e)

Explain how the force in (d)(ii) leads to a forward thrust on the aircraft.

(2)
(f)

The aircraft needs a total thrust of 3.2 N. Determine the electrical power that must be supplied to its thruster units, assuming that each unit performs like the one described.

(2)
(g)

Suggest one reason why ion-wind thrusters are not used to propel large passenger aircraft.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
lines leave W radially «at right angles to the wire» and are crowded close together near W✓ 1
lines spread out and end on C at right angles to its surface, with arrows from W to C✓ 1Lines must not cross.
Part (b)
the field lines are most closely spaced close to the thin wire, so the field strength is greatest there✓ 1
only there is the field strong enough to ionize the air «exceed the breakdown field strength»✓ 1MP2 only scores if MP1 scores.
Part (c)
E = V/d = 2.0 × 104/0.050 = 4.0 × 105 V m−1✓ 1
Part (d)(i)
time to cross the gap = 0.050/150 = 3.3 × 10−4 s✓ 1
charge in the gap = It = 0.60 × 10−3 × 3.33 × 10−4 = 2.0 × 10−7 C, so number = 2.0 × 10−7/1.60 × 10−19 = 1.25 × 1012✓ 1Must see full substitution OR answer to 3 s.f.
Part (d)(ii)
force on one ion = eE = 1.60 × 10−19 × 4.0 × 105 = 6.4 × 10−14 N✓ 1Allow ECF from (c).
total force = 1.25 × 1012 × 6.4 × 10−14 = 0.080 N✓ 1Award [2] for CNA. Accept 0.083 N from 1.3 × 1012.
Part (e)
the ions pass on the force to the air: in collisions they push the air molecules backwards «giving the air backward momentum»✓ 1
by Newton's third law the ions exert an equal and opposite «forward» electric force on the charges on the electrodes, which are fixed to the aircraft✓ 1OWTTE. Accept: the air exerts an equal forward force on the aircraft.
Part (f)
number of units = 3.2/0.080 = 40 AND power per unit = IV = 0.60 × 10−3 × 2.0 × 104 = 12 W✓ 1Allow ECF from (d)(ii).
total power = 40 × 12 = 480 W✓ 1Award [2] for CNA. ALTERNATIVE: thrust per watt = 0.080/12 = 6.7 × 10−3 N W−1, so P = 3.2/6.7 × 10−3.
Part (g)
the thrust per watt is very small, so a huge electrical power «and a heavy power supply» would be needed for the thrust of a large aircraft✓ 1OWTTE. Accept the idea that the thrust is small compared with the weight of the equipment.

Answers: (c) 4.0 × 105 V m−1  ·  (d)(i) 1.25 × 1012  ·  (d)(ii) 0.080 N  ·  (f) 480 W (the remaining parts are explanations — see the table above)

135IBP2S-030-01-12Quantization of charge Paper 2Medium5 marks

Fine dust grains on the Moon can become charged and float above the surface. Close to the surface, the electric field is uniform. It has strength 15 V m−1 and is directed vertically upwards. A dust grain has mass 1.0×10−15 kg. The gravitational field strength on the Moon is 1.6 N kg−1. The elementary charge is 1.60×10−19 C.

(a)

Calculate the weight of the dust grain on the Moon.

(1)
(b)

Determine the magnitude and sign of the charge needed for the grain to float at rest.

(3)
(c)

Calculate the number of elementary charges this represents.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = mg = 1.0×10−15 × 1.6 = 1.6×10−15 N .✓ 1
Part (b)
For the grain to float, the electric force balances the weight: qE = mg .✓ 1
q = 1.6×10−15 / 15 = 1.07×10−16 C ≈ 1.1×10−16 C . Allow error carried forward from the weight calculated in the first part.✓ 1
The charge is positive, because the electric force must be upwards, in the same direction as the field .✓ 1
Part (c)
n = 1.07×10−16 / 1.60×10−19 ≈ 670 (accept answers from 667 to 690, e.g. 690 from using 1.1×10−16 C) . Allow error carried forward from the charge found in the previous part.✓ 1

Common error: The sign of the charge depends on the field direction. A negative charge in an upward field would be pushed downwards and could not float.

136IBP2S-030-01-13Coulomb's law Paper 2Medium6 marks

A technician tests an air ioniser used in a clean room for electronics manufacture. Its charged spherical electrode behaves like a point charge at its centre. A field probe measures the electric field strength E at several distances r from the centre.

r / m : 0.20 | 0.30 | 0.40 | 0.60
E / 104 V m−1 : 9.0 | 4.0 | 2.25 | 1.0

Take k = 8.99×109 N m2 C−2.

00.10.20.30.40.50.60.70246810r / mE / 10⁴ V m⁻¹
Data from the question
(a)

Show that the data are consistent with an inverse-square relationship between E and r.

(2)
(b)

Determine the magnitude of the charge on the electrode.

(2)
(c)

Predict the electric field strength at a distance of 0.80 m from the centre.

(1)
(d)

State the quantities that should be plotted to give a straight-line graph through the origin.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Calculates E r2 for at least three data points, e.g. 0.040 × 9.0×104 = 3600, 0.090 × 4.0×104 = 3600 and 0.16 × 2.25×104 = 3600. Equivalent ratio checks for at least three points are also accepted, e.g. doubling r from 0.20 m to 0.40 m divides E by 4, and tripling r from 0.20 m to 0.60 m divides E by 9 .✓ 1
States that E r2 is constant, at 3.6×103 V m, for the data, so E ∝ 1/r2 .✓ 1
Part (b)
Q = E r2/k = 3600 / 8.99×109 .✓ 1
Q = 4.0×10−7 C . Allow error carried forward from the value of E r2 found in the first part.✓ 1
Part (c)
E = 3600 / 0.802 = 5.6×103 V m−1 . Allow error carried forward from the value of E r2. Also accept a scaling argument: doubling r from 0.40 m quarters 2.25×104 V m−1.✓ 1
Part (d)
Plot E against 1/r2, or equivalent .✓ 1

Exam tip: To test an inverse-square law, show that E × r² stays constant for at least three data points. A single pair of points cannot rule out other relationships.

137IBP2S-030-02-13Combining two electric fields Paper 2Medium8 marks

In a laboratory study of charged dust, two small charged spheres are held on insulating rods along a horizontal line. Sphere A carries +6.0 nC and is at x = 0. Sphere B carries −2.0 nC and is at x = 0.30 m. Take k = 8.99 × 109 N m2 C−2.

+−A +6.0 nCB −2.0 nCxx = 0.15 mx = 0x = 0.30 m
Diagram not to scale
(a)

Determine the magnitude and direction of the resultant electric field at x = 0.15 m.

(3)
(b)

Explain why the resultant field can be zero only at a point beyond B, on the side away from A.

(2)
(c)

Determine the position where the resultant field is zero.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Field due to A: 8.99 × 109 × 6.0 × 10−9/0.152 = 2.4 × 103 N C−1. Field due to B: 8.99 × 109 × 2.0 × 10−9/0.152 = 8.0 × 102 N C−1✓ 1
Both fields point in the +x direction (towards B), so they add✓ 1
Resultant = 3.2 × 103 N C−1 in the +x direction✓ 1
Part (b)
Between the spheres, both fields point towards B, so they cannot cancel✓ 1
The two fields are opposite in direction outside the spheres. They can be equal in magnitude only on B's side, because that is nearer the smaller charge. To the left of A, the larger charge is always nearer, so its field always dominates✓ 1
Part (c)
Sets the magnitudes equal: k(6.0 × 10−9)/x2 = k(2.0 × 10−9)/(x − 0.30)2✓ 1
Takes square roots: (x − 0.30) = x/√3✓ 1
x = 0.71 m, which is 0.41 m beyond B✓ 1

Exam tip: Take square roots before rearranging. This turns the equation into a linear one and avoids a quadratic.

138IBP2S-030-02-14Uniform fields between plates Paper 2Medium7 marks

A student investigates a small conducting sphere carrying a fixed charge. The sphere hangs midway between two large vertical parallel plates, well away from their edges. A constant potential difference of 3.0 kV is maintained across the plates. A sensitive force sensor measures the horizontal electric force F on the sphere for different plate separations d.

d / cm: 2.0, 3.0, 4.0, 5.0, 6.0
F / mN: 0.91, 0.59, 0.46, 0.36, 0.30

charged spheredp.d. = 3.0 kV
Diagram not to scale. d is varied.
(a)

Outline why F is expected to be inversely proportional to d.

(2)
(b)

Show that the data are consistent with F being inversely proportional to d.

(2)
(c)

State the graph the student should plot to obtain a straight line through the origin.

(1)
(d)

Determine the charge on the sphere.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The field between the plates is uniform with E = V/d✓ 1
F = qE = qV/d; since q and V are constant, F ∝ 1/d✓ 1
Part (b)
Calculates F × d for at least three rows, e.g. 1.82, 1.77, 1.84, 1.80, 1.80 mN cm✓ 1
States that the product is approximately constant, about 1.8 mN cm, within a few percent✓ 1
Part (c)
A graph of F against 1/d✓ 1
Part (d)
Uses q = Fd/V with mean Fd ≈ 1.8 × 10−5 N m, or uses gradient of F against 1/d = qV✓ 1
q ≈ 6.0 × 10−9 C✓ 1
Allow error carried forward from the values of F × d calculated earlier.—

Common error: Convert to SI units before calculating the charge: mN to N and cm to m. 1 mN cm = 1 × 10⁻⁵ N m.

139IBP2S-030-02-16Uniform fields between plates Paper 2Medium7 marks

A gridded ion thruster on a deep-space probe accelerates xenon ions between two parallel grids. The grids are 1.2 mm apart and the potential difference between them is 1.1 kV. Each ion has charge +1.60 × 10−19 C and mass 2.18 × 10−25 kg. The ions enter the gap between the grids with negligible speed.

(a)

Calculate the electric field strength between the grids.

(1)
(b)

Calculate the electric force on one ion between the grids.

(1)
(c)

Show that the kinetic energy gained by one ion is about 1.8 × 10−16 J.

(1)
(d)

Explain why the kinetic energy gained does not depend on the grid separation, provided the potential difference stays the same.

(2)
(e)

Calculate the speed of an ion as it leaves the grids.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = V/d = 1100/(1.2 × 10−3) = 9.2 × 105 V m−1✓ 1
Part (b)
F = qE = 1.60 × 10−19 × 9.2 × 105 = 1.5 × 10−13 N✓ 1
Allow error carried forward from the field strength calculated in the first part.—
Part (c)
Ek = qV = 1.60 × 10−19 × 1100 = 1.76 × 10−16 J✓ 1
Part (d)
A smaller separation d gives a larger field E = V/d and so a larger force, but the force acts over a proportionally shorter distance✓ 1
Kinetic energy gained = work done = Fd = qEd = qV, so d cancels and the energy depends only on q and V✓ 1
Part (e)
Uses ½mv2 = 1.76 × 10−16 J, so v = √(2 × 1.76 × 10−16/(2.18 × 10−25))✓ 1
v = 4.0 × 104 m s−1✓ 1
Allow error carried forward from the kinetic energy found earlier. A correct answer found using F = ma and v2 = 2ad earns full marks.—

Exam tip: For the exit speed, the energy route qV = ½mv² is quicker than finding the acceleration first. It also needs no information about the gap width, so there are fewer rounding steps.

140IBP2S-030-03-14Conservation of charge Paper 2Medium7 marks

A technician tests an induction-based charge sensor using two identical uncharged metal spheres, A and B. The spheres are mounted on insulating stands and touch each other. A positively charged acrylic rod is held close to A, without touching it. While the rod is still in place, B is moved away from A. The rod is then removed.

(a)

Explain why A becomes negatively charged and B becomes positively charged.

(3)
(b)

State the total charge on A and B together after they are separated.

(1)
(c)

Each sphere ends up with a charge of magnitude 4.0 nC, and their centres are 0.25 m apart. Calculate the magnitude of the electric force between them and state its direction.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The metal contains free electrons that can move through both spheres ; electrons are attracted towards the positive rod and gather on A ; B is left with a deficit of electrons, and separating the spheres while the rod is present traps these charges .✓ 3
Part (b)
Zero, because no charge is transferred from the rod and charge is conserved .✓ 1
Part (c)
F = kq1q2/r2 = 8.99×109 × (4.0×10−9)2 / 0.252 ; F = 2.3×10−6 N ; attractive, because the charges are opposite .✓ 3

Common error: Saying that positive charges move towards B is wrong. In metals only electrons move. B becomes positive because it loses electrons.

141IBP2S-030-00-13Magnetic field of a straight wire Paper 2Medium7 marks

A student uses a smartphone magnetometer to measure the magnetic field strength B at different perpendicular distances r from a long straight wire. The wire carries a constant current. The background field has already been subtracted from the readings.

r / cm: 2.0, 4.0, 5.0, 8.0, 10.0
B / μT: 151, 74, 61, 37, 30

(μ0 = 4π × 10−7 T m A−1)

0246810020406080100120140160r / cmB / μT
Data from the question
(a)

Show that the data are consistent with B being inversely proportional to r.

(2)
(b)

Determine the current in the wire.

(2)
(c)

Suggest which graph the student should plot to obtain a straight line through the origin, and state what its gradient represents.

(2)
(d)

Suggest why the readings at large r are the least reliable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Calculates Br for at least three data pairs, e.g. 3.02 × 10−6, 2.96 × 10−6, 3.05 × 10−6 T m. Alternatively, shows for at least two pairs that doubling r roughly halves B, e.g. 2.0 → 4.0 cm gives 151 → 74 μT, and 5.0 → 10.0 cm gives 61 → 30 μT✓ 1
States that Br is constant within about 2–3%, or that the halving holds within experimental uncertainty, so B ∝ 1/r✓ 1
Part (b)
Uses Br ≈ 3.0 × 10−6 T m, the mean or any single data pair, in I = 2πBr/μ0✓ 1
I = 15 A. Answers in the range 14.8–15.3 A are accepted✓ 1
Part (c)
Plot B against 1/r✓ 1
The gradient is μ0I/(2π)✓ 1
Part (d)
At large r the wire's field is small. The sensor's resolution and any error in subtracting the background field (for example, the Earth's field) are then a larger fraction of the reading, so the percentage uncertainty is larger✓ 1

Exam tip: To test an inverse proportion, calculate the product of the two variables for several data pairs. A ratio test is only appropriate for direct proportion.

142IBP2S-030-00-14Quantization of charge Paper 2Medium6 marks

In a classroom demonstration, a small charged sphere is fixed on an insulating stand. A second small sphere of mass 0.050 g floats at rest directly above it. The two centres are 3.0 cm apart. Both spheres carry the same charge q and can be treated as point charges. (k = 8.99 × 109 N m2 C−2, g = 9.81 m s−2)

insulating standfixed spheresphere, mass 0.050 g3.0 cm
Diagram not to scale. Distance is between the centres.
(a)

State the condition for the upper sphere to remain at rest.

(1)
(b)

Determine q.

(3)
(c)

The upper sphere is pushed slightly downwards and then released. Explain why it moves back towards its original position.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The upward electric repulsion equals the weight of the sphere, so the resultant force is zero✓ 1
Part (b)
Calculates the weight: mg = 5.0 × 10−5 × 9.81 = 4.9 × 10−4 N✓ 1
Sets kq2/r2 = mg and rearranges: q = √(mgr2/k), with r = 0.030 m✓ 1
q = 7.0 × 10−9 C✓ 1
Part (c)
The separation decreases, so the repulsion increases, because F ∝ 1/r2✓ 1
The repulsion now exceeds the weight, so there is a resultant upward force directed towards the original position. The equilibrium is therefore stable for vertical displacements, and the sphere oscillates about that position unless the motion is damped✓ 1

Common error: 0.050 g is 5.0 × 10⁻⁵ kg, not 5.0 × 10⁻² kg, and 3.0 cm is 0.030 m. Convert both to SI units before substituting.

143IBP2S-030-05-12Gravity compared with electrostatics Paper 2Medium7 marks

An engineering laboratory uses a torsion balance to measure the gravitational constant. Two lead spheres, each of mass 1.5 kg, have their centres 0.10 m apart. Any stray charge on the spheres would add an electric force to the gravitational force being measured.

(a)

Calculate the gravitational force between the spheres.

(2)
(b)

Each sphere carries the same charge q. Determine the value of q for which the electric force between the spheres equals the gravitational force.

(2)
(c)

A sphere carries a charge of −q, where q is the value you found in the previous part. Calculate the number of excess electrons on this sphere.

(1)
(d)

Explain, with reference to your answer for the number of excess electrons, why the spheres must be discharged (earthed) before measurements are taken.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
F = Gm2/r2 = 6.67×10−11 × 1.52 / 0.102✓ 1
F = 1.5×10−8 N✓ 1
Part (b)
kq2/r2 = 1.5×10−8 N, or q = m√(G/k)✓ 1
q = 1.3×10−10 C (allow error carried forward from the gravitational force)✓ 1
Part (c)
n = 1.29×10−10 / 1.60×10−19 = 8.1×108 (allow error carried forward from the value of q)✓ 1
Part (d)
Any valid argument that about 8×108 electrons, or about 10−10 C, is a very small amount of charge that is easily gained by handling, friction or contact. For example, it is a tiny fraction of the roughly 1026 electrons in each sphere, or it is far smaller than typical static charges, which are of order nC or more✓ 1
Such a stray charge would produce an electric force comparable to, or larger than, the gravitational force, so the measured value of G would be invalid✓ 1

Exam tip: The separation cancels when you equate the two forces, giving q = m√(G/k). Spotting this saves time and avoids rounding errors.

144IBP2S-030-05-13Quantization of charge Paper 2Medium7 marks

A pharmaceutical company tests inhaler drug particles in an aerosol charge analyser. The analyser measures the charge on individual particles. Five readings are shown.

Particle — Charge / ×10−19 C
A — 6.41
B — 9.65
C — 4.79
D — 11.18
E — 8.03

(a)

Outline how these data suggest that electric charge is quantized.

(2)
(b)

Determine, using all the data, a value for the elementary charge.

(3)
(c)

Another student records a charge of 5.6×10−19 C for a particle. Comment on this reading, using your value of the elementary charge.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Dividing each reading by a common value of about 1.6×10−19 C gives numbers very close to integers (about 4, 6, 3, 7 and 5), so each charge is a whole-number multiple of this value✓ 1
The charge therefore occurs only in discrete amounts (a basic unit of charge), with no fractional multiples observed. Equivalently, the differences between readings are also whole-number multiples of about 1.6×10−19 C✓ 1
Part (b)
Numbers of elementary charges: A 4, B 6, C 3, D 7, E 5✓ 1
e = total charge / total number = 40.06×10−19 C / 25. Averaging the five individual values of e is also accepted✓ 1
e = 1.60×10−19 C✓ 1
Part (c)
5.6×10−19 / 1.60×10−19 = 3.5, which is not a whole number. Allow error carried forward from the value of e found above✓ 1
The reading is therefore probably an experimental or recording error and should be checked or repeated; a single anomalous value is not evidence against quantization✓ 1

Exam tip: Dividing the total charge by the total number of elementary charges averages out random error. This takes all the data into account in one step.

145IBP2S-030-05-14Electric field strength Paper 2Medium6 marks

In fair weather, there is an electric field of strength 130 N C−1 near the surface of the Earth, directed vertically downwards. Model the Earth as a sphere of radius 6.37×106 m with its net charge on its surface.

(a)

Deduce the sign of the Earth's net charge.

(1)
(b)

Calculate the magnitude of the Earth's net charge.

(2)
(c)

Calculate the number of excess electrons on the Earth.

(1)
(d)

Determine the ratio of the electric force on an electron near the surface to the electron's weight.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The field points towards the surface (downwards) and field lines end on negative charge, so the net charge is negative✓ 1
Part (b)
E = kQ/R2, so Q = ER2/k = 130 × (6.37×106)2 / 8.99×109✓ 1
Q = 5.9×105 C✓ 1
Part (c)
n = Q/e = 5.87×105 / 1.60×10−19 = 3.7×1024 (allow error carried forward from the previous part)✓ 1
Part (d)
FE = eE = 1.60×10−19 × 130 = 2.1×10−17 N✓ 1
Ratio = 2.08×10−17 / (9.11×10−31 × 9.8) = 2.3×1012✓ 1

Common error: Field direction is the direction of the force on a positive charge. A downward field therefore means the surface charge is negative, and the field pushes electrons upwards.

146IBP2S-030-04-14Uniform fields between plates Paper 2Medium6 marks

A power-station chimney contains an electrostatic precipitator to remove smoke particles. Two large parallel metal plates are 0.15 m apart, with a potential difference of 30 kV between them. A smoke particle of mass 2.0×10−12 kg carries a charge of −8.0×10−16 C.

0.15 mparticle (−)gas flowp.d. = 30 kV
Diagram not to scale
(a)

Describe the electric field lines between the plates, both in the central region and near the edges.

(2)
(b)

Calculate the electric field strength in the central region.

(1)
(c)

Determine the ratio of the magnitude of the electric force on the particle to its weight.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
In the central region the lines are straight, parallel, equally spaced and perpendicular to the plates, running from the positive plate to the negative plate .✓ 1
Near the edges the lines bulge outwards and curve, so the field there is non-uniform .✓ 1
Part (b)
E = V/d = 30×103 / 0.15 = 2.0×105 N C−1 (V m−1)✓ 1
Part (c)
F = qE = 8.0×10−16 × 2.0×105 = 1.6×10−10 N . Allow error carried forward from the field strength calculated in the previous part.✓ 1
W = mg = 2.0×10−12 × 9.81 = 1.96×10−11 N .✓ 1
Ratio = 1.6×10−10 / 1.96×10−11 ≈ 8.2 . Accept 8.0 if g = 10 N kg−1 is used.✓ 1

Common error: Convert kV to V before using E = V/d; leaving it in kV makes the field 1000 times too small.

147D-2-07Quantization of charge Paper 2Hard13 marks

In a Millikan-type experiment, an oil drop of mass 2.40 × 10−15 kg is held stationary between two horizontal parallel plates 8.00 mm apart. The upper plate is positive and the potential difference between the plates is 235 V. Ignore the buoyancy force of the air.

Two horizontal parallel plates 8.00 mm apart, the upper plate positive and the lower plate negative, with an oil drop shown as a dot midway between them.upper plate (+)lower plate (−)oil drop8.00 mm
Diagram NOT to scale
(a)
(i)

Draw a labelled free-body diagram for the drop on the diagram.

(2)
(ii)

Show that the magnitude of the charge on the drop is about 8.0 × 10−19 C.

(2)
(iii)

Deduce the sign of the charge on the drop and the number of excess electrons on it.

(2)
(b)

While the drop is held stationary it captures one more electron from the air.

(i)

Determine the magnitude and direction of the initial acceleration of the drop.

(3)
(ii)

Calculate the potential difference that would hold the drop stationary again.

(1)
(iii)

Suggest why the acceleration of the drop does not remain at the value found in (b)(i).

(1)
(c)

Outline how measurements of the charge on many oil drops provide evidence that electric charge is quantized.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
downward arrow labelled weight «gravitational force» AND upward arrow labelled electric force✓ 1Do not award if a drag or buoyancy force is added.
arrows of equal length, both starting on the drop✓ 1
Part (a)(ii)
E = V/d = 235/8.00 × 10−3 = 2.94 × 104 V m−1✓ 1
q = mg/E = 2.40 × 10−15 × 9.81/2.94 × 104 = 8.01 × 10−19 C✓ 1Must see full substitution OR answer to at least 3 s.f.
Part (a)(iii)
negative, because the electric force is upwards, opposite to the downward field «from the positive upper plate»✓ 1A reason is needed.
8.0 × 10−19/1.60 × 10−19 = 5 «excess electrons»✓ 1
Part (b)(i)
the extra electron adds an upward force eE while the weight is unchanged, so the resultant force = eE = 1.60 × 10−19 × 2.94 × 104 = 4.7 × 10−15 N✓ 1ALTERNATIVE: the electric force becomes 6/5 of the weight, so the resultant force is mg/5.
a = 4.7 × 10−15/2.40 × 10−15 = 2.0 m s−2✓ 1Accept 1.9–2.0 m s−2.
upwards✓ 1The direction mark is independent of the magnitude.
Part (b)(ii)
V = 235 × 5/6 = 196 V✓ 1Accept 195–197 V. Allow ECF from (a)(iii).
Part (b)(iii)
as the drop speeds up, the air resistance «drag» on it increases, so the resultant force and the acceleration decrease «until it reaches terminal speed»✓ 1OWTTE
Part (c)
the charges found are always whole-number multiples of one smallest charge✓ 1
the smallest charge is 1.6 × 10−19 C «e», and no fraction of it is ever found✓ 1OWTTE

Answers: (a)(ii) 8.01 × 10−19 C  ·  (a)(iii) negative; 5 electrons  ·  (b)(i) 2.0 m s−2 upwards  ·  (b)(ii) 196 V (the remaining parts are explanations — see the table above)

148D-2-23Combining two electric fields Paper 2Hard13 marks

Two small charged spheres A and B are fixed 0.12 m apart in a vacuum. A carries a charge of +8.0 nC and B carries a charge of −2.0 nC. M is the midpoint of AB.

k = 8.99 × 109 N m² C−2.

Charge A of +8.0 nC and charge B of −2.0 nC 0.12 m apart, with midpoint M+A+8.0 nC−B−2.0 nCM0.12 m
Diagram NOT accurately drawn
(a)

Sketch, on the diagram, the electric field pattern around A and B.

(3)
(b)

Explain why the resultant electric field cannot be zero at any point between A and B.

(2)
(c)

Determine the position of the point on the line through A and B at which the resultant electric field is zero.

(3)
(d)

A dust particle carrying a charge of −3.0 nC is placed at M.

Determine the magnitude and direction of the electric force on the particle.

(3)
(e)

Calculate the magnitude of the electric force that A exerts on B, and state the magnitude and direction of the force that B exerts on A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
lines radial close to each charge, with arrows pointing away from A and towards B✓ 1
lines from A curve round to end on B, and no lines cross✓ 1
more lines at A than at B «about four times as many», with some lines from A not ending on B «going off to the edge of the diagram»✓ 1Do not award MP3 if equal numbers of lines are drawn at A and B.
Part (b)
between A and B the field due to A points away from A, towards B✓ 1
the field due to B also points towards B «B is negative», so the two fields are in the same direction and add✓ 1OWTTE
Part (c)
the point lies beyond B, on the side away from A «outside the pair, nearer the smaller charge»✓ 1
k × 8.0/(0.12 + x)² = k × 2.0/x² OR (0.12 + x)/x = 2✓ 1
x = 0.12 m beyond B «0.24 m from A»✓ 1Award [3] for CNA provided the side of B is clear.
Part (d)
field due to A = 2.0 × 104 N C−1 AND field due to B = 5.0 × 103 N C−1, both towards B✓ 1ALTERNATIVE: Coulomb forces on the particle 6.0 × 10−5 N and 1.5 × 10−5 N, both towards A.
resultant field = 2.50 × 104 N C−1 towards B✓ 1
F = 3.0 × 10−9 × 2.50 × 104 = 7.5 × 10−5 N, directed towards A✓ 1Direction needed for MP3: the force on a negative charge is opposite to the field. Allow ECF.
Part (e)
F = 8.99 × 109 × 8.0 × 10−9 × 2.0 × 10−9 / 0.12² = 1.0 × 10−5 N✓ 1
1.0 × 10−5 N on A, directed towards B «Newton's third law; the force is attractive»✓ 1Both magnitude and direction needed. Allow ECF.

Answers: (c) 0.12 m beyond B (0.24 m from A)  ·  (d) 7.5 × 10−5 N towards A  ·  (e) 1.0 × 10−5 N (the remaining parts are explanations — see the table above)

149D-2-60Uniform fields between plates Paper 2Hard14 marks

The membrane of a resting nerve cell is a layer of lipid 7.0 nm thick. Its outer surface carries a positive charge and its inner surface an equal negative charge, so that the potential difference across the membrane is 70 mV. Model the membrane as a pair of large parallel plates. Sodium ions, Na⁺, each of charge +e, can cross the membrane only through narrow channels (Figure 1).

The permittivity of the water in and around the cell is 80ε0; the permittivity of the lipid is about 2ε0.

k = 8.99 × 109 N m2 C−2; e = 1.60 × 10−19 C; kB = 1.38 × 10−23 J K−1.

Section through a cell membrane: a lipid layer 7.0 nm thick with positive charges on its outer surface and negative charges on its inner surface, and a channel through which a sodium ion can passlipidoutside the cellinside the cellchannel+−+−+−+−+−+−+−+−+−+−+−+−+−Na⁺7.0 nm
Figure 1 — Diagram NOT to scale
(a)

Draw, on Figure 1, the electric field lines in the lipid layer, away from the channel.

(2)
(b)

Calculate the electric field strength in the membrane.

(1)
(c)

A Na⁺ ion passes through a channel.

(i)

Calculate the magnitude of the electric force on the ion in the channel and state its direction.

(2)
(ii)

Calculate, in eV, the work done by the electric field on the ion as it crosses the membrane.

(2)
(d)

The temperature of the body is 37 °C. Determine whether the work done in (c)(ii) is large compared with the average random kinetic energy of an ion in the fluid around the cell.

(3)
(e)

In the water next to the membrane, a Na⁺ ion is 0.50 nm from a chloride ion, Cl⁻, of charge −e. Treat both ions as point charges.

(i)

Calculate the magnitude of the electric force between the two ions.

(2)
(ii)

State how the force between the ions would change if they were the same distance apart inside the lipid.

(1)
(iii)

The force in (e)(i) is larger than the force in (c)(i). Suggest why it is the field of the membrane, and not the forces from nearby ions, that determines the direction in which Na⁺ ions drift through the channels.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
straight, parallel, equally spaced lines perpendicular to the surfaces of the membrane✓ 1At least three lines, between the surfaces.
arrows pointing from the outer «positive» surface to the inner «negative» surface✓ 1
Part (b)
E = V/d = 0.070/7.0 × 10−9 = 1.0 × 107 V m−1✓ 1Accept N C−1.
Part (c)(i)
F = eE = 1.60 × 10−19 × 1.0 × 107 = 1.6 × 10−12 N✓ 1Allow ECF from (b).
into the cell «in the direction of the field, from the outer surface to the inner surface»✓ 1
Part (c)(ii)
W = Fd = 1.6 × 10−12 × 7.0 × 10−9 = 1.1 × 10−20 J✓ 1Allow ECF from (c)(i).
W = 1.12 × 10−20/1.60 × 10−19 = 0.070 eV✓ 1Award [2] for CNA. ALTERNATIVE: charge e moved through 70 mV gains 0.070 eV.
Part (d)
Ek = (3/2)kBT = 1.5 × 1.38 × 10−23 × 310 = 6.4 × 10−21 J✓ 1Temperature must be in kelvin. Award [0] for MP1 if 37 K is used.
= 0.040 eV OR work/average kinetic energy = 1.1 × 10−20/6.4 × 10−21 ≈ 1.7✓ 1Accept 1.7–1.8.
so the work is of the same order of magnitude as the random kinetic energy, not much larger: thermal motion is also important✓ 1Conclusion must follow from a comparison. Allow ECF.
Part (e)(i)
F = ke2/(80r2) = 8.99 × 109 × (1.60 × 10−19)2/(80 × (0.50 × 10−9)2)✓ 1k = 1/4πε; the factor 80 must be used.
F = 1.2 × 10−11 N✓ 1Award [1] max for 9.2 × 10−10 N «permittivity of water ignored».
Part (e)(ii)
about 40 times larger «permittivity 40 times smaller» OR 4.6 × 10−10 N✓ 1Allow ECF from (e)(i).
Part (e)(iii)
the nearby ions are at random positions on all sides, so their forces on a Na⁺ ion average to about zero over time, whereas the membrane field always acts in the same direction «into the cell»✓ 1OWTTE.

Answers: (b) 1.0 × 107 V m−1  ·  (c)(i) 1.6 × 10−12 N, into the cell  ·  (c)(ii) 0.070 eV  ·  (d) Ek = 0.040 eV; ratio ≈ 1.7  ·  (e)(i) 1.2 × 10−11 N (the remaining parts are explanations — see the table above)

150D-2-63Quantization of charge Paper 2Hard13 marks

In a modern version of Millikan's experiment, a single silica microsphere is held in a vacuum chamber at the focus of a laser beam (an optical trap), as shown in Figure 1. For small horizontal displacements x from the centre of the trap, the trap exerts a horizontal force −ktx on the sphere, where kt = 1.0 × 10−6 N m−1.

The sphere is midway between two vertical parallel plates that are 4.0 mm apart. The potential difference between the plates is 200 V and the left-hand plate is positive. Occasional pulses of ultraviolet light each remove one electron from the sphere. Figure 2 shows how the steady displacement x of the sphere changes with time t. Positive x is to the right.

e = 1.60 × 10−19 C.

Two vertical parallel plates 4.0 mm apart, the left plate positive; a microsphere held between them at the focus of a vertical laser beam; positive x is to the right+−laser beammicrosphere+x4.0 mm
Figure 1 — Diagram NOT to scale
Displacement x of the microsphere against time t: five equal upward steps of 8 nm, from −24 nm to +16 nm0102030405060t / s−32−24−16−8081624x / nm
Figure 2
(a)

Calculate the electric field strength between the plates.

(1)
(b)

At t = 0 the sphere is at rest with x = −24 nm.

(i)

Deduce the sign of the charge on the sphere at t = 0.

(1)
(ii)

Determine the magnitude of the charge on the sphere at t = 0.

(2)
(c)

Determine, using Figure 2, the magnitude of the change in the charge on the sphere at each step.

(2)
(d)

Explain how Figure 2 provides evidence that electric charge is quantized.

(2)
(e)

State the charge on the sphere at t = 60 s, in terms of e.

(1)
(f)

The mass of the sphere is 1.1 × 10−14 kg. Determine whether the weight of the sphere must be taken into account when the charge is calculated from x.

(2)
(g)

Even when the charge is constant, x fluctuates by about ±1 nm. Suggest one cause of these fluctuations.

(1)
(h)

Suggest one advantage of this method compared with measuring the charges on many different oil drops.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = V/d = 200/4.0 × 10−3 = 5.0 × 104 V m−1✓ 1Accept N C−1.
Part (b)(i)
negative, because the sphere is displaced to the left, opposite to the field «which points from the positive plate to the right»✓ 1A reason is needed.
Part (b)(ii)
the trap force balances the electric force: qE = kt|x| = 1.0 × 10−6 × 24 × 10−9 = 2.4 × 10−14 N✓ 1
q = 2.4 × 10−14/5.0 × 104 = 4.8 × 10−19 C✓ 1Allow ECF from (a). Award [2] for CNA.
Part (c)
each step is Δx = 8.0 nm, so ΔF = ktΔx = 8.0 × 10−15 N✓ 1Accept 7.8–8.2 nm.
Δq = 8.0 × 10−15/5.0 × 104 = 1.6 × 10−19 C✓ 1Allow ECF from (a).
Part (d)
x changes only in sudden, equal steps «never by a fraction of a step», so the charge changes only in equal amounts✓ 1
each step equals the elementary charge e «1.6 × 10−19 C», so the charge is always a whole-number multiple of e «−3e to +2e»✓ 1OWTTE. Allow ECF from (c).
Part (e)
+2e «five electrons removed from −3e»✓ 1Sign required.
Part (f)
weight = 1.1 × 10−14 × 9.81 = 1.1 × 10−13 N, larger than the electric force✓ 1
but it acts vertically, at right angles to x, so it has no effect on the horizontal displacement: it need not be taken into account✓ 1MP2 can be awarded for the direction argument alone.
Part (g)
random collisions of the remaining gas molecules with the sphere «Brownian motion» OR vibration of the apparatus OR noise in the position measurement✓ 1Any one.
Part (h)
the mass of the sphere does not need to be measured OR every step is measured on the same particle, so the step size gives e directly✓ 1OWTTE.

Answers: (a) 5.0 × 104 V m−1  ·  (b)(ii) 4.8 × 10−19 C  ·  (c) 1.6 × 10−19 C  ·  (f) weight 1.1 × 10−13 N (the remaining parts are explanations — see the table above)

151D-2-66Magnetic field of a straight wire Paper 2Hard13 marks

A steel pipeline is buried in level ground and runs in a straight line from north to south. To protect it from corrosion, a steady direct current I flows along the pipe. A surveyor walks along a line at right angles to the pipe, from west to east, carrying a magnetometer at ground level. The current is switched on and off repeatedly and the magnetometer records the change in its reading, which is the magnetic flux density due to the current alone.

Figure 1 is a vertical cross-section viewed from the south. The centre of the pipe is a depth h below the ground. Treat the pipe as a long straight wire. The magnetic flux density at a distance r from a long straight wire carrying a current I is B = μ0I/2πr. μ0 = 4π × 10−7 T m A−1.

Figure 2 shows how the magnitude of the horizontal component of the flux density due to the current varies with position x along the survey line.

Vertical cross-section viewed from the south: the ground surface, a magnetometer on the ground directly above a buried pipe at depth h, west on the left and east on the rightgroundmagnetometerpipeWESTEASThsurvey line, x
Figure 1 — Diagram NOT to scale
Horizontal component of the magnetic flux density due to the pipe current against position x along the survey line; maximum 600 nT at x = 0.5 m, 300 nT at x = −1.0 m and 2.0 m−4−3−2−1012345x / m (east positive)−400−300−200−1000100200300400500600700flux density / nT
Figure 2
(a)
(i)

Directly above the pipe, the magnetic field due to the current is horizontal and points west. Deduce the direction of the current in the pipe.

(1)
(ii)

Draw, on Figure 1, three magnetic field lines due to the current. Show the direction of the field.

(2)
(b)
(i)

Show that, at a horizontal distance s from the point directly above the pipe, the horizontal component of the field is Bh = μ0Ih/2π(s2 + h2).

(2)
(ii)

Hence show that Bh is half of its maximum value when s = h.

(1)
(c)

Use Figure 2 to determine the position x of the pipe, its depth h and the current I.

(4)
(d)

Sketch, on Figure 2, the variation with x of the vertical component of the flux density due to the current. Take upwards as positive.

(2)
(e)

The Earth's magnetic flux density is about 5 × 10−5 T. Suggest why the current is switched on and off during the survey.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
towards the south «out of the page in Figure 1, towards the viewer»✓ 1Right-hand grip rule.
Part (a)(ii)
three concentric circles centred on the pipe, spacing increasing with distance✓ 1Circles may be cut off by the ground surface.
arrows anticlockwise «as seen in Figure 1»✓ 1Allow ECF from (a)(i).
Part (b)(i)
r = √(s2 + h2) and the field is at right angles to r, so the horizontal component is B cos θ with cos θ = h/r✓ 1Accept a labelled diagram showing the similar triangles.
Bh = (μ0I/2πr) × (h/r) = μ0Ih/2π(s2 + h2)✓ 1Must see the component step.
Part (b)(ii)
at s = h: Bh = μ0Ih/(2π × 2h2) = ½ × μ0I/2πh, and μ0I/2πh is the value at s = 0 «the maximum»✓ 1Must see the comparison with the value at s = 0.
Part (c)
pipe at x = 0.5 m «where Bh is greatest»✓ 1Accept 0.4–0.6 m.
Bh = 300 nT «half of 600 nT» at x = −1.0 m and x = 2.0 m✓ 1Accept values within ±0.1 m.
h = (2.0 − (−1.0))/2 = 1.5 m✓ 1Allow ECF from MP2.
I = 2πhBmax/μ0 = 2π × 1.5 × 600 × 10−9/(4π × 10−7) = 4.5 A✓ 1Allow ECF from MP3. Accept 4.3–4.7 A.
Part (d)
curve passing through zero at x = 0.5 m, positive to the east «x > 0.5 m» and negative to the west✓ 1Allow ECF from (a).
extreme values of about ±300 nT at about x = 2.0 m and x = −1.0 m, tending to zero at large distances✓ 1Accept peaks within ±0.3 m and ±50 nT.
Part (e)
the Earth's field is about 100 times larger «and steel objects add their own constant fields»; switching the current lets these constant fields be removed by subtraction✓ 1OWTTE.

Answers: (c) x = 0.5 m; h = 1.5 m; I = 4.5 A (the remaining parts are explanations — see the table above)

152IBP2S-030-01-15Combining two electric fields Paper 2Hard8 marks

A teaching model of a dipole antenna uses two small spheres. Sphere A carries +3.0 nC and sphere B carries −3.0 nC. Their centres are 0.080 m apart. Point P lies on the perpendicular bisector of AB, 0.030 m from the midpoint of AB. Treat both spheres as point charges. Take k = 8.99×109 N m2 C−2 and e = 1.60×10−19 C.

+−A +3.0 nCB −3.0 nCP0.080 m0.030 m
Diagram not to scale
(a)

Show that the distance from A to P is 0.050 m.

(1)
(b)

Calculate the magnitude of the electric field strength at P due to sphere A alone.

(2)
(c)

Determine the magnitude and direction of the resultant electric field strength at P.

(3)
(d)

An electron is placed at P. Calculate the magnitude of the force on it and state its direction.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
AP = √(0.0402 + 0.0302) = √0.0025 = 0.050 m .✓ 1
Part (b)
E = kQ/r2 = 8.99×109 × 3.0×10−9 / 0.0502 .✓ 1
E = 1.1×104 V m−1 (1.08×104) .✓ 1
Part (c)
The field from A points away from A and the field from B points towards B. They have equal magnitude, and their components perpendicular to AB cancel .✓ 1
The components parallel to AB add: E = 2 × 1.08×104 × (0.040/0.050) . ECF from (b).✓ 1
E = 1.7×104 V m−1, directed parallel to AB, from A towards B .✓ 1
Part (d)
F = eE = 1.60×10−19 × 1.73×104 = 2.8×10−15 N . ECF from (c).✓ 1
The force is parallel to AB, pointing from B towards A, opposite to the field .✓ 1

Common error: Do not simply add the magnitudes of the two fields. Resolve each field into components, and check which components cancel and which add.

153IBP2S-030-02-15Combining two electric fields Paper 2Hard7 marks

In an atmospheric-physics model, two large horizontal parallel plates are 0.20 m apart. The upper plate is at +3.0 kV and the lower plate is earthed at 0 V. A small sphere with charge −2.0 nC is fixed at the point midway between the plates. Assume the charges on the plates are not affected by the sphere. Take k = 8.99 × 109 N m2 C−2.

+3.0 kV0 V−sphere, −2.0 nCP5.0 cm0.20 m
Diagram not to scale. Sphere is midway between the plates.
(a)

Calculate the uniform field strength E0 produced by the plates.

(1)
(b)

Determine the magnitude of the resultant field at a point 5.0 cm directly above the sphere.

(2)
(c)

Explain why the resultant field can be zero only at a point directly below the sphere.

(2)
(d)

Determine the distance of this zero-field point from the sphere.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E0 = 3000/0.20 = 1.5 × 104 V m−1 (directed downwards)✓ 1
Part (b)
Field due to the sphere: kQ/r2 = 8.99 × 109 × 2.0 × 10−9/0.0502 = 7.2 × 103 N C−1, directed downwards towards the sphere✓ 1
Both fields point downwards, so they add: resultant = 2.2 × 104 N C−1✓ 1
Allow error carried forward from (a).—
Part (c)
For a zero resultant, the sphere's field must be equal in magnitude and opposite in direction to E0, so it must point vertically upwards✓ 1
The field of the negative sphere points radially towards the sphere. It points vertically upwards only on the vertical line below the sphere. Above the sphere it points downwards and adds to E0. At points to the side it has a horizontal component that the vertical E0 cannot cancel✓ 1
Part (d)
Sets kQ/r2 = E0, so r = √(8.99 × 109 × 2.0 × 10−9/1.5 × 104)✓ 1
r = 3.5 × 10−2 m, i.e. 3.5 cm below the sphere. This is less than the 10 cm to the lower plate, so the point lies between the plates✓ 1
Allow error carried forward from (a).—

Exam tip: Sketch both field vectors at the point before combining them. The direction of the point-charge field changes with position, but the plate field does not.

154IBP2S-030-03-16Conservation of charge Paper 2Hard9 marks

A packaging engineer investigates static charge using two identical small metal-coated spheres, each of mass 2.0 g. Each sphere hangs from a common point on an insulating thread 0.50 m long. One sphere is charged by contact with a rubbed plastic rod and is then touched against the other sphere. The spheres repel and come to rest with each thread at 10° to the vertical. Treat the spheres as point charges and take g = 9.8 m s−2.

10°0.50 m
Diagram not to scale. Each thread makes 10° with the vertical.
(a)

Explain why the charge is shared equally between the spheres.

(2)
(b)

Show that the electric force on each sphere is about 3.5×10−3 N.

(3)
(c)

Determine the charge on each sphere.

(3)
(d)

State the charge given to the first sphere by the rod.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The spheres are identical conductors, so charge flows between them while they are in contact ; flow stops when both are at the same potential, which for identical spheres means equal charges .✓ 2
Part (b)
For equilibrium: T cos θ = mg and T sin θ = F ; F = mg tan θ ; F = 2.0×10−3 × 9.8 × tan 10° = 3.46×10−3 N .✓ 3
Part (c)
Separation d = 2 × 0.50 × sin 10° = 0.174 m ; q2 = Fd2/k, so q = d√(F/k) ; q = 1.08×10−7 ≈ 1.1×10−7 C . Allow error carried forward from the force found in the previous part.✓ 3
Part (d)
Total charge = 2q = 2.15×10−7 ≈ 2.2×10−7 C; accept 2.1–2.2×10−7 C, or error carried forward as twice the charge found in the previous part .✓ 1

Common error: Using the thread length as the separation is wrong. The separation is 2L sin θ, which is the distance between the sphere centres.

155IBP2S-030-00-15Magnetic field of a straight wire Paper 2Hard8 marks

A fast-charging cable for an electric vehicle contains two long parallel conductors. Their centres are 1.2 cm apart. Each carries a current of 250 A, and the currents are in opposite directions. (μ0 = 4π × 10−7 T m A−1)

(a)

Explain why the magnetic field at points far from the cable is much smaller than that of a single conductor carrying 250 A.

(2)
(b)

Calculate the magnetic field strength at the point midway between the conductors.

(2)
(c)

A point P lies in the plane of the conductors, on the line through both of them, 10.0 cm from the midpoint of the cable. Determine the magnetic field strength at P.

(3)
(d)

Suggest one design change to the cable that would further reduce the magnetic field at P.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Opposite currents produce fields in opposite (or, away from the plane of the conductors, nearly opposite) directions at points outside the pair✓ 1
At large distances the two conductors are almost equally far away, so the fields have nearly equal magnitudes and nearly cancel✓ 1
Part (b)
Each conductor gives B = 2 × 10−7 × 250/0.0060 = 8.3 × 10−3 T, and the two fields are in the same direction✓ 1
B = 1.7 × 10−2 T✓ 1
Part (c)
The distances from P to the conductors are 9.4 cm and 10.6 cm✓ 1
B = 2 × 10−7 × 250 × (1/0.094 − 1/0.106), subtracting because the fields are opposite✓ 1
B = 6.0 × 10−5 T✓ 1
Part (d)
Reduce the separation of the conductors, or twist them together, or use a coaxial arrangement, so that the opposing fields cancel more completely✓ 1

Exam tip: Before adding field magnitudes, decide whether the fields point the same way or opposite ways. Use the right-hand grip rule for each wire separately.

156IBP2S-030-05-15Gravity compared with electrostatics Paper 2Hard9 marks

On the Moon, ultraviolet light from the Sun releases electrons from dust grains. A grain that is charged in this way can levitate in the electric field above the lunar surface. Consider a spherical grain of radius 0.50 μm and density 3000 kg m−3. The gravitational field strength at the lunar surface is 1.62 N kg−1.

(a)

Calculate the mass of the grain.

(2)
(b)

The grain loses 500 electrons. State the sign of its charge and calculate the magnitude of the charge.

(2)
(c)

Determine the magnitude and direction of the electric field needed for the grain to levitate.

(3)
(d)

A grain of the same material with radius 1.0 μm is in the same field. Determine the number of electrons it must lose in order to levitate.

(2)
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Part (a)
V = (4/3)πr3 = (4/3)π(0.50×10−6)3 = 5.24×10−19 m3✓ 1
m = ρV = 3000 × 5.24×10−19 = 1.57×10−15 kg ≈ 1.6×10−15 kg✓ 1
Part (b)
Positive, because the grain has lost electrons✓ 1
q = 500 × 1.60×10−19 = 8.0×10−17 C✓ 1
Part (c)
Electric force balances weight: qE = mg✓ 1
E = (1.57×10−15 × 1.62) / (8.0×10−17) ≈ 32 N C−1. Allow error carried forward from the mass and charge calculated earlier✓ 1
Direction: vertically upwards, away from the surface, because the force on a positive charge is in the direction of the field✓ 1
Part (d)
Mass, and so weight, is proportional to r3, so the weight increases by a factor of 23 = 8✓ 1
At the same E, the charge must also increase by a factor of 8: n = 8 × 500 = 4000 electrons. Allow error carried forward from the number of electrons used earlier✓ 1

Exam tip: Use proportionality in the last part. Doubling the radius multiplies the mass by 8, so there is no need to repeat the whole calculation.