D.2 Electric and magnetic fields: IB Physics SL exam-style questions
D.2 covers forces between charges, Coulomb's law F = kq₁q₂/r², conservation of charge and Millikan's experiment as evidence for the quantisation of charge. You need charging by friction, contact and electrostatic induction, including earthing.
Electric fields are described by E = F/q and by field lines, including the uniform field E = V/d between parallel plates and its edge effects. Magnetic field patterns are needed for a bar magnet, a straight wire, a circular coil and a solenoid. Electric potential and equipotentials are Higher Level.
33 questions
218 marks
Paper 1A: 14
Paper 1B: 7
Paper 2: 12
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33 practice questions on D.2 Electric and magnetic fields
1D-1A-10
Coulomb's law·D.2 Electric and magnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
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?
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.
Syllabus understandingD.2 — the direction of forces between the two types of electric charge; Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0Command term: Calculate
2D-1A-11
Electric field strength·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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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Notes
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.
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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).
✓ 1
Answer 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.
Syllabus understandingD.2 — the electric field strength as given by E = F/qCommand term: Deduce
3D-1A-12
Uniform fields between plates·D.2 Electric and magnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksCalculate
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?
Diagram NOT accurately drawn
Field strength / V m−1Force on the electron / N
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Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d; the electric field strength as given by E = F/qCommand term: Calculate
4D-1A-13
Coulomb's law·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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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Step 1In Coulomb's law k = 1/4πε, so F ∝ 1/ε for fixed charges and separation.
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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.
✓ 1
Answer 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.
Syllabus understandingD.2 — Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; a range of permittivity values (guidance) Command term: Determine
5D-1A-14
Uniform fields between plates·D.2 Electric and magnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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?
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.
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All 2 steps must be completed — there is no mark for a part-answer.
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.
Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d; work done in electric fields in joules and electronvolts (guidance) Command term: Determine
6D-1A-15
Electric field lines·D.2 Electric and magnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Two parallel metal plates carry equal and opposite charges.
Which statement about the electric field lines is correct?
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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.
✓ 1
Answer 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.
Syllabus understandingD.2 — electric field lines; the relationship between field line density and field strength; two oppositely charged parallel plates, including edge effects (guidance) Command term: Identify
7D-1A-16
Quantization of charge·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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?
Diagram NOT accurately drawnShow mark scheme
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Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — Millikan's experiment as evidence for quantization of electric charge Command term: Determine
8D-1A-17
Combining two electric fields·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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
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Step 1Both positive: each charge pushes a positive test charge at M away from itself, so the two fields are opposite and cancel.
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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.
✓ 1
Answer 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.
Syllabus understandingD.2 — the electric field strength as given by E = F/q; the field between two point charges (guidance) Command term: Deduce
9D-1A-18
Charging by friction & induction·D.2 Electric and magnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksExplain
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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Notes
Step 1The negative rod repels free electrons in the metal to the far side of the sphere.
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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.
✓ 1
Answer 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.
Syllabus understandingD.2 — that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing) Command term: Explain
10D-1B-03
Coulomb's law·D.2 Electric and magnetic fields
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
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 / m
Δm / g
lg (r / m)
lg (Δm / g)
0.040
0.349
−1.398
−0.457
0.050
0.224
−1.301
−0.650
0.060
0.159
0.075
0.101
−1.125
−0.996
0.090
0.071
0.120
0.040
−0.921
−1.398
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)
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Part (a)
0.001/0.040 × 100 = 2.5 %
✓ 1
Accept 3 %. Accept ½ × resolution giving 1.3 %.
Part (b)
lg (r / m) = −1.222 and −1.046
✓ 1
Both needed. Accept 3 or 4 d.p.
lg (Δm / g) = −0.799 and −1.149
✓ 1
Both 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
✓ 1
Allow 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
✓ 1
Accept −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
✓ 1
Allow 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
✓ 1
OWTTE
The line is steeper: the gradient becomes more negative (magnitude greater than it should be)
✓ 1
Award 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
✓ 1
OWTTE
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)
Syllabus understandingD.2 — Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0Command term: Determine
11D-1B-04
Charging by friction & induction·D.2 Electric and magnetic fields
Paper 1BEasy8 marks
Data-based question8 steps to full marksDeduce
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.
Trial
Charge on strip / nC
Charge 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
✓ 1
Do 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
✓ 1
Absolute 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
✓ 1
Allow ECF from (b).
Part (d)
N = 12.4 × 10−9/1.60 × 10−19 = 7.8 × 1010
✓ 1
Accept 7.8 × 1010.
Part (e)
Some charge flows from the cloth through the student's hand/body to earth (the body is a conductor)
✓ 1
OWTTE
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)
Syllabus understandingD.2 — the conservation of electric charge; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing) Command term: Deduce
12D-1B-10
Magnetic field patterns·D.2 Electric and magnetic fields
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
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 / A
B / mT
0.30
0.96
0.60
1.66
0.90
2.44
1.20
3.17
1.50
3.95
1.80
4.64
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)
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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
✓ 1
Accept ±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
✓ 1
Do 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
✓ 1
OWTTE. 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)
✓ 1
Accept 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
✓ 1
Accept ±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
✓ 1
Allow 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
✓ 1
Fractional 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
✓ 1
Allow 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)
Syllabus understandingD.2 — magnetic field lines (field of an air-core solenoid; B = μ0nI is given) Command term: Determine
13D-2-03
Gravity compared with electrostatics·D.2 Electric and magnetic fields
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksDetermine
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.
this is about 10−6 of the elementary charge «1.60 × 10−19 C»
✓ 1
Allow 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
✓ 1
MP3 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
✓ 1
OWTTE
Answers: (a)(i) 1.24 × 1036 · (b)(ii) q = 2.6 × 10−25 C (the remaining parts are explanations — see the table above)
Syllabus understandingD.2 — Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; Millikan's experiment as evidence for quantization of electric charge (linked to D.1 — Newton's universal law of gravitation) Command term: Determine
14D-2-05
Coulomb's law·D.2 Electric and magnetic fields
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
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.
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)
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Part (a)(i)
F = 8.99 × 109 × (1.60 × 10−19)²/(6.0 × 10−6)²
✓ 1
F = 6.4 × 10−18 N
✓ 1
Award [2] for CNA.
Part (a)(ii)
the distance is doubled so the force is a quarter: 1.6 × 10−18 N
✓ 1
Allow 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
✓ 1
Allow 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»
✓ 1
Allow 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»
✓ 1
The field must give a force opposite to the resultant repulsion because the ion is positive.
Part (c)
zero
✓ 1
Award 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)
Syllabus understandingD.2 — the direction of forces between the two types of electric charge; Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; the electric field strength as given by E = F/qCommand term: Calculate
15D-2-06
Uniform fields between plates·D.2 Electric and magnetic fields
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksDraw
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.
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 point
Mark
Notes
Part (a)(i)
parallel, equally spaced lines perpendicular to the plates, with arrows from X to Y, in the central region
✓ 1
At least three lines.
lines near the edges curve «bulge» outwards and are further apart
✓ 1
Do 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
✓ 1
Accept N C−1.
Part (b)(i)
n = 4.8 × 10−16/1.60 × 10−19 = 3000
✓ 1
Must 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
✓ 1
Allow 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
✓ 1
Allow 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
✓ 1
Award 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
✓ 1
ALTERNATIVE: 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
✓ 1
Award [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
✓ 1
OWTTE
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)
Syllabus understandingD.2 — electric field lines; the uniform electric field strength between parallel plates as given by E = V/d; two oppositely charged parallel plates, including edge effects; work done in electric fields in joules and electronvolts (guidance) Command term: Draw
16D-2-07
Quantization of charge·D.2 Electric and magnetic fields
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
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.
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 point
Mark
Notes
Part (a)(i)
downward arrow labelled weight «gravitational force» AND upward arrow labelled electric force
✓ 1
Do not award if a drag or buoyancy force is added.
Must 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»
✓ 1
A 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
✓ 1
ALTERNATIVE: 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
✓ 1
Accept 1.9–2.0 m s−2.
upwards
✓ 1
The direction mark is independent of the magnitude.
Part (b)(ii)
V = 235 × 5/6 = 196 V
✓ 1
Accept 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»
✓ 1
OWTTE
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
✓ 1
OWTTE
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)
Syllabus understandingD.2 — Millikan's experiment as evidence for quantization of electric charge; the uniform electric field strength between parallel plates as given by E = V/d (linked to A.2 — free-body diagrams, Newton's second law and viscous drag) Command term: Determine
17D-2-14
Magnetic field patterns·D.2 Electric and magnetic fields
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksDraw
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.
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 point
Mark
Notes
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
✓ 1
Arrows needed on at least one loop around each wire or on the central line.
Part (a)(ii)
the right-hand face
✓ 1
Field 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
✓ 1
OWTTE
Part (b)(i)
the right-hand face of the coil behaves like a north pole «field lines leave the coil there»
✓ 1
Allow ECF from (a)(ii).
like poles repel, so the magnet is repelled «pushed to the right, away from the coil»
✓ 1
MP2 only scores if MP1 scores.
Part (b)(ii)
to the left «away from the magnet»
✓ 1
Newton'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
✓ 1
Accept: 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
Syllabus understandingD.2 — magnetic field lines; magnetic field patterns of a current-carrying circular coil and of a bar magnet (guidance); the determination of the direction of the magnetic field based on the current direction in a current-carrying straight wire (guidance) (linked to A.2 — Newton's third law) Command term: Draw
18D-1A-33
Conservation of charge·D.2 Electric and magnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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
Show mark scheme
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Mark
Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — the conservation of electric charge; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact Command term: Determine
19D-1A-35
Charged conducting sphere·D.2 Electric and magnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDescribe
A hollow metal sphere carries a positive charge.
Which row describes the electric field inside the sphere and just outside it?
InsideJust outside
Show mark scheme
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Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — electric field lines; the field inside and outside a single spherical conducting body (guidance) Command term: Describe
20D-1A-36
Magnetic field of a straight wire·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
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?
Diagrams NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — magnetic field lines; the determination of the direction of the magnetic field based on the current direction in a current-carrying straight wire (guidance) Command term: Identify
21D-1A-37
Electric field lines·D.2 Electric and magnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
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)?
Show mark scheme
Marking point
Mark
Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — electric field lines; the relationship between field line density and field strength Command term: Deduce
22D-1A-38
Coulomb's law·D.2 Electric and magnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
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 point
Mark
Notes
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.
✓ 1
Answer 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.
Syllabus understandingD.2 — Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0Command term: Determine
23D-1B-13
Uniform fields between plates·D.2 Electric and magnetic fields
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
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 / kV
x / mm
1.0
5.5
2.0
10.0
3.0
15.5
4.0
20.0
5.0
26.0
6.0
30.5
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 point
Mark
Notes
Part (a)
F = 0.250 × 10−3 × 9.81 × 0.0305/0.300 = 2.49 × 10−4 N
✓ 1
Must 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)
✓ 1
Accept 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
Recharge the ball to the same charge before each reading OR take all the readings quickly OR carry out the experiment in dry air
✓ 1
OWTTE
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)
Syllabus understandingD.2 — the electric field strength as given by E = F/q; the uniform electric field strength between parallel plates as given by E = V/d Command term: Determine
24D-1B-16
Magnetic field of a straight wire·D.2 Electric and magnetic fields
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
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 / mm
Bon / μT
Boff / μT
B / μT
lg (r / mm)
lg (B / μT)
10
179.0
18.6
160.4
1.000
2.205
15
124.8
18.4
106.4
1.176
2.027
20
98.9
18.7
80.2
1.301
1.904
30
71.5
18.5
45
54.5
18.6
35.9
1.653
1.555
60
44.9
18.4
26.5
1.778
1.423
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 point
Mark
Notes
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
✓ 1
OWTTE
Part (b)
B = 71.5 − 18.5 = 53.0 μT
✓ 1
lg (r / mm) = 1.477 AND lg (B / μT) = 1.724
✓ 1
Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (c)
0.5 + 0.5 = ±1.0 μT
✓ 1
Absolute 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
✓ 1
Accept −0.96 to −1.04. No unit.
Part (e)
gradient ≈ −1, so B ∝ r−1: B is inversely proportional to r, as predicted
✓ 1
Allow 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
✓ 1
Accept 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
✓ 1
OWTTE. 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)
Syllabus understandingD.2 — magnetic field lines (the field of a current-carrying straight wire; B = μ0I/2πr is given) Command term: Determine
25D-1B-18
Quantization of charge·D.2 Electric and magnetic fields
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
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.
Drop
m / 10−15 kg
V / V
q / 10−19 C
A
1.54
186
4.87
B
0.86
320
C
2.39
216
6.51
D
1.16
217
E
3.09
225
(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.
= 4.1 %, so Δq = 4.1 % × 1.58 × 10−19 ≈ 0.07 × 10−19 C
✓ 1
Accept 0.06–0.07 × 10−19 C.
q = (1.58 ± 0.07) × 10−19 C
✓ 1
MP3 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
✓ 1
Both needed.
Part (c)
Numbers of elementary charges: A 3, B 1, C 4, D 2, E 5
✓ 1
Allow ECF from (b).
e = total charge/total number = 24.2 × 10−19/15 = 1.61 × 10−19 C
✓ 1
Accept 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
✓ 1
Allow 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)
✓ 1
Award 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
✓ 1
OWTTE
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)
Syllabus understandingD.2 — Millikan's experiment as evidence for quantization of electric charge Command term: Determine
26D-1B-20
Uniform fields between plates·D.2 Electric and magnetic fields
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
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 / mm
1/d / m−1
E / kV m−1
20
50.0
100.2
30
33.3
66.1
40
25.0
50.0
60
16.7
33.1
80
12.5
24.2
100
10.0
18.3
125
8.00
13.7
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 point
Mark
Notes
Part (a)
Straight line through the origin passing within ±0.5 kV m−1 of the four points with d ≤ 60 mm
✓ 1
Line need not pass through the remaining three points.
Part (b)
Gradient calculated from a large triangle
✓ 1
gradient = 2.00 kV (= 2000 V)
✓ 1
Accept 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 %)
✓ 1
Allow 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
✓ 1
OWTTE
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
✓ 1
OWTTE
Part (f)
Use plates that are much larger (wider) compared with the largest separation
✓ 1
Accept: 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)
Syllabus understandingD.2 — the uniform electric field strength between parallel plates as given by E = V/d Command term: Determine
27D-2-23
Combining two electric fields·D.2 Electric and magnetic fields
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
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.
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)
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Notes
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»
✓ 1
Do 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
✓ 1
OWTTE
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»
✓ 1
Award [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
✓ 1
ALTERNATIVE: 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
✓ 1
Direction 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»
✓ 1
Both 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)
Syllabus understandingD.2 — the electric field strength as given by E = F/q; electric field lines; the relationship between field line density and field strength; Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0Command term: Determine
28D-2-24
Charging by friction & induction·D.2 Electric and magnetic fields
Paper 2Medium11 marks
Short answer & extended response11 steps to full marksExplain
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.
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)
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Notes
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
✓ 1
Do 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
the «free» electrons in the sphere are repelled by the negative rod, and the wire lets them move further away from it
✓ 1
OWTTE
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
✓ 1
OWTTE
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)
Syllabus understandingD.2 — that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing); Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; the conservation of electric charge Command term: Explain
29D-2-25
Conservation of charge·D.2 Electric and magnetic fields
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksDetermine
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.
(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)
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Part (a)
F = 8.99 × 109 × 12.0 × 10−9 × 4.0 × 10−9 / 0.060² = 1.2 × 10−4 N
✓ 1
Accept 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
✓ 1
Allow 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
✓ 1
Award MP1 for use of a charge transfer of 8.0 nC.
from Y to X
✓ 1
Do 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
✓ 1
Allow 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
✓ 1
OWTTE. 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)
Syllabus understandingD.2 — the conservation of electric charge; Millikan's experiment as evidence for quantization of electric charge; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing); Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0Command term: Determine
30D-2-26
Charged conducting sphere·D.2 Electric and magnetic fields
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksSketch
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.
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)
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Part (a)
E = 0 for r < R
✓ 1
E rises to its maximum value Es at r = R
✓ 1
Accept a vertical step at R.
for r > R, a decreasing curve passing through about Es/4 at 2R «and Es/9 at 3R»
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
✓ 1
Accept 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
✓ 1
ALTERNATIVE: 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
✓ 1
Conclusion 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
✓ 1
OWTTE
Part (f)
otherwise the charge would flow through the student to the ground «the student would be earthed»
✓ 1
OWTTE
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)
Syllabus understandingD.2 — the electric field strength as given by E = F/q (field inside and outside a spherical conducting body); Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing) Command term: Sketch
31D-2-27
Electric field lines·D.2 Electric and magnetic fields
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksExplain
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.
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)
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Notes
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»
✓ 1
OWTTE
Part (b)
away from the building: straight, parallel, equally spaced vertical lines
✓ 1
arrows pointing upwards, from the ground to the cloud
✓ 1
Do 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
✓ 1
Conclusion 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
✓ 1
MP2 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
✓ 1
OWTTE
Answers: (c) 2.5 × 104 V m−1; no (the remaining parts are explanations — see the table above)
Syllabus understandingD.2 — electric field lines; the relationship between field line density and field strength; the uniform electric field strength between parallel plates as given by E = V/d; that the electric charge can be transferred between bodies using friction, electrostatic induction and by contact, including the role of grounding (earthing). Link: B.5 — resistance and P = I²RCommand term: Explain
32D-2-28
Coulomb's law·D.2 Electric and magnetic fields
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksExplain
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.
(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)
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Part (a)(i)
F = 8.99 × 109 × (1.60 × 10−19)² / (5.3 × 10−11)²
✓ 1
F = 8.2 × 10−8 N
✓ 1
Award [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»
✓ 1
OWTTE
Part (b)
E = F/e OR ke/r² = 5.1 × 1011 N C−1
✓ 1
Allow 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
✓ 1
Allow ECF from (a)(i).
T = 2πr/v = 1.53 × 10−16 s
✓ 1
number per second = 1/T = 6.6 × 1015
✓ 1
Accept 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
✓ 1
OWTTE
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
✓ 1
Accept 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»
✓ 1
OWTTE
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)
Syllabus understandingD.2 — Coulomb's law as given by F = kq1q2/r² for charged bodies treated as point charges where k = 1/4πε0; the electric field strength as given by E = F/q. Links: A.2 — centripetal force; E.1 — emission spectra as evidence for discrete atomic energy levels Command term: Explain
33D-2-29
Magnetic field patterns·D.2 Electric and magnetic fields
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
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.
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)
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Notes
Part (a)
three concentric circles centred on W, with the spacing increasing with distance from W
✓ 1
arrows anticlockwise «as seen from above»
✓ 1
Right-hand grip rule with the current out of the page.
Part (b)
east «to the right»
✓ 1
Allow 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»
✓ 1
Allow 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
✓ 1
Accept 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)
Syllabus understandingD.2 — magnetic field lines (field pattern of a current-carrying straight wire; direction of the field from the current direction) Command term: Determine
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