120 free, high-quality past-paper-style questions on Electric and magnetic fields for IB Physics HL, each with a full mark scheme. Covers: Conservation of charge; Magnetic field of a straight wire; Electric potential and work; and more. For when you have finished the past papers, or want to practise one topic at a time.D.2 at Standard 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.539 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1D-1A-41Conservation of chargePaper 1AEasy1 mark
Three identical small conducting spheres X, Y and Z are mounted on insulating handles. At first X has a charge of +8.0 nC, Y has a charge of −2.0 nC and Z is uncharged.
X is touched against Y and they are then separated. Y is next touched against Z and they are separated.
What is the final charge on Z?
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Step 1Charge is conserved, and identical conducting spheres in contact share the total charge equally.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2X and Y: total = +8.0 + (−2.0) = +6.0 nC, so each has +3.0 nC after they are separated.
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Step 3Y and Z: total = +3.0 + 0 = +3.0 nC, so Z ends with +1.5 nC.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: +6.0 nC is shared between X and Y, then Y's +3.0 nC is shared with Z.
BThis shares the total +6.0 nC equally among all three spheres, as if all three had touched at once. The order of the contacts is ignored.
CThis adds the magnitudes (8.0 + 2.0 = 10 nC) and ignores the sign of the charge on Y: 10/2 = 5.0 nC, then 5.0/2 = 2.5 nC.
DThis assumes that all of Y's charge (+3.0 nC) passes to Z. Identical spheres in contact share the charge equally.
2D-1A-46Magnetic field of a straight wirePaper 1AEasy1 mark
A long straight vertical wire passes through a hole in a horizontal card. There is a steady current upwards in the wire. A small plotting compass is placed on the card due east of the wire. The Earth's magnetic field can be ignored.
In which direction does the north pole of the compass needle point?
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Step 1The field lines of a straight wire are concentric circles around the wire in a plane perpendicular to it.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Right-hand grip rule: with the thumb pointing up along the current, the fingers curl anticlockwise when the card is viewed from above.
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Step 3At a point due east of the wire the anticlockwise circle points north, and the compass lines up with the field.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the field as radial, pointing away from the wire like the electric field of a positive charge. Magnetic field lines of a wire go round the wire.
BThis treats the field as radial, pointing into the wire. The field lines of a wire are circles around it, not lines towards it.
CThis applies the grip rule with the wrong hand or with the current reversed, so the field direction comes out the wrong way round the circle.
DCorrect: the field circulates anticlockwise when viewed from above, which is northwards at a point east of the wire.
3D-1A-49Electric potential and workPaper 1AEasy1 mark
An alpha particle (charge +2e) moves in an electrostatic analyser from a point where the electric potential is +1200 V to a point where the electric potential is −300 V. Only the electric force acts on it.
What is the increase in the kinetic energy of the alpha particle?
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Step 1Work done by the field = q × (fall in potential): W = qΔVe.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Fall in potential = +1200 − (−300) = 1500 V.
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Step 3Gain in kinetic energy = 2e × 1500 V = 3000 eV = 3.0 keV (4.8 × 10−16 J).
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses a charge of e instead of 2e, which gives 1500 eV.
BThis ignores the negative sign of the final potential and takes the difference as 1200 − 300 = 900 V, which gives 2 × 900 = 1800 eV.
CThis uses only the starting potential (1200 V), as if the final point were at zero potential: 2 × 1200 = 2400 eV.
4D-1A-63Electric field and potentialPaper 1AEasy1 mark
At point P, a distance r from an isolated point charge Q, the electric field strength is E and the electric potential is V.
The charge is replaced by a point charge 2Q and P is moved to a distance r/2 from it. Which row gives the new electric field strength and the new electric potential at P?
new field strengthnew potential
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Step 1E = kQ/r²: doubling the charge doubles E, and halving the distance multiplies it by 4, so the new field is 8E.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2V = kQ/r: doubling the charge doubles V, and halving the distance doubles it again, so the new potential is 4V.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis changes the distance only and ignores the doubling of the charge.
BThis treats the field strength as inversely proportional to r, like the potential.
CCorrect: E ∝ Q/r² and V ∝ Q/r.
DThis treats the potential as inversely proportional to r², like the field strength.
5IBP2H-031-01-01Charging by induction and earthingPaper 1AEasy1 mark
An uncharged metal sphere stands on an insulating base. A negatively charged plastic rod is held close to the sphere without touching it. With the rod still in place, the sphere is briefly connected to earth and then disconnected. The rod is then taken away. What is the final charge on the sphere?
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The negative rod repels free electrons to the far side of the sphere. While the sphere is earthed, these repelled electrons flow to earth. The earth link is broken before the rod is removed, so the electrons cannot return. The sphere is left with an electron deficit, which is a net positive charge. Once the rod is gone, this charge spreads over the whole surface. A assumes charge passes from the rod. B forgets that the rod is still present while the sphere is earthed. D describes the separated charge before earthing, not the final state.
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Answer C
Answer: C
Common error: In induction the sphere always ends up with the opposite sign to the inducing rod. Only electrons move, and they move through the earth connection.
6IBP2H-031-01-02Charged conducting spherePaper 1AEasy1 mark
A hollow metal sphere of radius R carries charge +Q. Which row gives the electric field strength E and the electric potential V at a point a distance R/2 from the centre?
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The charge on a conductor sits on its outer surface, so the field inside is zero. Zero field means zero potential gradient, so the potential is the same everywhere inside and equals the surface value kQ/R. A wrongly assumes zero field implies zero potential. C treats the charge as a point charge at the centre and uses r = R/2. D correctly gives E = 0 but still uses r = R/2 in the potential formula.
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Answer B
Answer: B
Exam tip: E = −ΔV/Δr. Wherever E is zero, V is constant, but not necessarily zero.
7IBP2H-031-00-01Electric field and potentialPaper 1AEasy1 mark
Two small charged dust grains in a planetary ring exert an electrostatic force of magnitude F on each other. The charge on one grain doubles and the separation of the grains becomes three times larger. What is the new magnitude of the force?
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Coulomb's law gives \(F = \frac{kq_1q_2}{r^2}\). Doubling one charge multiplies the force by 2. Tripling the separation divides it by 32 = 9. The new force is \(\frac{2}{9}F\). Option B forgets to square the distance. Option C wrongly squares the charge factor. Option D multiplies by the distance factor instead of dividing.
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Answer A
Answer: A
Common error: Only the separation is squared in Coulomb's law. Each charge appears to the first power.
8IBP2H-031-00-08Millikan's experimentPaper 1AEasy1 mark
Which unit is equivalent to the unit of electric field strength, N C−1?
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1 V = 1 J C−1 = 1 N m C−1. So 1 V m−1 = 1 N C−1, which is consistent with E = V/d. Option A is the unit of potential. Option C multiplies by distance instead of dividing. Option D is the reciprocal of the unit.
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Answer B
Answer: B
Exam tip: Field strength is force per unit charge and also potential gradient. Both give the same unit.
9IBP2H-031-02-01Conservation of chargePaper 1AEasy1 mark
In a tumble dryer a nylon sock and a cotton shirt rub together. The two items are initially uncharged and together form an isolated system. Afterwards the sock has a charge of +5.0 nC. Which row gives the charge on the shirt and the particles transferred?
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Charge is conserved. The total charge was zero before rubbing, so it is zero afterwards. The shirt must therefore carry −5.0 nC. In solids only electrons move; protons stay bound in nuclei. The sock lost electrons and became positive, and the shirt gained them. Option B moves protons. Option C creates charge. Option D gives a non-zero total charge of +10 nC.
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Answer A
Answer: A
Common error: A positively charged insulator has lost electrons. It has not gained protons. Only electrons move during charging by friction.
10IBP2H-031-02-02Magnetic field of a solenoidPaper 1AEasy1 mark
Which statement about magnetic field lines is correct?
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For a straight wire, B = μ0I/(2πr). The field lines are circles around the wire, and they spread out as B decreases with r. Magnetic field lines always form closed loops, so they do not start or end on poles (A is wrong). Radial lines describe the electric field of a point charge, not a magnetic field (B is wrong). Inside a long solenoid the field is almost uniform, and it weakens near the ends (D is wrong).
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Answer C
Answer: C
Exam tip: Line spacing shows field strength. For a wire the circles get further apart with distance, because B ∝ 1/r.
Two small ice crystals on a mountain ridge carry charges and repel each other with a force F when their separation is r. One charge is then doubled and the separation is increased to 3r. What is the new force between them?
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Coulomb's law gives F ∝ q1q2/r2, so the new force is 2/32 = 2/9 of F (C). A (F/9) ignores the doubled charge. B (2F/3) uses 1/r instead of 1/r2. D (9F/2) inverts the factor.
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Answer C
Answer: C
Common error: Force between point charges follows an inverse-square law, so a tripled separation divides the force by 9, not 3.
12IBP3H-D2-20r8-05Magnetic field of a straight wirePaper 1AEasy1 mark
The magnetic field at a distance r from a long straight current-carrying wire has magnitude B. What is the magnitude of the field at a distance 3r from the same wire when the current is halved?
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For a long straight wire B ∝ I/r, so the new field is (1/2)(1/3)B = B/6 (B). A uses an inverse-square law. C ignores the change in current. D treats the current as doubled.
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Answer B
Answer: B
Common error: The field of a long straight wire falls off as 1/r, not 1/r².
13IBP3H-D2-20r11-01Conservation of chargePaper 1AEasy1 mark
In a data centre, a neutral polyester cloth is rubbed on a neutral plastic casing. Afterwards the casing carries a charge of −4.8 nC. Which row correctly describes the cloth and the charge transfer?
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Charge is conserved, so the cloth gains +4.8 nC. The casing became negative, so electrons moved to it from the cloth, and the number is 4.8×10−9/1.60×10−19 = 3.0×1010. B breaks conservation of charge (the total would be −9.6 nC). C reverses the direction of transfer. D multiplies by e instead of dividing.
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Answer A
Answer: A
Common error: Only electrons move when objects are charged by rubbing; the charge on the other object is equal in size and opposite in sign.
14IBP3H-D2-20r11-02Coulomb's lawPaper 1AEasy1 mark
Two small charged grains of dust lie a distance r apart and exert an electrostatic force of magnitude F on each other. The charge on one grain is doubled, the other charge is unchanged, and the separation is increased to 3r. What is the new magnitude of the force?
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F ∝ q1q2/r2, so the factor is 2/32 = 2/9, giving 2F/9. A multiplies by 32 instead of dividing. B squares the factor of 2. D uses 1/r instead of 1/r2.
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Answer C
Answer: C
Exam tip: Treat each change as a separate factor and multiply the factors together.
15IBP3H-D2-20r11-04Uniform electric fieldsPaper 1AEasy1 mark
Two parallel metal plates carry equal and opposite charges. In the region between the plates, away from the edges, the electric field lines are drawn as equally spaced parallel lines. What does this show about the field in that region?
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Equal spacing of field lines indicates a uniform field, so the field strength is constant and a given charge experiences the same force anywhere. A is wrong because the potential changes steadily from one plate to the other. B confuses a uniform field with a point of zero field. C contradicts the constant field strength.
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Answer D
Answer: D
Exam tip: Line spacing shows field strength; line direction shows field direction.
16D-1A-04Electric field and potentialPaper 1AMedium1 mark
Two point charges, each +Q, are fixed at two vertices of an equilateral triangle of side a. Point P is the third vertex of the triangle.
Which row gives the magnitude of the electric field strength E and the electric potential Ve at P?
Diagram not to scale
E at PVe at P
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Step 1Each charge is a distance a from P and gives a field of magnitude kQ/a², directed away from that charge along the side of the triangle.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two sides meet at P at 60°, so the two field vectors are at 60° to each other: E = 2(kQ/a²) cos 30° = √3 kQ/a².
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Step 3Potential is a scalar, so the two contributions simply add: Ve = kQ/a + kQ/a = 2kQ/a.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis adds the two field magnitudes as if both fields pointed the same way. The fields are at 60° to each other, so the resultant is less than their sum.
BCorrect: vector addition of two equal fields at 60° gives √3 kQ/a², and the scalar potentials add to 2kQ/a.
CThe field is right, but this resolves the potential as if it were a vector. Potential is a scalar: the two contributions add directly.
DThis takes the angle between the two field vectors as 120° (the exterior angle at P), so that 2 cos 60° = 1. The fields point away from the charges along the two sides, which meet at 60°.
17D-1A-18Millikan's experimentPaper 1AMedium1 mark
In an experiment of Millikan's type, the charges on four oil drops are measured as 4.8 × 10−19 C, 8.0 × 10−19 C, 11.2 × 10−19 C and 14.4 × 10−19 C.
What is the largest value of the basic unit of charge that is consistent with these data?
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Step 1Every measured charge must be a whole-number multiple of the basic unit of charge, so the unit must divide exactly into all four values.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Test the candidate values: with 1.6 × 10−19 C the drops carry 3, 5, 7 and 9 units; 3.2 × 10−19 C gives 1.5, 2.5, 3.5, 4.5 and 4.8 × 10−19 C gives 8.0/4.8 = 1.67 — not whole numbers.
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Step 3The largest unit that divides all four values exactly is 1.6 × 10−19 C (smaller units such as 0.80 × 10−19 C also divide them, but are not required by the data).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is consistent with the data (6, 10, 14 and 18 units), but it is not the largest such value: 1.6 × 10−19 C also fits every drop.
BCorrect: 4.8, 8.0, 11.2 and 14.4 are 3, 5, 7 and 9 times 1.6 (× 10−19 C).
CThis takes the constant difference between successive values (8.0 − 4.8 = 11.2 − 8.0 = 3.2) as the unit; but 4.8/3.2 = 1.5, so the charges are not whole multiples of 3.2 × 10−19 C.
DThis takes the smallest measured charge as the unit; but 8.0/4.8 = 1.67 is not a whole number.
Two small charged spheres in air exert an electric force F on each other. The permittivity of air may be taken as ε0. The spheres are then immersed in an oil of permittivity 3.0ε0 and their separation is halved.
What is the new electric force between the spheres?
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Step 1F = q1q2/(4πεr²), so F ∝ 1/(εr²).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Halving r multiplies the force by 4; increasing the permittivity to 3.0ε0 divides it by 3.
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Step 3New force = 4F/3.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis inverts the effect of the separation (dividing by 4 instead of multiplying) and also divides by 3.
BThis allows for the permittivity but ignores the halving of the separation.
CThis treats the force as inversely proportional to r rather than r²: 2/3.
DCorrect: × 4 for the separation and ÷ 3 for the permittivity.
19D-1A-26Equipotential surfacesPaper 1AMedium1 mark
The diagram shows the cross-sections of five plane equipotential surfaces in a region of electric field, with their potentials and a scale. Point P is midway between the 60 V and 40 V surfaces.
What is the magnitude of the electric field strength at P?
Cross-sections of equipotential surfaces (dashed) in a region of electric field.Show mark scheme
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Step 1E = −ΔVe/Δr: the field strength is the potential gradient at right angles to the equipotentials, so use the surfaces on either side of P.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2ΔV = 60 V − 40 V = 20 V across 4.5 cm − 2.5 cm = 2.0 cm.
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Step 3E = 20/0.020 = 1.0 × 103 V m−1, directed from the 60 V surface towards the 40 V surface.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis divides the 20 V by the distance of P from the 100 V surface (3.5 cm) instead of by the spacing of the surfaces on either side of P.
BCorrect: 20 V/0.020 m = 1.0 × 103 V m−1.
CThis divides the potential at P (50 V) by its distance from the 100 V surface, as if E = V/d could be applied from a reference surface: 50/0.035.
DThis uses the 1.0 cm spacing of the first two surfaces. The surfaces are not equally spaced, so the field is not uniform and must be found near P.
20D-1A-40Charging by induction and earthingPaper 1AMedium1 mark
An uncharged aluminium sphere stands on an insulating base. A negatively charged PVC rod is held close to the sphere but never touches it. With the rod still in place, the sphere is briefly connected to earth by a wire. The wire is then removed, and finally the rod is taken away.
I. While the sphere is connected to earth, electrons flow from the sphere to earth.
II. After the rod is taken away, the sphere is positively charged.
III. If the rod had been taken away before the earth wire was removed, the sphere would also have been left positively charged.
Which statements are correct?
Diagram not to scale
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Step 1The negative rod repels free electrons in the sphere; when the sphere is earthed, these electrons flow from the sphere to earth: I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The wire is removed while the rod is still there, so the electrons cannot return. The sphere has a deficit of electrons and is left positive: II is correct.
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Step 3If the rod were removed first, with the earth wire still connected, electrons would flow back from earth and the sphere would be left uncharged: III is wrong.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: I and II only.
BThis accepts III. While the earth wire is still connected, removing the rod lets electrons flow back from earth, so the sphere ends uncharged. It also rejects II, which is correct.
CThis rejects I, as if electrons were attracted on to the sphere from earth. The negative rod repels electrons, so they flow away to earth.
DThis accepts III. The order of the last two steps matters: electrons can return through the earth wire for as long as it stays connected.
Two small beads P and R are fixed 0.200 m apart on a horizontal insulating thread in air. P has a charge of +3.0 nC and R has a charge of −5.0 nC. A third bead S with a charge of −2.0 nC is placed on the line between them, 0.080 m from P.
What is the magnitude of the resultant electric force on S?
Diagram not to scale
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Step 1P is positive and attracts S towards P. R and S are both negative, so R repels S — away from R, which is also towards P.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Force from P = 8.99 × 109 × 3.0 × 10−9 × 2.0 × 10−9/0.080² = 8.43 × 10−6 N; force from R (0.120 m away) = 8.99 × 109 × 5.0 × 10−9 × 2.0 × 10−9/0.120² = 6.24 × 10−6 N.
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Step 3Both forces act towards P, so they add: 8.43 + 6.24 = 14.7 × 10−6 N ≈ 1.5 × 10−5 N.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis subtracts the forces, as if R attracted S towards R: 8.43 − 6.24 = 2.2 μN. R and S have the same sign, so R repels S.
BThis includes only the force from the nearer bead P.
CThis uses 0.200 m, the distance from P to R, as the distance from S to R: 8.43 + 2.25 = 10.7 μN.
DCorrect: the attraction towards P and the repulsion from R are in the same direction.
22D-1A-43Electric field strengthPaper 1AMedium1 mark
In an experiment of Millikan's type, a charged oil drop is held stationary between two horizontal parallel plates. The potential difference between the plates is then tripled and, at the same moment, the separation of the plates is halved. Air resistance is negligible at the instant of the change.
What is the magnitude of the acceleration of the drop immediately after the change?
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Step 1Initially the electric force balances the weight: qE = qV/d = mg.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2E = V/d: tripling V and halving d multiply E, and so the electric force, by 6. The electric force becomes 6mg upwards.
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Step 3Resultant force = 6mg − mg = 5mg upwards, so the acceleration is 5g.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis triples the field but ignores the halving of the separation: 3mg − mg = 2mg.
BCorrect: (6mg − mg)/m = 5g.
CThis forgets that the weight still acts: 6mg/m.
DThis adds the weight to the electric force, as if both acted in the same direction.
24D-1A-45Uniform electric fieldsPaper 1AMedium1 mark
One cell of an electrostatic air cleaner consists of two parallel metal plates with a potential difference between them. The plates are long compared with their separation. Point P is midway between the plates near their centre. Point Q is also near the centre, a quarter of the way from the negative plate to the positive plate. Point R is level with P but just beyond the edge of the plates, as shown.
An electron is placed in turn at P, Q and R. Which describes the magnitudes FP, FQ and FR of the electric force on it?
Electric field between two oppositely charged parallel plates, including the edge region.Show mark scheme
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Step 1Well inside the plates the field is uniform, E = V/d, and has the same value at every point, so FP = FQ.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the edges the field lines bow outwards; just beyond the edge they are further apart, so the field is weaker there but not zero.
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Step 3F = eE, so FP = FQ > FR > 0.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: uniform field between the plates and a weaker, non-zero field just beyond the edge.
BThis ignores the edge effect: E = V/d applies only well inside the plates, and beyond the edge the field lines spread out.
CThis assumes that the field is confined exactly to the space between the plates. The field lines bulge beyond the edges, so the field there is not zero.
DThis treats each plate like a point charge, with a stronger field closer to it. Between large parallel plates the field is uniform, so Q is no different from P.
25D-1A-47Magnetic field of a solenoidPaper 1AMedium1 mark
A long air-core solenoid has ends P and Q. When the solenoid is viewed along its axis from end P, the current in the turns is clockwise.
Which row gives the magnetic polarity of end P and the direction of the magnetic field lines inside the solenoid?
Polarity of end PField inside the solenoid
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Step 1Right-hand grip rule for a solenoid: curl the fingers in the direction of the current and the thumb points along the field inside the solenoid.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With a clockwise current seen from P, the thumb points away from the observer, so the field inside runs from P towards Q.
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Step 3Field lines leave the solenoid at its north pole, so Q is a north pole and P, where the lines enter, is a south pole.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the field inside runs from P to Q, and the lines enter the solenoid at P, which is therefore a south pole.
BP is a south pole, but inside a solenoid (as inside a bar magnet) the field lines run from the south pole to the north pole. They form closed loops and do not run from N to S inside.
CThis contradicts itself: lines leave a north pole, so if the field inside ran from P to Q, P would have to be a south pole.
DThis applies the grip rule backwards (clockwise gives north). The answer is self-consistent, but both parts are reversed.
26D-1A-48Field and potential of several chargesPaper 1AMedium1 mark
Four point charges are fixed at the corners of a square of side a. The two upper corners each carry +q and the two lower corners each carry −q. M is the midpoint of the upper side of the square.
What is the electric potential at M?
Four point charges at the corners of a square of side a. M is the midpoint of the upper side.Show mark scheme
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Step 1Potential is a scalar: add kq/r for every charge, with its sign. Each +q is a/2 from M; each −q is √(a² + (a/2)²) = (√5/2)a from M.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two positive charges give 2 × kq/(a/2) = 4kq/a; the two negative charges give −2 × kq/((√5/2)a) = −(4/√5)kq/a.
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Step 3Ve = 4(1 − 1/√5)kq/a ≈ 2.2kq/a.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis carries over the result for the centre O, where the four contributions cancel because all four charges are equidistant. M is closer to the positive charges, so the potential there is positive.
BThis uses the side a, instead of the slant distance (√5/2)a, as the distance from M to each negative charge: 4kq/a − 2kq/a.
CCorrect: 4kq/a from the near positive charges minus (4/√5)kq/a from the far negative charges.
DThis ignores the signs of the lower charges and adds all four contributions as if every charge were positive.
28D-1A-65Magnetic field patternsPaper 1AMedium1 mark
The diagram shows the magnetic field lines in a plane perpendicular to two long, straight, parallel wires X and Y. The wires carry currents of equal magnitude.
Which row gives the directions of the currents in X and in Y?
Magnetic field lines (with arrows showing their direction) in the plane of the page. The wires are perpendicular to the page.
current in Xcurrent in Y
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Step 1Midway between the wires there is a point of zero field, and far away the lines loop around both wires: the fields of the two wires oppose each other between them, so the currents are in the same direction.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The lines are directed clockwise around each wire (and around both).
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Step 3Right-hand grip rule: a current into the page gives clockwise field lines, so both currents are into the page.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: equal currents in the same direction, into the page, give clockwise field lines.
BThis applies the grip rule the wrong way round: a current out of the page gives anticlockwise field lines.
COpposite currents would make the two fields add between the wires (crowded lines, no zero-field point) and no line would loop around both wires.
DOpposite currents would make the two fields add between the wires, and the line directions around X and Y would be opposite.
29IBP2H-031-01-03Electric field strengthPaper 1AMedium1 mark
In a Millikan-type experiment, a charged oil drop is held stationary between horizontal parallel plates when the potential difference across them is V. The plate separation is halved and the drop's charge does not change. What potential difference is now needed to hold the drop stationary?
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For balance, qV/d = mg. The weight and the charge are unchanged, so the field V/d must stay the same. Halving d therefore requires halving V. D inverts the relationship. A wrongly uses an inverse-square dependence on d. C assumes the field between plates does not depend on their separation.
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Answer B
Answer: B
Exam tip: In Millikan problems, first identify what must stay constant. Here it is the field strength E = V/d.
30IBP2H-031-01-04Field and potential of several chargesPaper 1AMedium1 mark
Three point charges +q, +q and −q are fixed at the corners of an equilateral triangle of side a. What is the electric potential energy of this system?
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Add the energy of each of the three pairs: k(q)(q)/a + k(q)(−q)/a + k(q)(−q)/a = −kq2/a. A ignores the signs of the charges. B gets the sign of the total wrong. C assumes that equal amounts of positive and negative charge give zero energy.
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Answer D
Answer: D
Common error: Count each pair once and keep the sign of each charge. Like charges give positive terms; opposite charges give negative terms.
31IBP2H-031-01-05Equipotential surfacesPaper 1AMedium1 mark
In the electrode region of a medical defibrillator test rig, the field is uniform. Two points P and Q lie on the same field line, 4.0 cm apart. The potential at P is 120 V and at Q is 40 V. What are the magnitude and direction of the electric field?
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The field magnitude is E = ΔV/Δx = (120 − 40)/0.040 = 2.0 × 103 V m−1. The field points from high potential to low potential, so it points from P to Q. B has the direction reversed. C forgets to convert 4.0 cm into metres. D uses the potential at a single point instead of the potential difference.
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Answer A
Answer: A
Exam tip: Field lines always point down the potential slope, from higher V to lower V.
32IBP2H-031-01-06Electric potential and workPaper 1AMedium1 mark
An electron starts from rest at a point where the electric potential is −200 V. It moves, under the electric force only, to a point where the potential is +300 V. What kinetic energy does the electron gain?
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The change in potential energy is ΔEp = qΔV = (−e)(+500 V) = −500 eV. The electron's potential energy falls by 500 eV, so its kinetic energy rises by 500 eV. A adds the magnitudes of the potentials instead of subtracting them. B uses only the final potential. D applies the rule for positive charges to a negative charge.
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Answer C
Answer: C
Common error: Negative charges speed up when they move towards higher potential. Always use ΔEp = qΔV with the sign of q included.
33IBP2S-030-01-08Equipotential surfacesPaper 1AMedium1 mark
Beneath a thundercloud, the electric potential increases with height, rising by 1.5×106 V for every 100 m gained. The field is approximately uniform. Which row gives the magnitude and direction of the electric field strength?
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The field strength is the negative of the potential gradient: E = −ΔV/Δx. Its magnitude is 1.5×106 V ÷ 100 m = 1.5×104 V m−1. The field points from high potential to low potential. The potential is higher further up, so the field points downwards. The distractors come from these mistakes: • 'Upwards' ignores the minus sign. • 1.5×108 V m−1 multiplies instead of dividing. • 1.5×106 V m−1 forgets to divide by the distance.
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Answer A
Answer: A
Exam tip: Electric field lines always point 'downhill' in potential, from higher V to lower V.
34IBP2H-031-00-02Charged conducting spherePaper 1AMedium1 mark
The dome of a Van de Graaff generator is an isolated conducting sphere of radius R carrying charge Q. Point P is inside the dome, a distance R/2 from its centre. Which row gives the electric field strength and the electric potential at P?
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All the charge sits on the outer surface of a conductor, and the field inside is zero. Because E = −ΔV/Δr = 0 inside, the potential does not change across the interior. It therefore equals the surface value kQ/R. Option A assumes zero field means zero potential. Option B treats the charge as a point charge at the centre with P outside it. Option D uses the result for a uniformly charged insulating sphere.
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Answer C
Answer: C
Exam tip: Zero field means constant potential, not zero potential. Inside a charged conductor, V equals its surface value.
35IBP2H-031-00-03Equipotential surfacesPaper 1AMedium1 mark
An engineer maps the equipotential surfaces above a charged circuit board. The potential difference between adjacent equipotential surfaces is always the same. Near point P the surfaces are 2.0 mm apart, and near point Q they are 6.0 mm apart. What is \(\frac{E_P}{E_Q}\), the ratio of the electric field strength at P to that at Q?
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The field strength is the potential gradient, E = ΔV/Δx. With equal ΔV, E is inversely proportional to the spacing, so EP/EQ = 6.0/2.0 = 3. Option A inverts the relationship. Option C wrongly applies an inverse-square law to the spacing. Option D assumes the field is the same everywhere on surfaces with equal ΔV.
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Answer B
Answer: B
Exam tip: Crowded equipotentials mean a strong field, just as crowded contour lines on a map mean a steep slope.
36IBP2H-031-00-04Electric field strengthPaper 1AMedium1 mark
In a test of an electrostatic crop sprayer, a charged droplet is held stationary between two horizontal parallel plates. The plates are a distance d apart and have a constant potential difference across them. The plate separation is then increased to 2d, with the potential difference unchanged. What is the motion of the droplet immediately afterwards?
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Initially qE = mg, with the electric force acting upwards. Since E = V/d, doubling d at constant V halves E. The electric force falls to mg/2 upwards. The resultant force is mg − mg/2 = mg/2 downwards, so the acceleration is g/2 downwards. Option A assumes the field does not depend on the plate separation. Option B assumes the electric force disappears completely. Option D wrongly applies an inverse-square law to a uniform field, which would give E/4 and a resultant of 3mg/4.
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Answer C
Answer: C
Common error: The field between parallel plates is uniform and equals V/d. It does not follow an inverse-square law.
37IBP2H-031-00-07Electric potential and workPaper 1AMedium1 mark
In a spacecraft ion thruster, a singly charged xenon ion moves from an equipotential surface at +250 V to one at −150 V. No other forces act on it. What is the increase in its kinetic energy?
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The potential difference is 250 − (−150) = 400 V. The work done by the field is qΔV = 1.6 × 10−19 × 400 = 6.4 × 10−17 J, and this equals the gain in kinetic energy. The path taken does not matter. Option A subtracts the magnitudes and ignores the sign. Option B uses only the starting potential. Option D omits the charge of the ion.
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Answer C
Answer: C
Common error: Keep the signs of the potentials. The change from +250 V to −150 V is 400 V, not 100 V.
38IBP2S-030-03-04Charged conducting spherePaper 1AMedium1 mark
The dome of an electrostatic generator is a conducting sphere of radius 0.20 m charged to a potential of 300 kV. What is the electric potential at a point inside the dome, 0.10 m from its centre?
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The field inside the conductor is zero. No work is therefore done moving a charge within it, so the potential is the same everywhere inside as on the surface: 300 kV. Option A confuses zero field with zero potential. Option B assumes the potential falls linearly towards the centre. Option D applies V ∝ 1/r inside the sphere.
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Answer C
Answer: C
Exam tip: Inside a charged conducting sphere, E = 0 but V = kQ/R. Zero field means constant potential, not zero potential.
39IBP2S-030-03-08Charged conducting spherePaper 1AMedium1 mark
Two isolated conducting spheres, of radii r and 2r, are far apart and joined by a long thin conducting wire. Charge is placed on the system. What is the ratio \(\frac{\text{field strength at surface of small sphere}}{\text{field strength at surface of large sphere}}\)?
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The connected spheres reach the same potential V = kQ/R, so Q ∝ R and the small sphere carries half the charge of the large one. The surface field is E = kQ/R2 = V/R, so E ∝ 1/R, giving a ratio of 2r/r = 2. Option A uses the charge ratio. Option B assumes equal fields. Option D uses E ∝ 1/R2 with equal charges.
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Answer C
Answer: C
Exam tip: For a conducting sphere, Esurface = V/R. At equal potential, the smaller sphere has the stronger surface field. This is why sharp points discharge.
40IBP2S-030-03-09Charged conducting spherePaper 1AMedium1 mark
Air breaks down and conducts when the electric field strength exceeds about 3.0×106 V m−1. What is the approximate maximum potential that a conducting sphere of radius 0.15 m can reach in air?
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For a conducting sphere, V = kQ/R and E = kQ/R2, so V = ER. This gives V = 3.0×106 × 0.15 = 4.5×105 V. Option D divides E by R. Option A multiplies E by R2. Option B uses ER/2.
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Answer C
Answer: C
Exam tip: V = ER at the surface of a sphere is a quick check that links field-strength and potential questions.
41IBP2H-031-02-03Electric potential and workPaper 1AMedium1 mark
In the retarding-grid stage of an electron energy analyser, an electron moves from a point where the electric potential is +300 V to a point where the electric potential is −200 V. What is the change in the electric potential energy of the electron?
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Use ΔEp = qΔV. Here ΔV = (−200) − (+300) = −500 V. So ΔEp = (−1.60×10−19)(−500) = +8.0×10−17 J, which is an increase. A negative charge moving to a lower potential gains potential energy. Option B ignores the negative sign of the electron's charge. Options C and D use a potential difference of 100 V instead of 500 V.
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Answer A
Answer: A
Common error: Always include the sign of the charge in ΔEp = qΔV. An electron gains potential energy when it moves towards a lower potential.
42IBP2H-031-02-04Magnetic field of a straight wirePaper 1AMedium1 mark
A cable for an electric-vehicle charger contains two long parallel conductors a distance d apart. Each carries a current I, 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). With opposite currents, the right-hand grip rule shows that both fields point the same way between the wires. They therefore add, giving 2μ0I/(πd). Option A is correct only for currents in the same direction. Option B uses distance d and only one wire. Option C counts only one wire.
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Answer D
Answer: D
Exam tip: Find the direction of each wire's field at the point before combining them. For opposite currents the fields add between the wires and largely cancel far away.
43IBP2H-031-02-05Field and potential of two point chargesPaper 1AMedium1 mark
A fixed sphere carries a charge of +5.0 μC. How much work must be done to bring a small sphere carrying +2.0 μC from infinity to a point 0.30 m from the centre of the fixed sphere? (Coulomb constant k ≈ 9×109 N m2 C−2)
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The work done equals the gain in electric potential energy, which is qV = kQq/r. This gives (9×109)(5.0×10−6)(2.0×10−6)/0.30 = 0.09/0.30 = 0.30 J. Option D divides by r2, confusing potential energy with force. Option B halves the correct value without reason. Option A assumes that no work is needed because the start is at infinity.
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Answer C
Answer: C
Common error: Potential and potential energy go as 1/r. Field strength and force go as 1/r². Check which one the question asks for.
44IBP2H-031-02-06Conservation of chargePaper 1AMedium1 mark
Three identical metal spheres, P, Q and R, sit on insulating stands. Initially P has +12 nC, Q has −4 nC and R is uncharged. P is touched to Q and separated. Then Q is touched to R and separated. What are the final charges?
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When P and Q touch, the total (+12 − 4) = +8 nC is shared equally, giving +4 nC each. When Q and R touch, Q's +4 nC is shared equally, giving +2 nC each. The final total is 4 + 2 + 2 = +8 nC, so charge is conserved. Option B shares the charge as if all three spheres touched at once. Option C gives a total of 12 nC, so charge is created. Option D assumes Q passes all its charge to R.
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Answer A
Answer: A
Exam tip: After each contact, check that the total charge is unchanged. Identical conductors share their combined charge equally.
45IBP2S-030-00-10Electric field and potentialPaper 1AMedium1 mark
A point charge +Q and a point charge −Q are fixed a short distance apart, forming a model of a polar molecule. Which describes the electric potential V and the electric field strength E at the midpoint between them?
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Potential is a scalar. The contributions +kQ/r and −kQ/r cancel, so V = 0. Field strength is a vector. At the midpoint, the field from +Q points away from +Q, and the field from −Q points towards −Q. Both point the same way, towards −Q, so they add and E ≠ 0. Option A treats both quantities as scalars. Option C reverses the roles of the two quantities.
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Answer B
Answer: B
Common error: Zero potential does not mean zero field. The field depends on how the potential changes with position, not on its value.
46IBP3H-D2-20r8-02Magnetic field of a straight wirePaper 1AMedium1 mark
Two long straight parallel wires in a telecoms cable duct are 0.10 m apart and carry currents of 3.0 A in opposite directions. The permeability of free space is μ0 = 4π × 10−7 N A−2. What is the magnitude of the resultant magnetic field at the point midway between the wires?
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Each wire gives B = μ0I/(2πr) = 2×10−7 × 3.0/0.050 = 1.2 × 10−5 T. With opposite currents the two fields point the same way at the midpoint, so they add: 2.4 × 10−5 T (A). B assumes the fields cancel, which happens for currents in the same direction. C uses a misremembered formula with a missing factor 2. D counts only one wire.
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Answer A
Answer: A
Exam tip: Use the right-hand grip rule for each wire separately; the fields add only if they point the same way at the point in question.
47IBP3H-D2-20r8-03Electric field and potentialPaper 1AMedium1 mark
Two fixed point charges of equal magnitude, +Q and −Q, are a distance d apart. Which statement is correct for the point midway between them?
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The potentials +kQ/(d/2) and −kQ/(d/2) are scalars and cancel, so V = 0. The fields are vectors: +Q pushes away from itself and −Q pulls towards itself, both towards −Q, so they add (D). A treats both quantities as cancelling. B and C claim a non-zero potential, which ignores the opposite signs.
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Answer D
Answer: D
Common error: Potential is a scalar and field strength is a vector; cancellation of one does not imply cancellation of the other.
48IBP3H-D2-20r8-04Electric potential energy of a system of chargesPaper 1AMedium1 mark
Two fixed point charges +q are a distance a apart. How much work must be done by an external agent to bring a third charge +q from infinity to the third vertex of the equilateral triangle of side a?
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The potential at the vertex due to the two fixed charges is 2kq/a, so the work is qΔV = 2kq2/a (A). B counts only one pair. C is the total electric potential energy of the three-charge system, not the work for the third charge. D halves the single-pair energy.
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Answer A
Answer: A
Exam tip: Work to add one charge to existing charges is the charge times the potential already present at its final position.
49IBP3H-D2-20r8-06Field and potential of several chargesPaper 1AMedium1 mark
Two point charges, +4Q and +Q, are fixed 3.0 m apart in a telecoms test range. At what distance from the +Q charge, on the line between them, is the resultant electric field zero?
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Fields cancel when 4kQ/x12 = kQ/x22, so x1 = 2x2. With x1 + x2 = 3.0 m, x2 = 1.0 m (C). A takes the ratio of distances as 4 instead of 2. B is the midpoint. D is the distance from the +4Q charge.
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Answer C
Answer: C
Exam tip: Set the two field magnitudes equal and take square roots; the distance ratio is the square root of the charge ratio.
50IBP3H-D2-20r11-03Field and potential of several chargesPaper 1AMedium1 mark
Two small fixed spheres carry charges +3.0 nC and +1.0 nC. Their centres are 0.40 m apart in air. What is the electric field strength at the midpoint of the line joining them, and in which direction does the field point? (Use k = 8.99×109 N m2 C−2.)
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At the midpoint each charge is 0.20 m away. The 3.0 nC charge pushes a positive test charge towards the 1.0 nC side more strongly than the 1.0 nC charge pushes it back, so E = k(3.0−1.0)×10−9/0.202 ≈ 450 N C−1 towards the 1.0 nC charge. A has the wrong direction. B adds the magnitudes instead of subtracting. D uses the full 0.40 m separation instead of 0.20 m.
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Answer C
Answer: C
Common error: Fields are vectors: for like charges the contributions at a point between them point in opposite directions and subtract.
51IBP3H-D2-20r11-05Charged conducting spherePaper 1AMedium1 mark
An isolated conducting sphere of radius 0.20 m carries a positive charge. The electric potential at its surface is +900 V, with zero potential at infinity. What is the electric potential at a point 0.60 m from the centre of the sphere?
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Outside the sphere V = kQ/r, so V ∝ 1/r. V = 900 × 0.20/0.60 = 300 V. A uses V ∝ 1/r2 (that is the field). B uses r = 0.40 m, the distance from the surface. D uses V ∝ r.
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Answer C
Answer: C
Common error: Potential falls as 1/r but field strength falls as 1/r²; do not mix them up.
52IBP3H-D2-20r11-07Equipotential surfacesPaper 1AMedium1 mark
Near the ground, the atmosphere has a uniform electric field of strength 130 V m−1 directed vertically downwards. A cyclist's head is 1.8 m above the ground. Which statement about the potential difference between the cyclist's head and the ground is correct?
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Field lines point from high to low potential, so with the field pointing downwards the head is at the higher potential. For a uniform field ΔV = Ed = 130 × 1.8 = 234 V. A divides instead of multiplying. B reverses the direction of potential change. C gives the field strength rather than a potential difference.
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Answer D
Answer: D
Common error: The electric field points towards decreasing potential.
53D-1A-12Electric potential and workPaper 1AHard1 mark
The graph shows how the electric potential V varies with distance x along a straight line in a region of electric field. An electron at x = 0 is to move along the line to x = 6.0 cm. Only the electric force acts on the electron.
What is the minimum kinetic energy that the electron must have at x = 0 to reach x = 6.0 cm?
Variation of electric potential with distance along a line.Show mark scheme
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Step 1The electron has charge −e, so its electric potential energy is −eV: it is greatest where V is lowest.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2From x = 0 to x = 2.0 cm, V falls from +10 V to −20 V, so the electron's potential energy rises by e × 30 V = 30 eV.
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Step 3Beyond 2.0 cm V rises and the electron speeds up again, so it reaches 6.0 cm provided it gets past the minimum of V: minimum kinetic energy 30 eV.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis compares only the end points: V at 6.0 cm (+25 V) is higher than at x = 0, so the electron loses potential energy overall. It must first climb the potential-energy barrier at 2.0 cm.
BThis uses the potential difference between the end points, 25 V − 10 V, ignoring the minimum of V in between and the sign of the electron's charge.
CThis uses the potential at the minimum, −20 V, measured from zero instead of from the starting potential of +10 V.
DCorrect: the electron must climb through a potential difference of 30 V to reach the minimum of V.
54D-1A-15Charged conducting spherePaper 1AHard1 mark
A hollow conducting sphere of radius R carries a positive charge Q. Point X is inside the sphere, a distance R/2 from its centre. Point Y is outside the sphere, a distance 2R from its centre.
What is the work done by the electric field on a small positive charge q as it moves from X to Y?
Diagram not to scale
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Step 1Inside a charged conducting sphere the field strength is zero, so the potential is constant and equal to its value at the surface: VX = kQ/R.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Outside, the sphere acts as a point charge at its centre: VY = kQ/(2R).
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Step 3Work done by the field = q(VX − VY) = kQq/R − kQq/(2R) = kQq/(2R).
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis takes VX = 0 because the field strength inside the sphere is zero. Zero field means a constant potential, not a zero potential.
BCorrect: VX = kQ/R and VY = kQ/(2R).
CThis uses the correct VX but ignores the potential at Y, as if Y were at infinity.
DThis treats X as if it were outside a point charge at the centre: VX = kQ/(R/2) = 2kQ/R. Inside the conductor there is no field, so no work is done between X and the surface.
55D-1A-19Magnetic field of a straight wirePaper 1AHard1 mark
Two long straight parallel wires P and Q are perpendicular to the page and a distance d apart. P carries a current 2I out of the page and Q carries a current I into the page. The magnetic field strength at a distance r from a long straight wire carrying current I is proportional to I/r.
Points J, K, L and M lie on the line through the wires, as shown: K is midway between the wires, J is between the wires a distance d/3 from Q, L is a distance d beyond Q and M is a distance d beyond P.
At which point is the resultant magnetic field zero?
Two long parallel wires perpendicular to the page (⊙ current out of the page, ⊗ current into the page) and four points on the line through them.Show mark scheme
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Step 1Right-hand grip rule: between the wires the fields of the two opposite currents point the same way, so they add. The resultant cannot be zero at J or K.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Outside the pair the two fields are opposite. They can be equal in magnitude only on the side of the smaller current, Q.
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Step 3At L: P gives a field ∝ 2I/(2d) = I/d and Q gives a field ∝ I/d. These are equal and opposite, so the resultant is zero. At M, P gives ∝ 2I/d and Q only ∝ I/(2d), so they do not balance.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is where the fields would cancel if both currents were in the same direction (2I/(2d/3) = I/(d/3)). With opposite currents the fields add between the wires.
BThis carries over "zero at the midpoint" from two equal currents in the same direction. Here the currents are opposite and unequal, and at the midpoint the two fields add.
CM is on the side of the larger current, where the field of P (∝ 2I/d) is always larger than the field of Q (∝ I/2d).
DCorrect: at L the fields of P and Q are equal in magnitude and opposite in direction.
56IBP2H-031-01-07Field and potential of two point chargesPaper 1AHard1 mark
A point charge +4Q is fixed at x = 0 and a point charge −Q is fixed at x = d. At which position on the x-axis is the resultant electric field strength zero?
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Between the charges the two fields point the same way (towards −Q), so they cannot cancel there. The null point must lie outside the pair, nearer the smaller charge, so x > d. Setting 4kQ/x2 = kQ/(x − d)2 gives 2(x − d) = x, so x = 2d. A solves the equation for the region between the charges and ignores the field directions. B and C are the points where the potential, not the field, is zero.
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Answer D
Answer: D
Exam tip: For opposite charges, the field null point lies outside the pair, on the side of the smaller charge.
57IBP2H-031-01-08Charged conducting spherePaper 1AHard1 mark
A model shows why sharp points on a lightning conductor discharge first. Two isolated metal spheres of radii R and 2R are far apart and joined by a thin conducting wire. A total charge is shared between them. What is the ratio (field strength at the surface of the small sphere) / (field strength at the surface of the large sphere)?
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The joined spheres reach the same potential, so kQ1/R = kQ2/(2R). This gives Q2 = 2Q1. The surface field is E = kQ/r2 = V/r, so E ∝ 1/r at equal potential, and the small sphere has twice the surface field. A uses the ratio of the charges. B assumes equal potential means equal field. D assumes equal charges, giving E ∝ 1/r2.
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Answer C
Answer: C
Exam tip: Connected conductors share potential, not charge. At the same potential, a smaller radius means a stronger surface field.
58IBP2S-030-01-09Charged conducting spherePaper 1AHard1 mark
The dome of a science-museum Van de Graaff generator can be treated as a point charge at its centre for points outside it. At a point P outside the dome, the electric potential is 600 V and the electric field strength is 200 V m−1. How far is P from the centre of the dome?
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For a point charge, V = kQ/r and E = kQ/r2. Dividing the first by the second gives V/E = r, so r = 600/200 = 3.0 m. The distractors come from these mistakes: • 0.33 m inverts the ratio. • 1.7 m takes a square root, wrongly treating the ratio as r2. • 9.0 m squares the ratio.
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Answer A
Answer: A
Exam tip: When both V and E are given for a point charge, divide one by the other to eliminate kQ. You can then find r without knowing the charge.
59IBP2S-030-02-08Electric field and potentialPaper 1AHard1 mark
Two identical positive point charges +Q are fixed a distance d apart. Which correctly describes the electric field strength E and the electric potential V at the midpoint between them? (V is zero at infinity.)
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Field strength is a vector. At the midpoint, the two fields are equal in magnitude and opposite in direction, so they cancel and E = 0. Potential is a scalar. Each charge contributes +kQ/(d/2), and these add to V = 4kQ/d, which is not zero. The answer is therefore C. Options A, B and D all treat E and V as combining in the same way. The pattern in B is what happens for a +Q, −Q pair, not for two like charges.
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Answer C
Answer: C
Exam tip: Add fields as vectors and potentials as signed scalars. Zero field does not imply zero potential, and zero potential does not imply zero field.
60IBP2H-031-00-05Field and potential of several chargesPaper 1AHard1 mark
In a model of an ionised molecule, a point charge +4q is fixed a distance d from a point charge −q. On the straight line between the two charges, at what distance from the +4q charge is the electric potential zero?
Show mark scheme
Marking point
Mark
Notes
Potential is a scalar, so the two contributions simply add. Setting \(\frac{k(4q)}{x} - \frac{kq}{d-x} = 0\) gives 4(d − x) = x, so x = 4d/5. Option C comes from equating inverse-square field magnitudes instead of potentials. Option B assumes the midpoint. Option A measures the distance from the wrong charge.
✓ 1
Answer D
Answer: D
Common error: V depends on 1/r and carries the sign of the charge. Do not use field strengths (1/r²) when looking for a point of zero potential.
61IBP2H-031-00-06Electric potential energy of a system of chargesPaper 1AHard1 mark
In a medical isotope facility, a proton and an alpha particle are each accelerated from rest through the same potential difference in a uniform electric field. What is the ratio \(\frac{\text{speed of proton}}{\text{speed of alpha particle}}\)?
Show mark scheme
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Mark
Notes
The kinetic energy gained is ½mv2 = qV, so v = √(2qV/m). The proton has q/m = e/mp. The alpha particle has (2e)/(4mp), which is half of this. The speed ratio is therefore √2. Option C forgets the square root. Option A assumes the same potential difference gives the same speed. Option D inverts the ratio.
✓ 1
Answer B
Answer: B
Exam tip: Write ½mv² = qV first, then compare q/m for the two particles. Speed depends on √(q/m).
62IBP2H-031-02-07Field and potential of several chargesPaper 1AHard1 mark
A point charge +Q is fixed at a distance d from a point charge −2Q. At which point on the straight line between the two charges is the electric potential zero?
Show mark scheme
Marking point
Mark
Notes
Potential is a scalar, so the two contributions add algebraically. At distance x from +Q: kQ/x − 2kQ/(d − x) = 0. This gives d − x = 2x, so x = d/3. The zero lies closer to the smaller charge. Option B assumes equal magnitudes. Option C places the point nearer the larger charge. Option D uses an inverse-square relation, which is the condition for zero field strength, not zero potential.
✓ 1
Answer A
Answer: A
Common error: Zero potential and zero field strength are different conditions. Potential uses 1/r with signs. Field strength uses 1/r² and vector directions.
63IBP2H-031-02-08Magnetic field patternsPaper 1AHard1 mark
A long vertical wire carries a steady current of 10 A upwards. The horizontal component of the Earth's magnetic field is 2.0×10−5 T, directed north. Ignore the vertical component. At which point in a horizontal plane is the resultant horizontal magnetic field zero?
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Notes
The neutral point is where the wire's field is equal and opposite to the Earth's field. Its distance is r = μ0I/(2πB) = (2×10−7×10)/(2.0×10−5) = 0.10 m. Viewed from above, an upward current produces an anticlockwise field. West of the wire this field points south, so it cancels the Earth's northward field. East of the wire it points north and adds to the Earth's field (A). North and south of the wire the wire's field is east–west, so it cannot cancel a north–south field (C, D).
✓ 1
Answer B
Answer: B
Exam tip: The wire's field is always tangential, never radial. Use the right-hand grip rule to find where it opposes the external field.
64IBP2S-030-04-10Electric field linesPaper 1AHard1 mark
In fair weather, the electric field near the Earth's surface is about 130 N C−1, directed vertically downwards. Which statement is correct?
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Notes
The field lines point down towards the ground and end there, so the surface carries negative charge. Field lines point in the direction of decreasing potential. Potential therefore falls going downwards, which means it increases with height. The equipotential surfaces are horizontal, perpendicular to the field lines.
✓ 1
Answer D
Answer: D
Common error: A negatively charged surface does not mean the potential becomes more negative as you move away from it. Follow the field direction: field lines always point from high potential to low potential.
65IBP3H-D2-20r8-07Electric potential and workPaper 1AHard1 mark
A particle of mass m and charge +q is released from rest at a distance r from a fixed point charge +Q. What is its speed when its distance from +Q has increased to 2r? (Ignore gravity.)
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Notes
Energy conservation: ½mv2 = kQq(1/r − 1/(2r)) = kQq/(2r), so v = √(kQq/(mr)) (D). A omits the ½ in the kinetic energy. B uses the full initial energy kQq/r, ignoring the energy remaining at 2r. C adds the two potential-energy terms instead of subtracting.
✓ 1
Answer D
Answer: D
Common error: The kinetic energy gained is the decrease in potential energy between the two positions, not the potential energy at the start.
66IBP3H-D2-20r8-08Field and potential of several chargesPaper 1AHard1 mark
A charge of +2q is fixed at x = 0 and a charge of −q is fixed at x = 3.0 m. At which positions on the x-axis is the electric potential zero (taking V = 0 at infinity)?
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Notes
Between the charges: 2/x = 1/(3 − x) gives x = 2.0 m. Beyond the −q charge: 2/x = 1/(x − 3) gives x = 6.0 m (C). Nothing works for x < 0 because the +2q charge is both larger and closer there. A uses the midpoint, which is only correct for equal and opposite charges. B misses the second point. D has a sign error on the second point.
✓ 1
Answer C
Answer: C
Exam tip: Potential is a scalar, so zero-potential points can occur on both sides of the weaker opposite charge; test each region of the axis.
67IBP3H-D2-20r11-08Conservation of chargePaper 1AHard1 mark
Two identical small conducting spheres carry charges +6.0 nC and −2.0 nC. Their centres are a fixed distance apart, and they attract each other with a force of magnitude F. The spheres are touched together and returned to their original separation. How does the force between them now compare with F?
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Notes
Total charge is conserved: +4.0 nC shared equally gives +2.0 nC on each sphere. Before, F ∝ 6.0 × 2.0 = 12; after, F' ∝ 2.0 × 2.0 = 4, so F' = F/3, and like charges repel. A keeps the attraction. B shares the magnitudes (4.0 nC each) instead of the net charge. D assumes the charges cancel.
✓ 1
Answer C
Answer: C
Common error: When identical conductors touch, the net charge (with signs) is shared equally, not the magnitudes.
68D-1B-07Millikan's experimentPaper 1BMedium7 marks
In a Millikan-type experiment, oil drops of density 875 kg m−3 are viewed between two horizontal metal plates 5.00 mm apart. With no potential difference between the plates, drop A falls at constant (terminal) speed. A student times it with a hand-held stopwatch over 1.00 mm of the eyepiece scale and records 8.35 s.
A potential difference V is then applied and adjusted until drop A is stationary: V = 457 V. The viscosity of air is 1.82 × 10−5 Pa s, and the buoyancy force of the air on a drop may be neglected. The charges on four other drops, B to E, were found in the same way.
Drop
B
C
D
E
q / 10−19 C
3.19
7.93
6.42
9.50
Diagram not to scale
(a)
Drop A.
(i)
Show that the radius of drop A is about 1.1 × 10−6 m.
(2)
(ii)
Determine the charge on drop A.
(2)
(b)
Evaluation.
(i)
Millikan's hypothesis is that every charge is a whole-number multiple of e = 1.60 × 10−19 C. Evaluate this hypothesis using drop A and at least two other drops.
(2)
(ii)
Suggest how the student could reduce the percentage uncertainty in the speed of drop A, and hence in its radius, while still using the same stopwatch.
(1)
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Notes
Part (a)(i)
At terminal speed the drag balances the weight: 6πηrv = (4/3)πr³ρg, with v = 1.00 × 10−3/8.35 = 1.20 × 10−4 m s−1
Accept 4.7–4.9 × 10⁻¹⁹ C; using 1.1 × 10⁻⁶ m gives 5.2 × 10⁻¹⁹ C (allow ECF).
Part (b)(i)
q/e for at least three drops: A 3.00, B 1.99, C 4.96, D 4.01, E 5.94
✓ 1
Allow ECF from (a)(ii). Award only if drop A and at least two other drops are used.
Every ratio is within about 1 % of a whole number (3, 2, 5, 4, 6), so the data support the hypothesis
✓ 1
The conclusion must be consistent with the candidate's ratios; with 5.2 × 10⁻¹⁹ C (3.3 e) for A, "A does not fit; the radius was rounded" is acceptable.
Part (b)(ii)
Time the drop over a longer distance (several millimetres of the scale): the reaction-time uncertainty stays the same, so it is a smaller percentage of the longer time; r ∝ √v, so the percentage uncertainty in r is half that in v
✓ 1
Accept: time several successive falls of the same drop and use the mean. Do not accept "use light gates" (a different instrument).
Answers: (a)(ii) 4.8 × 10−19 C (the remaining parts are explanations — see the table above)
69D-1B-15Potential around a charged spherePaper 1BMedium7 marks
A hollow aluminium sphere of radius 0.060 m stands on an insulating base and carries a charge Q. A student measures the electric potential V at distances r from the centre of the sphere using a flame probe connected to an electrostatic voltmeter. Before taking the readings, the student touches the probe against an earthed metal plate: the voltmeter then reads −45 V.
The student's hypothesis is that V is inversely proportional to r, as it would be for a point charge at the centre of the sphere. The uncorrected voltmeter readings are shown. k = 8.99 × 109 N m² C−2.
r / m
0.080
0.100
0.150
0.200
0.250
0.300
Voltmeter reading / V
3330
2650
1760
1300
1040
850
Diagram not to scale
(a)
Testing the hypothesis.
(i)
Determine the potential at r = 0.100 m, corrected for the zero error.
(1)
(ii)
Test the hypothesis, using at least three corrected readings.
(2)
(b)
The charge and its limit.
(i)
Determine Q.
(2)
(ii)
The air next to the sphere becomes conducting, and the sphere discharges, when the electric field strength reaches 3.0 × 106 V m−1. Determine the field strength at the surface of the sphere and hence the factor by which Q could be increased before the sphere discharges.
(2)
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Notes
Part (a)(i)
V = 2650 − (−45) = 2695 V ≈ 2.70 × 103 V
✓ 1
Subtracting 45 V (2605 V) scores 0.
Part (a)(ii)
Corrected Vr for at least three readings, e.g. 3375 × 0.080 = 270, 2695 × 0.100 = 270, 895 × 0.300 = 269 (V m)
✓ 1
Allow ECF from (a)(i). Uncorrected readings give Vr falling from 266 to 255 V m: [1 max] for the part with a consistent conclusion.
Vr is constant to within about 1 %, so V ∝ 1/r and the hypothesis is supported
✓ 1
Part (b)(i)
V = kQ/r, so Q = Vr/k with Vr = 270 V m (mean)
✓ 1
Q = 270/8.99 × 109 = 3.0 × 10−8 C
✓ 1
Allow ECF from (a)(ii). Accept 2.95–3.05 × 10⁻⁸ C.
Part (b)(ii)
Outside the sphere the field is that of a point charge at its centre: E = kQ/R² = 8.99 × 109 × 3.0 × 10−8/0.060² = 7.5 × 104 V m−1
✓ 1
Allow ECF from (b)(i).
Factor = 3.0 × 106/7.5 × 104 = 40
✓ 1
Allow ECF. Accept 39–41.
Answers: (a)(i) 2.70 × 103 V · (a)(ii) Vr ≈ 270 V m (constant) · (b)(i) 3.0 × 10−8 C · (b)(ii) 7.5 × 104 V m−1; 40 (the remaining parts are explanations — see the table above)
70D-1B-16Mapping a uniform fieldPaper 1BMedium7 marks
Two long, straight, parallel copper strips are painted on a sheet of conducting paper 8.0 cm apart and connected to a 6.00 V supply; the negative strip is at 0 V. A probe connected to a voltmeter is used to measure the potential V on the paper at distances x from the negative strip, along a line halfway between the ends of the strips.
The graph shows the data with the line of best fit, extended to both strips.
x / cm
1.0
2.0
3.0
4.0
5.0
6.0
7.0
V / V
0.92
1.57
2.31
3.03
3.68
4.42
5.09
Graph drawn to scale.
(a)
The field.
(i)
Determine the electric field strength between the strips.
(2)
(ii)
Calculate the percentage difference between your answer to (a)(i) and the value predicted by E = V/d.
(1)
(b)
Explaining the difference.
(i)
Use the graph to explain the difference in (a)(ii).
(2)
(ii)
The two contacts and the paper are in series with the supply. Estimate the ratio (resistance of one contact)/(resistance of the paper between the strips).
(2)
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Notes
Part (a)(i)
Large triangle on the line, e.g. (0, 0.21) to (8.0, 5.80): gradient = 5.59/8.0 = 0.70 V cm−1
✓ 1
E = 70 V m−1
✓ 1
Accept 68–72 V m⁻¹. The conversion to V m⁻¹ is required for this mark.
Part (a)(ii)
E = 6.00/0.080 = 75 V m−1; difference = (75 − 70)/75 × 100 = 7 %
✓ 1
Allow ECF from (a)(i). Accept 6–8 %.
Part (b)(i)
The line meets the V axis at about 0.21 V, not 0, and reaches only about 5.8 V at x = 8.0 cm
✓ 1
Accept 0.18–0.24 V and 5.75–5.85 V.
About 0.2 V is lost across the contact between each strip and the paper (contact resistance), so only about 5.6 V is across the 8.0 cm of paper: 5.6/0.080 = 70 V m−1, less than 75 V m−1
✓ 1
Allow ECF from (a)(i). "Voltmeter zero error" explains the intercept but not the smaller gradient: do not award the second mark for it.
Part (b)(ii)
The current is the same in components in series, so their resistances are in the ratio of the potential differences across them
✓ 1
Ratio ≈ 0.2/5.6 ≈ 0.04
✓ 1
Allow ECF from (b)(i). Accept 0.03–0.05.
Answers: (a)(i) 70 V m−1 · (a)(ii) 7 % · (b)(ii) ≈ 0.04 (the remaining parts are explanations — see the table above)
71D-1B-17Magnetic field of a circular coilPaper 1BMedium6 marks
A flat circular coil of N turns is sealed inside a plastic housing, so its radius R cannot be measured directly. The coil carries a constant current of 1.50 A. A magnetic field sensor measures the field strength B at distances x from the centre of the coil along its axis.
The field on the axis is given by B = μ0NIR²/[2(R² + x²)3/2], and at the centre by B0 = μ0NI/(2R). These can be combined to give (B0/B)2/3 = 1 + x²/R². The graph shows six of the seven points.
x / m
0.000
0.040
0.060
0.080
0.100
0.120
0.140
B / mT
0.471
0.341
0.237
0.170
0.112
0.082
0.057
x² / 10−3 m²
0
1.6
3.6
6.4
10.0
14.4
19.6
(B0/B)2/3
1.000
1.240
1.581
1.973
2.605
4.087
Graph drawn to scale.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient, including its unit.
(2)
(b)
The coil.
(i)
Hence determine R.
(1)
(ii)
Determine N.
(2)
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Notes
Part (a)(i)
(0.471/0.082)2/3 = 3.21
✓ 1
Accept 3.2.
Part (a)(ii)
Point plotted correctly; gradient from a large triangle, e.g. (0, 1.00) to (20.0 × 10−3, 4.13): 3.13/0.0200
72D-1B-22Uniform electric fieldsPaper 1BMedium7 marks
A small insulated sphere carrying a fixed charge q hangs between two large, vertical, parallel metal plates. The potential difference between the plates is kept at 4.00 kV. A force sensor measures the horizontal electric force F on the sphere. The plates are moved apart in steps, keeping the sphere midway between them, and for each position the student records the reading ds of a scale fixed to one of the plates.
The student expects that F = qV/d, where d is the separation of the plates, and plots a graph of 1/F against ds. The graph shows five of the six points.
ds / cm
F / mN
(1/F) / mN−1
2.0
0.405
2.47
3.0
0.283
3.53
4.0
0.224
4.46
5.0
0.181
6.0
0.155
6.45
7.0
0.133
7.52
Graph drawn to scale. The dashed line marks ds = 0.
(a)
The graph.
(i)
Complete the table.
(1)
(ii)
Plot the missing point, draw the line of best fit and determine its gradient.
(2)
(b)
The charge and the scale.
(i)
Determine q.
(2)
(ii)
The line does not pass through the origin. Determine the intercept of the line on the ds axis and deduce what this shows about the readings ds.
(2)
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Notes
Part (a)(i)
1/0.181 = 5.52 mN−1
✓ 1
Accept 5.5 mN⁻¹.
Part (a)(ii)
Point plotted at (5.0, 5.52) and a single straight line of best fit drawn
✓ 1
Allow ECF from (a)(i).
Gradient = 1.00 mN−1 cm−1
✓ 1
Accept 0.96–1.04 mN⁻¹ cm⁻¹.
Part (b)(i)
1/F = d/(qV), so the gradient is 1/(qV) = 1.00 mN−1 cm−1 = 1.00 × 105 N−1 m−1
✓ 1
Allow ECF from (a)(ii). The conversion 1 mN⁻¹ cm⁻¹ = 10⁵ N⁻¹ m⁻¹ is required for this mark.
q = 1/(1.00 × 105 × 4.00 × 103) = 2.5 × 10−9 C
✓ 1
Accept 2.4–2.6 × 10⁻⁹ C.
Part (b)(ii)
Line extrapolated: intercept on the ds axis = −0.5 cm
✓ 1
Allow ECF from the candidate's line. Accept −0.4 to −0.6 cm.
1/F should be zero when the true separation is zero, so d = ds + 0.5 cm: every reading is about 0.5 cm less than the true separation (a systematic, zero-offset error that does not affect the gradient or q)
✓ 1
Accept: the scale zero is about 0.5 cm away from the surface of the plate. Do not accept "random error" or "parallax".
Answers: (a)(i) 5.52 mN−1 · (a)(ii) 1.00 mN−1 cm−1 · (b)(i) 2.5 × 10−9 C · (b)(ii) −0.5 cm (the remaining parts are explanations — see the table above)
Two identical small conducting spheres, each of radius 0.50 cm, are given equal charges. One sphere is fixed on an insulating stand; the other is mounted on an insulating rod attached to a digital force sensor, which measures the electrostatic force F between the spheres. The separation r of their centres is set with a scale, and its uncertainty is negligible. Each value of F has an absolute uncertainty of ±0.10 mN.
To test whether F ∝ rn with n = −2, the student plots lg(F / mN) against lg(r / cm). The graph shows five of the six points with their error bars.
r / cm
F / mN
lg(r / cm)
lg(F / mN)
2.0
13.24
0.301
1.122
3.0
6.11
0.477
0.786
4.0
3.42
0.602
0.534
5.0
2.19
0.699
6.0
1.58
0.778
0.199
8.0
0.89
0.903
−0.051
Graph drawn to scale. Error bars show ±0.10 mN.
(a)
The line of best fit.
(i)
Complete the table.
(1)
(ii)
Plot the missing point and draw the line of best fit. Determine the gradient of the line.
(2)
(b)
Testing the law.
(i)
Draw the lines of maximum and minimum gradient consistent with the error bars. Hence state n with its uncertainty.
(2)
(ii)
Discuss whether these data support Coulomb's law.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
lg 2.19 = 0.340
✓ 1
Accept 0.34.
Part (a)(ii)
Point plotted at (0.699, 0.340) and a single straight line of best fit drawn through all six points
✓ 1
Allow ECF from (a)(i) for the plotted point.
Gradient = −1.95, from a large triangle
✓ 1
Accept −1.92 to −1.98. A positive value scores 0.
Part (b)(i)
Steepest line ≈ −1.97 and least steep line ≈ −1.91, each passing through every error bar (the bars at small r are very short, so both lines must pass almost exactly through the first two points)
✓ 1
Accept −1.96 to −1.99 and −1.89 to −1.92. Lines that miss any error bar (e.g. lines joining opposite ends of the bars at r = 2.0 cm and 8.0 cm) score 0 for this mark.
n = −1.95 ± 0.03
✓ 1
Uncertainty = half the difference of the two gradients. Allow ECF from (a)(ii) and from the candidate's lines. Accept ±0.02 to ±0.05; the uncertainty to 1 s.f. and the value to the same decimal place.
Part (b)(ii)
−2 lies outside the range found in (b)(i) (about −1.97 to −1.91), so the data do not follow an exact inverse-square law for these spheres
✓ 1
The conclusion must agree with the candidate's own range (ECF from (b)(i)): a candidate whose range includes −2 may conclude that the data are consistent with an inverse-square law.
This does not disprove Coulomb's law, which applies to point charges: at small separations the like charges repel each other to the far sides of the spheres, so the effective separation is larger than r, the force at small r is less than the point-charge value and the graph is less steep than −2
✓ 1
Accept: the spheres cannot be treated as point charges when r is only a few radii. Do not accept "human error" or "air resistance".
Answers: (a)(i) 0.340 · (a)(ii) −1.95 · (b)(i) −1.95 ± 0.03 (the remaining parts are explanations — see the table above)
74IBP2H-031-01-09Charged sphere in a uniform fieldPaper 2Easy5 marks
An electrostatic precipitator removes soot from the flue gas of a power station. Two vertical parallel metal plates are 0.25 m apart and have a potential difference of 40 kV between them. A soot particle carries a charge of −3.2 × 10−17 C.
Diagram not to scale
(a)
Calculate the electric field strength between the plates.
(2)
(b)
State the direction of the electric force on the soot particle.
(1)
(c)
Calculate the magnitude of the electric force on the soot particle.
(2)
Show mark scheme
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Mark
Notes
Part (a)
E = V/d = 40 × 103/0.25
✓ 1
E = 1.6 × 105 V m−1 (accept N C−1)
✓ 1
Part (b)
Towards the positive plate, i.e. opposite to the field direction
✓ 1
Part (c)
F = qE = 3.2 × 10−17 × 1.6 × 105 (allow ECF from the field strength calculated in the first part)
✓ 1
F = 5.1 × 10−12 N
✓ 1
Exam tip: Convert kV to V before calculating. The field between parallel plates is uniform, so the force on the particle is the same everywhere between them.
75IBP2H-031-00-09Electrostatic powder coatingPaper 2Easy4 marks
An electrostatic precipitator removes soot from the flue gas of a power station. Gas flows between two large parallel metal plates 0.15 m apart. The potential difference between the plates is 30 kV.
Diagram not to scale
(a)
Calculate the electric field strength between the plates.
(2)
(b)
A soot particle carries a charge of 8.0 × 10−17 C. Calculate the electric force on the particle.
(1)
(c)
Outline why the force on the particle is the same wherever it is between the plates, away from their edges.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses E = V/d = 30 × 103/0.15
✓ 1
E = 2.0 × 105 V m−1
✓ 1
Part (b)
F = qE = 8.0 × 10−17 × 2.0 × 105 = 1.6 × 10−11 N (ECF from (a))
✓ 1
Part (c)
The field between the plates is uniform (same magnitude and direction everywhere), so F = qE is constant
✓ 1
Exam tip: Convert kV to V before substituting. A factor of 10³ error here carries through every later part.
76IBP2H-031-02-09Electric fields in the atmospherePaper 2Easy6 marks
A child slides down a plastic playground slide. Afterwards the child carries a charge of −45 nC. Treat the child and the slide as an isolated system that was initially uncharged. The elementary charge is e = 1.60×10−19 C.
(a)
State the principle of conservation of charge.
(1)
(b)
State the charge now on the slide.
(1)
(c)
Calculate the number of electrons transferred, and state the direction of transfer.
(2)
(d)
The child's hair stands on end. Outline why.
(2)
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Notes
Part (a)
The total (net) charge of an isolated system remains constant
✓ 1
Part (b)
+45 nC
✓ 1
Part (c)
n = Q/e = 45×10−9/1.60×10−19
✓ 1
n = 2.8×1011 electrons, transferred from the slide to the child
✓ 1
Part (d)
Each hair gains charge of the same sign (negative)
✓ 1
Like charges repel, so the hairs push away from each other and from the head
✓ 1
Common error: A negative charge means the object gained electrons. Say that electrons moved, not that positive charge moved.
78IBP3H-D2-20r8-10Magnetic field patternsPaper 2Easy5 marks
A high-voltage cable crossing a mountain pass carries a steady current of 270 A, flowing from west to east. A walker holds a compass 6.0 m directly below the cable. Use μ0 = 4π × 10−7 N A−2. Treat the cable as a long straight wire.
Cable and compass; not to scale
(a)
Calculate the magnitude of the magnetic field of the cable at the compass.
(2)
(b)
The horizontal component of the Earth's field is 2.0 × 10−5 T. Determine the distance below the cable at which the cable's field equals this value.
(2)
(c)
State the direction of the cable's field at the compass 6.0 m below the cable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses B = μ0I/(2πr)
✓ 1
B = 9.0 × 10−6 T
✓ 1
Part (b)
Rearranges r = μ0I/(2πB)
✓ 1
r = 2.7 m
✓ 1
Part (c)
North (right-hand grip rule: thumb east, field below the wire points north)
✓ 1
Exam tip: Check the direction with the right-hand grip rule: thumb along the current, fingers curl in the direction of the field lines.
79D-2-07Electric fields in the atmospherePaper 2Medium12 marks
During a thunderstorm the base of a cloud carries a large negative charge, and the ground below it becomes positively charged. The base of the cloud is 1.2 km above the ground, and the region between them may be modelled as the uniform field between two parallel plates. The electric field strength in this region is 5.0 × 104 V m−1.
Water droplets are spheres of density 1000 kg m−3. Viscosity of air = 1.8 × 10−5 Pa s; density of air = 1.2 kg m−3.
Diagram not to scale
(a)
The field below the cloud.
(i)
Outline how the ground below the cloud becomes positively charged.
(2)
(ii)
Calculate the potential difference between the base of the cloud and the ground.
(1)
(b)
A droplet of radius 10 μm in this region carries a positive charge of 4.8 × 10−16 C.
(i)
Calculate the number of electrons that the droplet has lost.
(1)
(ii)
Determine the magnitude and the direction of the electric force on the droplet.
(2)
(iii)
Determine the terminal speed of the droplet.
(3)
(c)
A lightning flash transfers a charge of 20 C between the cloud and the ground.
(i)
Estimate the energy transferred by the flash.
(2)
(ii)
State the assumption made in your estimate.
(1)
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Notes
Part (a)(i)
The negative charge on the cloud repels free electrons in the ground (electrostatic induction)
✓ 1
The electrons move away through the conducting ground to the rest of the Earth (earthing), leaving the surface below the cloud with a net positive charge
✓ 1
Do not accept "positive charges move towards the cloud".
Part (a)(ii)
V = Ed = 5.0 × 104 × 1200 = 6.0 × 107 V
✓ 1
Part (b)(i)
N = 4.8 × 10−16/1.60 × 10−19 = 3.0 × 103
✓ 1
Part (b)(ii)
F = qE = 4.8 × 10−16 × 5.0 × 104 = 2.4 × 10−11 N
✓ 1
Upwards: the field points from the positive ground to the negative cloud, and the force on a positive charge is along the field
Accept 4.9–5.2 × 10⁻³ m s⁻¹. Award [2 max] for 1.2 × 10⁻² m s⁻¹ (electric force ignored) or 1.9 × 10⁻² m s⁻¹ (forces added).
Part (c)(i)
W = QV with V from (a)(ii)
✓ 1
Allow ECF from (a)(ii).
W = 20 × 6.0 × 107 = 1.2 × 109 J
✓ 1
ALT: if the potential difference is taken to fall to zero during the flash, ½QV = 6 × 10⁸ J; accept either value when it is consistent with the assumption in (c)(ii).
Part (c)(ii)
The potential difference stays constant during the flash (the charge transferred is a small fraction of the charge on the cloud) / the whole charge moves through the full potential difference
✓ 1
Must be consistent with the value in (c)(i).
Answers: (a)(ii) 6.0 × 107 V · (b)(i) 3.0 × 103 · (b)(ii) 2.4 × 10−11 N upwards · (b)(iii) 5.0 × 10−3 m s−1 · (c)(i) 1.2 × 109 J (the remaining parts are explanations — see the table above)
80D-2-15Field and potential of two point chargesPaper 2Medium11 marks
Two point charges, A of +4.00 μC and B of −2.00 μC, are fixed 0.300 m apart. Point P is 0.250 m from each charge, and M is the midpoint of AB, as shown.
Not to scale. M is the midpoint of AB. (k = 8.99 × 10⁹ N m² C⁻²)
(a)
The field at P.
(i)
Calculate the electric field strength at P due to charge A alone, and state its direction.
(1)
(ii)
Show that the resultant electric field strength at P is about 5.7 × 105 N C−1.
(2)
(iii)
Determine the angle between the resultant field at P and the line AB.
(1)
(b)
Potential and work.
(i)
Calculate the electric potential at P.
(1)
(ii)
Determine the work done by the electric field when a charge of +1.0 nC moves from P to M.
(2)
(c)
Zero potential.
(i)
Determine the position on the line AB, between the charges, at which the electric potential is zero.
(2)
(ii)
Explain why the electric field strength is not zero at this point, and state, with a reason, on which side of B the field strength is zero.
(2)
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Notes
Part (a)(i)
EA = kq/r² = 8.99 × 109 × 4.00 × 10−6/0.250² = 5.75 × 105 N C−1, directed away from A along AP
✓ 1
The direction is required for the mark.
Part (a)(ii)
EB = 2.88 × 105 N C−1 towards B; with cos θ = 0.600 and sin θ = 0.800, the components parallel to AB are (5.75 + 2.88) × 0.600 = 5.18 × 105 and perpendicular to AB (5.75 − 2.88) × 0.800 = 2.30 × 105
✓ 1
Allow ECF from (a)(i).
E = √(5.18² + 2.30²) × 105 = 5.67 × 105 N C−1
✓ 1
The answer must be given to at least 3 s.f.
Part (a)(iii)
tan−1(2.30/5.18) = 24°, pointing away from AB on the side of P and in the direction from A to B
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
Ve = k(qA + qB)/r = 8.99 × 109 × 2.00 × 10−6/0.250 = 7.19 × 104 V
W = q(VeP − VeM) = 1.0 × 10−9 × (7.19 × 104 − 1.20 × 105) = −4.8 × 10−5 J: the field does negative work
✓ 1
Allow ECF from (b)(i). Accept +4.8 × 10⁻⁵ J only if it is clearly stated as the work done by an external agent.
Part (c)(i)
k(4.00)/x = k(2.00)/(0.300 − x)
✓ 1
x = 0.200 m from A (0.100 m from B)
✓ 1
Part (c)(ii)
Between the charges both fields point the same way (away from A and towards B), so they add: zero potential does not imply zero field
✓ 1
Beyond B, on the side away from A: there the fields are opposite and the smaller charge B is nearer, so the two can be equal
✓ 1
Answers: (a)(i) 5.75 × 105 N C−1 · (a)(ii) 5.67 × 105 N C−1 · (a)(iii) 24° · (b)(i) 7.19 × 104 V · (b)(ii) −4.8 × 10−5 J · (c)(i) 0.200 m from A (the remaining parts are explanations — see the table above)
81D-2-22Electrostatic powder coatingPaper 2Medium12 marks
In electrostatic powder coating, paint powder is blown through a spray gun lined with PTFE (a plastic). The particles are charged by friction as they rub against the lining and leave the gun positively charged. They then travel between a flat metal grid at a potential of +30 kV and a flat earthed metal panel that is to be coated. The grid and the panel are parallel and 0.25 m apart.
A typical powder particle has a mass of 2.0 × 10−10 kg and a charge of +2.4 × 10−13 C.
Diagram not to scale
(a)
Charging.
(i)
State the direction in which electrons are transferred when a particle is charged by friction against the lining.
(1)
(ii)
Explain, with reference to the conservation of charge, why the lining of the gun must be earthed.
(2)
(b)
The field between the grid and the panel.
(i)
Calculate the number of electrons removed from a typical particle.
(1)
(ii)
Determine the electric force on a typical particle between the grid and the panel.
(2)
(iii)
Compare this force with the weight of the particle and comment on the path of the particle.
(2)
(c)
Coating the panel.
(i)
Determine the speed with which a particle that starts from rest at the grid would reach the panel if air resistance were negligible.
(2)
(ii)
The coating is found to be thicker along the edges of the panel. Explain this with reference to the electric field lines near the edges.
(2)
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Notes
Part (a)(i)
From the powder particles to the PTFE lining
✓ 1
Part (a)(ii)
Charge is conserved, so the lining gains a negative charge equal in magnitude to the total positive charge carried away by the powder
✓ 1
Without earthing this negative charge would build up (until sparking occurs, or it attracts the particles back and stops further charging); the earth connection lets the excess electrons flow away
✓ 1
Part (b)(i)
N = 2.4 × 10−13/1.60 × 10−19 = 1.5 × 106
✓ 1
Part (b)(ii)
E = V/d = 30 × 103/0.25 = 1.2 × 105 V m−1
✓ 1
F = qE = 2.4 × 10−13 × 1.2 × 105 = 2.9 × 10−8 N
✓ 1
Part (b)(iii)
Weight = 2.0 × 10−10 × 9.81 = 2.0 × 10−9 N, so the electric force is about 15 times larger
✓ 1
Allow ECF from (b)(ii). Accept 14–15.
Gravity has little effect: the particles follow paths close to the electric field lines, from the grid to the panel
✓ 1
Part (c)(i)
Work done by the field = qV = 2.4 × 10−13 × 3.0 × 104 = 7.2 × 10−9 J = gain in kinetic energy
✓ 1
Allow ECF from the charge and potential difference used in (b)(ii).
v = √(2 × 7.2 × 10−9/2.0 × 10−10) = 8.5 m s−1
✓ 1
ALT: a = F/m and v² = 2as.
Part (c)(ii)
Near the edges of the panel the field lines crowd together, so the field line density and hence the field strength are greatest there
✓ 1
More field lines end on the edges and the force on the particles is larger, so more particles are guided there and the coating is thicker
✓ 1
Answers: (b)(i) 1.5 × 106 · (b)(ii) 2.9 × 10−8 N · (b)(iii) about 15 × the weight · (c)(i) 8.5 m s−1(the remaining parts are explanations — see the table above)
82D-2-25Magnetic field patternsPaper 2Medium12 marks
A long, straight, horizontal overhead cable of a direct-current power line runs north–south, 10.0 m above level ground. It carries a steady current of 600 A towards the north.
The magnetic field strength at a distance r from a long straight wire carrying current I is B = μ0I/(2πr), with μ0 = 4π × 10−7 T m A−1. The horizontal component of the Earth's magnetic field at the site is 1.8 × 10−5 T, directed towards geographic north.
(a)
The field of the cable.
(i)
Sketch the magnetic field lines of the cable in a vertical plane perpendicular to it, as seen by an observer standing to the south of the cable and looking north.
(2)
(ii)
Determine the magnitude and the direction of the magnetic field due to the cable at the ground directly below it.
(2)
(b)
A compass under the cable.
(i)
A compass is placed on the ground directly below the cable. Determine the angle between the compass needle and geographic north.
(2)
(ii)
The current in the cable is reversed. State and explain the new reading of the compass.
(1)
(c)
Other field patterns.
(i)
A long air-core solenoid carries a steady current. Sketch the magnetic field lines inside and outside the solenoid.
(2)
(ii)
State one similarity and one difference between the field of the solenoid and the field of a bar magnet.
(2)
(iii)
A single flat circular coil is viewed from one side, and the current in it is anticlockwise. State the magnetic polarity of the face of the coil nearest the observer.
(1)
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Notes
Part (a)(i)
Concentric circles centred on the cable, with the spacing between the circles increasing with distance from the cable
✓ 1
Clockwise, as seen by the observer (the current flows away from the observer)
✓ 1
Part (a)(ii)
B = 4π × 10−7 × 600/(2π × 10.0) = 1.2 × 10−5 T
✓ 1
Horizontal, towards the west
✓ 1
By the right-hand grip rule: clockwise as seen from the south means westwards below the cable.
Part (b)(i)
The needle lines up with the resultant of 1.8 × 10−5 T north and 1.2 × 10−5 T west: tan θ = 1.2/1.8
✓ 1
θ = 34° west of north
✓ 1
Accept 33.7°. Direction (west) required. Allow ECF from (a)(ii).
Part (b)(ii)
34° east of north: reversing the current reverses the cable's field (now eastwards) without changing its magnitude, while the Earth's field is unchanged
✓ 1
Allow ECF from (b)(i).
Part (c)(i)
Inside: straight, parallel, equally spaced lines along the axis (uniform field), with the line density greatest inside
✓ 1
Outside: lines leave one end and loop round to the other end, spreading out (weaker field), forming closed loops; direction consistent with the grip rule
✓ 1
Part (c)(ii)
Similarity: the external field pattern is the same as that of a bar magnet, with a north pole at one end and a south pole at the other
✓ 1
Difference: e.g. the field of the solenoid can be switched off or reversed and its strength changed by changing the current; inside the solenoid there is a uniform field in an accessible (air) space
✓ 1
Any one valid difference.
Part (c)(iii)
North: by the right-hand grip rule, with the fingers following the anticlockwise current the thumb points towards the observer, so the field lines emerge from the near face
✓ 1
Accept a correct argument from the "anticlockwise = N" end rule.
Answers: (a)(ii) 1.2 × 10−5 T, towards the west · (b)(i) 34° west of north · (b)(ii) 34° east of north (the remaining parts are explanations — see the table above)
Model sketch
Model sketch: concentric circles (clockwise) centred on the cable, further apart at greater distance. Seen from the south, looking north
83D-2-34Electric potential energy in the hydrogen atomPaper 2Medium14 marks
In a simple model of the hydrogen atom, the electron (charge −e) moves in a circular orbit of radius 5.29 × 10−11 m around a stationary proton (charge +e). Only the electric force between them is considered.
Diagram not to scale
(a)
Electric potential energy.
(i)
Outline why the electric potential energy of the electron–proton system is negative.
(1)
(ii)
Show that the electric potential energy of the system is about −27 eV.
(2)
(b)
Energy of the electron in its orbit.
(i)
Show that the kinetic energy of the electron is ke²/(2r).
(2)
(ii)
Hence determine, in eV, the total energy of the atom.
(2)
(iii)
State the ionisation energy of the atom in this model.
(1)
(iv)
Calculate the speed of the electron.
(2)
(c)
Ionisation and gravity.
(i)
Determine the longest wavelength of electromagnetic radiation that can ionise the atom, and identify the region of the electromagnetic spectrum to which it belongs.
(3)
(ii)
Outline why the gravitational force between the electron and the proton can be ignored in this model.
(1)
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Notes
Part (a)(i)
Ep is the work done to assemble the system from infinite separation; the force between the opposite charges is attractive, so this work is negative
✓ 1
Accept: Ep = kq₁q₂/r with q₁q₂ < 0, zero at infinity.
Must be consistent with the candidate's wavelength.
Part (c)(ii)
The ratio of the electric force to the gravitational force, ke²/(Gmemp), is about 2 × 1039, so the gravitational force is negligible
✓ 1
A comparison of the two forces (by calculation or by order of magnitude) is required; do not accept "gravity is weak" on its own.
Answers: (a)(ii) −27.2 eV · (b)(ii) −13.6 eV · (b)(iii) 13.6 eV · (b)(iv) 2.2 × 106 m s−1 · (c)(i) 9.1 × 10−8 m, ultraviolet (the remaining parts are explanations — see the table above)
84IBP2H-031-01-10Millikan's experimentPaper 2Medium8 marks
In a Millikan-type experiment, an oil drop of radius 1.2 µm is held stationary between horizontal plates 6.0 mm apart. The potential difference between the plates is 780 V. The density of the oil is 880 kg m−3. The upthrust from the air is negligible. Take g = 9.8 m s−2. Other drops gave these charges (all ×10−19 C): Drop A: 6.4 Drop B: 11.2 Drop C: 3.2 Drop D: 9.6 Drop E: 14.4
Diagram not to scale
(a)
Outline the condition for the drop to remain stationary.
(2)
(b)
Calculate the magnitude of the charge on the drop.
(3)
(c)
Determine, using the data for drops A–E, a value for the elementary charge. Explain your reasoning.
(2)
(d)
Deduce the number of excess electrons on the drop in part 2.
(1)
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Notes
Part (a)
The electric force on the drop is upwards and equal in magnitude to its weight
✓ 1
so qV/d = mg and the resultant force on the drop is zero
✓ 1
Part (b)
m = ρ × (4/3)πr3 = 880 × (4/3)π(1.2 × 10−6)3 = 6.37 × 10−15 kg
✓ 1
q = mgd/V = 6.37 × 10−15 × 9.8 × 6.0 × 10−3 / 780
✓ 1
q = 4.8 × 10−19 C (accept use of g = 9.81 m s−2)
✓ 1
Part (c)
Every charge is a whole-number multiple of 1.6 × 10−19 C (4, 7, 2, 6 and 9 times). These multiples share no common factor greater than 1, so 1.6 × 10−19 C is the largest value that divides all the charges
✓ 1
Charge is quantised, so the elementary charge is taken as this highest common factor: e = 1.6 × 10−19 C
✓ 1
Part (d)
n = 4.8 × 10−19 / 1.6 × 10−19 = 3 electrons (ECF from parts 2 and 3)
✓ 1
Common error: Cube the radius, not the diameter. Convert micrometres to metres before cubing, or the mass will be wrong by a factor of 10¹⁸.
85IBP2H-031-01-11Potential gradient and permittivityPaper 2Medium7 marks
In a quantum-computing ion trap, three singly charged calcium ions (each of charge +e, mass 6.64 × 10−26 kg) are held at rest in a straight line. Adjacent ions are 5.0 µm apart. Treat the ions as point charges.
Diagram not to scale
(a)
Calculate the electric potential energy of the system of three ions.
(3)
(b)
The trapping field is switched off for the middle ion only, and the two outer ions are held fixed. The middle ion is nudged sideways and moves far away. Determine its kinetic energy when it is far away.
(2)
(c)
Calculate the final speed of the middle ion.
(2)
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Notes
Part (a)
Recognises three pairs: two at separation d and one at 2d
Only the two pair terms involving the middle ion are lost; the outer-pair energy ke2/2d is unchanged. So Ek = 2ke2/d, or equivalently Ek = (answer to first part) − ke2/2d
✓ 1
Ek = 2 × 8.99 × 109 × (1.60 × 10−19)2/5.0 × 10−6 = 9.2 × 10−23 J (allow ECF from the first part if the subtraction route is used)
✓ 1
Part (c)
v = √(2Ek/m) = √(2 × 9.2 × 10−23/6.64 × 10−26)
✓ 1
v = 53 m s−1 (allow ECF from the second part)
✓ 1
Exam tip: For a system of charges, list every pair once. Removing one charge removes only the pair terms that involve that charge.
86IBP2H-031-00-10Charged conducting spherePaper 2Medium8 marks
A science museum uses a Van de Graaff generator with a spherical conducting dome of radius 0.20 m. Dry air breaks down and sparks when the electric field strength reaches 3.0 × 106 V m−1.
(a)
Show that the maximum charge the dome can hold is about 1.3 × 10−5 C.
(2)
(b)
Calculate the maximum electric potential of the dome.
(2)
(c)
State the electric field strength and the electric potential at the centre of the dome when it holds its maximum charge.
(2)
(d)
Explain why a dome of larger radius can reach a higher potential before sparking occurs.
(2)
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Part (a)
Surface field of the sphere E = kQ/R2, so Q = ER2/k
✓ 1
Q = 3.0 × 106 × 0.202/8.99 × 109 = 1.33 × 10−5 C
✓ 1
Part (b)
V = kQ/R, or V = ER
✓ 1
V = 6.0 × 105 V (allow ECF from (a) if V = kQ/R is used)
✓ 1
Part (c)
E = 0, because the field is zero everywhere inside a charged conductor
✓ 1
V = 6.0 × 105 V, the same as the surface value (ECF from (b))
✓ 1
Part (d)
At the surface E = V/R, so for the same potential a larger radius gives a smaller surface field strength
✓ 1
Sparking occurs at a fixed breakdown field, so the maximum potential Vmax = EbreakdownR increases with R (it is proportional to R)
✓ 1
Exam tip: For a conducting sphere, combining E = kQ/R² and V = kQ/R gives V = ER at the surface. This shortcut saves time.
87IBP2H-031-00-11Coulomb's lawPaper 2Medium7 marks
In an electrostatic plastic-sorting machine, two small charged beads are held on an insulating rod. They can be treated as point charges. Bead A carries +3.0 nC and bead B carries −2.0 nC. Their centres are 0.40 m apart.
Diagram not to scale
(a)
Calculate the magnitude of the electrostatic force between the beads.
(2)
(b)
Determine the distance from A of the point on the line through A and B where the resultant electric field strength is zero.
(3)
(c)
Explain why the resultant field cannot be zero at any point between A and B.
Between the beads, the field due to A points away from A, towards B
✓ 1
The field due to the negative B also points towards B, so the two fields are in the same direction and add
✓ 1
Common error: For unlike charges, the zero-field point lies outside the pair, nearer the smaller charge. Check that the root you choose lies in that region.
88IBP2H-031-00-13Equipotential surfacesPaper 2Medium11 marks
A student uses a potential probe to measure the electric potential V at distance r from the centre of a charged conducting sphere of radius 5.0 cm. The results are shown below.
r / m: 0.10 | 0.20 | 0.30 | 0.40 V / kV: 36.0 | 18.0 | 12.0 | 9.0
(a)
Show that the data are consistent with V ∝ 1/r.
(2)
(b)
Determine the charge on the sphere.
(2)
(c)
Estimate the electric field strength at r = 0.30 m using the potential gradient between r = 0.20 m and r = 0.40 m.
(2)
(d)
Calculate the field strength at r = 0.30 m from Coulomb's law. Explain why the estimate in (c) is larger than this value.
(3)
(e)
Equipotential surfaces around the sphere are drawn at equal intervals of potential. Explain why adjacent surfaces become further apart as r increases.
(2)
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Part (a)
Calculates Vr for at least three data points (for example 0.10 × 36.0 × 103, 0.30 × 12.0 × 103)
✓ 1
Vr is constant at 3.6 × 103 V m, so V ∝ 1/r
✓ 1
Part (b)
Q = Vr/k = 3600/8.99 × 109
✓ 1
Q = 4.0 × 10−7 C
✓ 1
Part (c)
E ≈ −ΔV/Δr = (18.0 − 9.0) × 103/0.20
✓ 1
E ≈ 4.5 × 104 V m−1
✓ 1
Part (d)
E = kQ/r2 = 3600/0.302 = 4.0 × 104 V m−1 (ECF from (b))
✓ 1
The estimate in (c) is an average gradient over the interval, the gradient of a chord
✓ 1
The potential gradient falls non-linearly (as 1/r2), so the steeper inner region, near 0.20 m, dominates and the average exceeds the value at 0.30 m
✓ 1
Part (e)
The field strength decreases with distance (E ∝ 1/r2)
✓ 1
The spacing Δr ≈ ΔV/E, so for a fixed ΔV the spacing increases as E falls
✓ 1
Exam tip: A gradient taken over a wide interval of a curved V–r graph only approximates the field. Compare it with kQ/r² to judge its accuracy.
89IBP2H-031-00-15Electric fields in the atmospherePaper 2Medium8 marks
Below a thundercloud, the electric field between the cloud base and the ground can be modelled as uniform. It has strength 1.2 × 104 V m−1 and is directed vertically upwards.
Diagram not to scale
(a)
Describe the equipotential surfaces in this region.
(2)
(b)
Calculate the vertical separation of two equipotential surfaces that differ in potential by 500 V.
(2)
(c)
State and explain whether the ground is at a higher or a lower potential than the cloud base.
(2)
(d)
A raindrop carrying a charge of +3.0 × 10−11 C falls vertically through 250 m in this region. Determine the change in its electric potential energy.
(2)
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Part (a)
Horizontal planes, perpendicular to the vertical field lines
✓ 1
Equally spaced for equal potential intervals, because the field is uniform
✓ 1
Part (b)
Δx = ΔV/E = 500/1.2 × 104
✓ 1
Δx = 4.2 × 10−2 m (4.2 cm)
✓ 1
Part (c)
The field points from high to low potential, and here it points upwards
It is an increase, because the positive drop moves against the electric force towards higher potential (ECF from (c))
✓ 1
Exam tip: Always state whether potential energy increases or decreases. The sign follows from the charge and the direction of motion relative to the field.
90IBP2S-030-03-13Electric potential and workPaper 2Medium9 marks
A satellite passing through Earth's shadow becomes charged by electrons from the surrounding plasma. It is modelled as a conducting sphere of radius 1.0 m. A probe measures the electric potential at distances r from the centre of the satellite:
91IBP2H-031-02-10Magnetic field patternsPaper 2Medium7 marks
A single long overhead power cable runs horizontally 12 m above flat ground. At times of high demand it carries an alternating current of frequency 50 Hz and root-mean-square (rms) value 400 A. Treat the cable as a long straight wire, and ignore the fields of other conductors. In parts (a), (b) and (d), use the rms current, so that all magnetic field values are rms values. μ0 = 4π×10−7 T m A−1.
Cross-section; not to scale
(a)
Calculate the magnitude of the magnetic field due to the cable at ground level directly below the cable.
(2)
(b)
The Earth's magnetic field at this location has magnitude 5.0×10−5 T. Determine the ratio of the cable's field at ground level to the Earth's field.
(1)
(c)
Outline why a compass held below the cable shows no noticeable deflection caused by the cable.
(2)
(d)
A sensor is raised vertically from the ground directly below the cable. Determine the height above the ground at which the cable's field has magnitude 1.0×10−5 T.
(2)
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Part (a)
B = μ0I/(2πr) = (4π×10−7×400)/(2π×12)
✓ 1
B = 6.7×10−6 T
✓ 1
Part (b)
6.7×10−6 / 5.0×10−5 = 0.13 (no unit); allow ECF from (a)
✓ 1
Part (c)
The current, and so the cable's field, reverses direction 100 times per second
✓ 1
The needle has too much inertia to follow such rapid changes. The cable's field averages to zero over time, so the needle stays aligned with the Earth's field
✓ 1
Part (d)
r = μ0I/(2πB) = (2×10−7×400)/(1.0×10−5) = 8.0 m from the cable
✓ 1
Height = 12 − 8.0 = 4.0 m above the ground
✓ 1
Exam tip: The formula B = μ₀I/(2πr) gives the distance from the wire. If a question asks for a height above the ground, convert your answer.
92IBP2H-031-02-12Charged conducting spherePaper 2Medium8 marks
A spherical high-voltage terminal on a particle-accelerator test rig is isolated in dry air. A probe measures the electric potential V at distance r from the centre of the terminal. All readings are taken outside the terminal: r / m: 0.20, 0.30, 0.40, 0.60 V / kV: 180, 120, 90, 60 k = 8.99×109 N m2 C−2.
(a)
Show that the data are consistent with the terminal behaving as a point charge at its centre.
(2)
(b)
Determine the charge on the terminal.
(2)
(c)
Calculate the work done by an external agent to move a +3.0 nC test charge slowly from r = 0.60 m to r = 0.20 m.
(2)
(d)
Determine the electric field strength at r = 0.30 m.
(2)
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Notes
Part (a)
Calculates V×r for at least three data pairs, each equal to 36 kV m
✓ 1
V×r is constant, so V ∝ 1/r, as expected outside a point charge
✓ 1
Part (b)
Q = Vr/k = 36×103/8.99×109
✓ 1
Q = 4.0×10−6 C
✓ 1
Part (c)
ΔV = 180 − 60 = 120 kV
✓ 1
W = qΔV = 3.0×10−9×1.2×105 = 3.6×10−4 J
✓ 1
Part (d)
Uses E = kQ/r2, or the equivalent E = V/r for a point charge, with the charge found earlier allowed as carried forward
✓ 1
E = 120×103/0.30 = 4.0×105 V m−1
✓ 1
Exam tip: To test for an inverse relation, multiply the variables for each data pair and check that the product stays constant.
93IBP2H-031-02-15Magnetic field of a solenoidPaper 2Medium7 marks
An electromagnetic door lock uses a long solenoid. When there is a current, the solenoid pulls a steel bolt into its core. One end of the solenoid is labelled P. Viewed from end P, the current in the turns flows anticlockwise.
(a)
Outline one similarity and one difference between the magnetic field of the solenoid and that of a bar magnet.
(2)
(b)
State the magnetic polarity of end P, and outline how it is determined.
(2)
(c)
A small compass is placed first inside the solenoid near its centre, then outside alongside the middle of the solenoid. Explain why the needle points in opposite directions in the two positions.
(2)
(d)
State one change that would increase the field strength inside the solenoid without increasing the current.
(1)
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Notes
Part (a)
Similarity: the field outside the solenoid has the same shape as a bar magnet's field, with lines leaving one end (N) and entering the other (S)
✓ 1
Difference, any one of: the solenoid's field can be switched off or varied in strength by changing the current; its polarity can be reversed by reversing the current; its hollow interior holds a strong, nearly uniform field along the axis that can be accessed
✓ 1
Part (b)
End P is a north pole
✓ 1
Right-hand grip rule: fingers curl in the direction of the current and the thumb points along the field inside, towards P, so field lines emerge from P
✓ 1
Part (c)
Inside, the field runs along the axis towards P (from the S end to the N end), so the needle points towards P
✓ 1
Field lines are closed loops, so outside they return from P around to the other end; alongside the middle they point away from P, the opposite way
✓ 1
Part (d)
Either increase the number of turns per unit length, or insert a soft-iron core
✓ 1
Exam tip: Field lines form closed loops. They run S to N inside a solenoid or magnet and N to S outside it.
94IBP2S-030-04-15Charged conducting spherePaper 2Medium7 marks
A spacecraft passing through a dense plasma collects electrons and reaches a potential of −2.0 kV relative to the distant surroundings. Model the spacecraft as an isolated conducting sphere of radius 1.5 m. (k = 8.99×109 N m2 C−2, e = 1.60×10−19 C)
(a)
Describe the electric field lines just outside the spacecraft.
(2)
(b)
Outline how the equipotential surfaces around the spacecraft relate to the field lines. Explain why equipotentials drawn at equal potential intervals become further apart with distance.
(2)
(c)
Calculate the charge on the spacecraft.
(2)
(d)
Calculate the number of excess electrons on the spacecraft.
(1)
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Part (a)
The lines are radial, meeting the surface at right angles .
✓ 1
They point inwards, towards the spacecraft, because it is negatively charged .
✓ 1
Part (b)
The equipotentials are concentric spheres, everywhere perpendicular to the field lines .
✓ 1
The field strength decreases with distance, so a greater distance is needed for the same change in potential .
✓ 1
Part (c)
V = kQ/r, so Q = Vr/k = (−2.0×103 × 1.5) / 8.99×109
✓ 1
Q = −3.3×10−7 C
✓ 1
Part (d)
N = |Q|/e = 3.34×10−7 / 1.60×10−19 = 2.1×1012 . Allow error carried forward from the charge calculated in the previous part.
✓ 1
Exam tip: Field lines and equipotentials are always perpendicular. Equipotentials that are closer together mean a stronger field.
95IBP3H-D2-20r8-11Magnetic field of a straight wirePaper 2Medium7 marks
Two long straight parallel conductors in a telecoms duct are 0.50 m apart. They carry currents of 4.0 A and 6.0 A in the same direction. Use μ0 = 4π × 10−7 N A−2.
(a)
Explain why a point of zero resultant magnetic field exists between the conductors but would not if the currents were in opposite directions.
(2)
(b)
Calculate the distance of the zero-field point from the conductor carrying 4.0 A.
(3)
(c)
Calculate the force per unit length between the conductors and state whether it is attractive or repulsive.
(2)
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Part (a)
Same-direction currents give fields in opposite directions between the wires, so they can cancel (1 mark for direction argument)
✓ 1
Opposite-direction currents give fields in the same direction between the wires, so they add and cannot cancel
✓ 1
Part (b)
Sets μ0(4.0)/(2πx) = μ0(6.0)/(2π(0.50 − x))
✓ 1
4.0(0.50 − x) = 6.0x
✓ 1
x = 0.20 m
✓ 1
Part (c)
F/L = μ0I1I2/(2πd) = 9.6 × 10−6 N m−1 (0.0000096 N m−1)
✓ 1
Attractive (currents in the same direction)
✓ 1
Common error: The zero-field point lies closer to the conductor with the smaller current, since the field of each wire falls off as 1/r.
96IBP3H-D2-20r8-12Field and potential of two point chargesPaper 2Medium9 marks
Charges of +8.0 nC and −8.0 nC are fixed at the points (0, 0) and (0.60 m, 0) respectively. Point P is at (0.30 m, 0.40 m). Use k = 8.99 × 109 N m2 C−2.
Diagram not to scale
(a)
Show that P is 0.50 m from each charge, and calculate the magnitude of the field at P due to one charge.
(2)
(b)
Determine the magnitude and direction of the resultant electric field at P.
(3)
(c)
State the electric potential at P and justify your answer.
(2)
(d)
Point R is at a potential of +120 V. Calculate the work done by an external agent in moving a +3.0 nC charge from P to R.
(2)
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Part (a)
Distance = √(0.302 + 0.402) = 0.50 m for each charge
✓ 1
E = kQ/r2 = 288 N C−1
✓ 1
Part (b)
Vertical components cancel (each field makes the same angle with the x-axis, with opposite vertical directions)
✓ 1
Horizontal component of each = 288 × 0.60 added together: 2 × 287.7 × 0.60
✓ 1
E = 345 N C−1 in the +x direction
✓ 1
Part (c)
V = 0
✓ 1
The two potentials have equal magnitude (equal distances and charges) and opposite sign, so they cancel
✓ 1
Part (d)
W = qΔV with ΔV = 120 − 0
✓ 1
W = 3.6 × 10−7 J (0.00000036 J)
✓ 1
Exam tip: Add electric fields as vectors by resolving into components, but add potentials as plain numbers with their signs.
98IBP3H-D2-20r8-14Charged sphere in a uniform fieldPaper 2Medium7 marks
In a food-processing electrostatic precipitator, two parallel plates 4.0 cm apart have a potential difference of 12 kV across them, creating a uniform field. A dust grain of mass 5.0 × 10−12 kg and charge +8.0 × 10−15 C starts from rest next to the positive plate.
Diagram not to scale
(a)
Calculate the electric field strength between the plates.
(2)
(b)
Use energy considerations to calculate the speed of the grain after it has moved 1.0 cm along the field. Give your answer to 3 significant figures.
(3)
(c)
Show that the weight of the grain can be neglected, using g = 9.8 m s−2.
Weight = 5.0 × 10−12 × 9.8 = 4.9 × 10−11 N; electric force qE = 2.4 × 10−9 N
✓ 1
Weight is about 2% of the electric force (49 times smaller), so negligible
✓ 1
Exam tip: When a charge moves through a potential difference from rest, equate qΔV to its kinetic energy; this avoids resolving forces and accelerations.
99IBP3H-D2-20r11-15Electric fields in the atmospherePaper 2Medium8 marks
During a thunderstorm the base of a cloud is 2.0 km above flat ground. The potential difference between the cloud base and the ground is 1.5×108 V. Dry air breaks down when the electric field strength exceeds about 3.0×106 V m−1.
Diagram not to scale
(a)
Calculate the average electric field strength between the cloud base and the ground.
(2)
(b)
The average field strength is much less than the breakdown value, yet lightning occurs. Suggest why.
(2)
(c)
A lightning strike transfers 20 C of charge through the full potential difference. Calculate the energy transferred.
(2)
(d)
A household uses 12 kWh of electrical energy each day. Calculate, to 3 significant figures, the number of days for which the energy of the strike would supply the household. (1 kWh = 3.6×106 J)
(2)
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Notes
Part (a)
M1 \(E=\frac{\Delta V}{d}\)
✓ 1
A1 \(\frac{1.5\times10^{8}}{2000}=7.5\times10^{4}\) V m−1
✓ 1
Part (b)
The field is not uniform, so the average value hides much larger local fields
✓ 1
Near charge concentrations (for example pointed objects or the cloud's charge regions) the field exceeds \(3.0\times10^{6}\) V m−1, so breakdown begins there
✓ 1
Part (c)
M1 \(E=q\Delta V\)
✓ 1
A1 \(20\times1.5\times10^{8}=3.0\times10^{9}\) J
✓ 1
Part (d)
M1 \(12\times3.6\times10^{6}=4.32\times10^{7}\) J per day
✓ 1
A1 \(\frac{3.0\times10^{9}}{4.32\times10^{7}}=69.4\) days
✓ 1
Exam tip: Convert kWh to joules before comparing energies: 1 kWh = 3.6×10⁶ J.
100D-2-02Electric potential and field from a graphPaper 2Hard13 marks
Two small charged spheres are fixed on a straight, horizontal insulating track: P, of charge +3.0 nC, at x = 0, and Q, of unknown positive charge, at x = 0.30 m. The graph shows how the electric potential V varies with x between x = 0.03 m and x = 0.24 m. Ignore gravitational effects.
Graph drawn to scale. P is at x = 0 and Q at x = 0.30 m.
(a)
The minimum of the potential.
(i)
State and explain what the graph shows about the electric field strength at x = 0.10 m.
(2)
(ii)
Hence deduce, without calculating any potential, that the charge on Q is 12 nC.
(2)
(b)
The field at x = 0.20 m.
(i)
By drawing a tangent, determine the magnitude of the electric field strength at x = 0.20 m and state its direction.
(2)
(ii)
Confirm your answer to (b)(i) by calculation.
(2)
(c)
A charged bead of charge +2.0 nC, free to slide along the track without friction, is released from rest at x = 0.05 m.
(i)
Use the graph to determine the maximum kinetic energy of the bead.
(2)
(ii)
Identify the other position at which the bead is momentarily at rest.
(1)
(iii)
The bead oscillates between these two positions. Explain why its motion is not simple harmonic.
(2)
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Part (a)(i)
The gradient of the graph is zero at the minimum, so E = −ΔV/Δx = 0
✓ 1
The fields of P and Q are equal in magnitude and opposite in direction there
✓ 1
Accept "the field of P (in the +x direction) cancels the field of Q (in the −x direction)".
Part (a)(ii)
kqP/0.10² = kqQ/0.20²
✓ 1
Allow ECF from the position of the minimum read in (a)(i). The distance from Q (0.20 m) must be used.
qQ = qP(0.20/0.10)² = 4 × 3.0 nC = 12 nC
✓ 1
Part (b)(i)
Tangent drawn at x = 0.20 m; |E| = |gradient| ≈ 1.0 × 104 V m−1
✓ 1
Accept 0.85–1.2 × 10⁴ V m⁻¹.
In the −x direction, towards P (the potential falls towards P)
✓ 1
Direction must be justified from the gradient or from the larger, nearer charge Q.
E = −1.0 × 104 V m−1 (in the −x direction), in agreement with (b)(i)
✓ 1
The comparison with (b)(i) is required.
Part (c)(i)
ΔV = V(0.05) − V(0.10) = 971 − 809 = 162 V
✓ 1
Accept readings within ±10 V of each value.
Ek = qΔV = 2.0 × 10−9 × 162 = 3.2 × 10−7 J
✓ 1
Accept 2.8–3.6 × 10⁻⁷ J.
Part (c)(ii)
Where V again equals V(0.05) ≈ 971 V: x ≈ 0.17 m
✓ 1
Allow ECF from the reading in (c)(i). Accept 0.16–0.18 m.
Part (c)(iii)
For simple harmonic motion the restoring force (acceleration) must be proportional to the displacement from equilibrium, a = −ω²x
✓ 1
The graph is not symmetrical about x = 0.10 m (it is steeper on the side of Q), so the force (gradient) does not increase in proportion to the displacement
✓ 1
Accept: the turning points are at different distances (0.05 m and about 0.07 m) from the equilibrium position.
Answers: (a)(ii) 12 nC · (b)(i) 1.0 × 104 V m−1, towards P · (b)(ii) −1.0 × 104 V m−1 · (c)(i) 3.2 × 10−7 J · (c)(ii) x ≈ 0.17 m (the remaining parts are explanations — see the table above)
101D-2-12Charged sphere in a uniform fieldPaper 2Hard13 marks
A small sphere of mass 2.00 × 10−4 kg, carrying a positive charge q, hangs from an insulating thread of length 4.0 cm. The top of the thread is fixed midway between two vertical parallel plates that are 5.00 cm apart. When a potential difference of 2.50 kV is applied to the plates, the sphere comes to rest with the thread at 30.0° to the vertical. The radius of the sphere is negligible.
Diagram not to scale
(a)
Equilibrium.
(i)
Calculate the electric field strength between the plates.
(1)
(ii)
Show that q is about 2.3 × 10−8 C.
(2)
(iii)
Calculate the tension in the thread.
(1)
(b)
The thread is cut.
(i)
Determine the magnitude and direction of the acceleration of the sphere immediately after the thread is cut.
(2)
(ii)
Explain why the sphere then moves in a straight line.
(2)
(iii)
Determine the time taken for the sphere to reach the plate.
(3)
(c)
(i)
The sphere touches the negative plate. Explain why it is then pushed back across the gap.
(2)
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Part (a)(i)
E = V/d = 2.50 × 103/0.0500 = 5.00 × 104 V m−1
✓ 1
Part (a)(ii)
T sin 30° = qE and T cos 30° = mg, so qE = mg tan 30°
✓ 1
q = 2.00 × 10−4 × 9.81 × tan 30.0°/5.00 × 104 = 2.27 × 10−8 C
✓ 1
Allow ECF from (a)(i). Full substitution or an answer to at least 3 s.f. is required.
Part (a)(iii)
T = mg/cos 30° = 1.962 × 10−3/0.866 = 2.27 × 10−3 N
✓ 1
Part (b)(i)
The resultant of the weight and the electric force is equal and opposite to the former tension, 2.27 × 10−3 N, so a = 2.27 × 10−3/2.00 × 10−4 = 11.3 m s−2
✓ 1
Allow ECF from (a)(iii). ALT: a = g/cos 30°.
At 30° to the vertical, downwards and towards the negative plate (along the line of the thread)
✓ 1
Part (b)(ii)
Both forces (weight and electric force) are constant, so the resultant force and the acceleration are constant in magnitude and direction
✓ 1
The sphere starts from rest, so it moves along the line of the resultant force
✓ 1
Do not accept "there is no horizontal force".
Part (b)(iii)
The sphere starts L sin 30° = 2.0 cm from the midline, so it is 2.5 − 2.0 = 0.50 cm from the plate
✓ 1
This geometrical step is required.
Horizontal acceleration = qE/m = 5.66 m s−2
✓ 1
Allow ECF from (a)(ii). ALT: a distance of 1.0 cm along the line at 11.3 m s⁻².
t = √(2 × 0.0050/5.66) = 0.042 s
✓ 1
Award [2 max] for 0.094 s (2.5 cm used).
Part (c)(i)
Charge is transferred by contact: electrons flow from the plate onto the sphere, which becomes negatively charged
✓ 1
The electric force on a negative charge is opposite to the field, towards the positive plate
✓ 1
Answers: (a)(i) 5.00 × 104 V m−1 · (a)(ii) 2.27 × 10−8 C · (a)(iii) 2.27 × 10−3 N · (b)(i) 11.3 m s−2 · (b)(iii) 0.042 s (the remaining parts are explanations — see the table above)
102D-2-23Electric potential energy of a system of chargesPaper 2Hard13 marks
In the alpha-cluster model of the carbon-12 nucleus, the nucleus is pictured as three alpha particles at the corners of an equilateral triangle of side 3.0 × 10−15 m. Each alpha particle is treated as a point charge +2e. Consider only the electric interactions.
Diagram not to scale
(a)
Electric potential energy.
(i)
Outline what is meant by the electric potential energy of a system of charges.
(1)
(ii)
Show that the electric potential energy of one pair of alpha particles in this model is about 1.9 MeV.
(2)
(iii)
Determine the total electric potential energy of the three-alpha system, in MeV.
(1)
(b)
The centre of the triangle.
(i)
Calculate the electric potential at the centre of the triangle due to the three alpha particles.
(2)
(ii)
State the electric field strength at the centre of the triangle, and explain your answer.
(1)
(iii)
The potential at the centre is not zero although the field strength there is zero. Explain this, and explain, with reference to E = −ΔVe/Δr, why the potential at the centre is a maximum along the line through the centre perpendicular to the plane of the triangle.
(2)
(c)
A proton probe.
(i)
A proton approaches the nucleus from a large distance along the line through the centre of the triangle, perpendicular to its plane. Determine, in MeV and in J, the minimum initial kinetic energy the proton needs to reach the centre. Assume the alpha particles stay fixed.
(2)
(d)
Stability.
(i)
The answer to (a)(iii) is positive. Explain what this implies about the forces needed to hold the nucleus together.
(2)
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Part (a)(i)
The work done (by an external agent) to assemble the system by bringing the charges from infinite separation to their positions
Accept 4.98 × 106 V. Using the side length gives 2.9 × 106 V: award the second mark only.
Part (b)(ii)
Zero: the three fields are equal in magnitude and point away from the three corners at 120° to one another, so their vector sum is zero
✓ 1
Part (b)(iii)
Potential is a scalar: the three positive contributions add and cannot cancel, while field strength is a vector and the three contributions cancel by symmetry
✓ 1
Along the axis the field at the centre is zero, so the potential gradient −ΔVe/Δr is zero there; on either side of the plane the field points away from the triangle, so V falls with distance from the centre in both directions: the centre is a maximum of V along the axis
✓ 1
Part (c)(i)
The proton must gain electric potential energy eV: along this line the potential rises continuously up to its value at the centre, so this is the minimum energy needed
A positive electric potential energy means the electric forces are repulsive: work would have to be done to push the alpha particles together, and on their own they would fly apart
✓ 1
Allow ECF from the sign of the answer to (a)(iii).
An additional attractive force must act that is stronger than the electric repulsion at these separations: the short-range strong nuclear force between the nucleons
✓ 1
Answers: (a)(ii) 1.92 MeV · (a)(iii) 5.8 MeV · (b)(i) 5.0 × 106 V · (b)(ii) 0 · (c)(i) 5.0 MeV; 8.0 × 10−13 J (the remaining parts are explanations — see the table above)
103D-2-24Charged conducting spherePaper 2Hard14 marks
A hollow, thin-walled metal sphere A of radius 0.120 m is mounted on an insulating stand and carries a charge of +36.0 nC. (k = 8.99 × 109 N m² C−2)
Spheres far apart; not to scale
(a)
The isolated sphere.
(i)
Calculate the electric potential of sphere A.
(1)
(ii)
Sketch graphs to show how the electric potential and the electric field strength vary with distance r from the centre of A, from r = 0 to r = 0.36 m. Label the values at r = 0.120 m.
(2)
(b)
Connecting a second sphere.
(i)
Sphere A is now connected by a long thin wire to a distant, uncharged metal sphere B of radius 0.030 m. Explain why charge flows along the wire and why the flow stops.
(2)
(ii)
Determine the charge on each sphere after the flow stops.
(2)
(iii)
Determine the common potential of the spheres.
(1)
(iv)
Show that the electric field strength at the surface of B is four times that at the surface of A.
(2)
(v)
Air near a conductor becomes ionised when the field strength there is very large. Use (b)(iv) to explain why charge leaks into the air most readily from sharply pointed parts of a charged conductor.
(2)
(c)
Work.
(i)
State and explain the work done in moving a small positive test charge from the surface of B to the surface of A.
(2)
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Part (a)(i)
Ve = kQ/R = 8.99 × 109 × 36.0 × 10−9/0.120 = 2.70 × 103 V
✓ 1
Part (a)(ii)
Potential: constant at 2.70 kV from r = 0 to 0.120 m, then falling as 1/r (a curve, not a straight line) to 0.90 kV at 0.36 m, approaching zero at infinity
✓ 1
Field strength: zero inside, jumping to a maximum of V/R = 2.25 × 104 V m−1 at the surface, then falling as 1/r² (to 2.5 × 103 V m−1 at 0.36 m)
✓ 1
Values at 0.36 m not required; shape and labelled surface values are.
Part (b)(i)
Charge flows because there is a potential difference between the spheres (A at 2.70 kV, B at 0 V), so there is a field along the wire that moves the free electrons
✓ 1
The flow stops when the two spheres (joined as one conductor) are at the same potential, so the potential difference and the field along the wire are zero
✓ 1
Part (b)(ii)
Same potential: kQA/0.120 = kQB/0.030, so QA = 4QB; charge is conserved: QA + QB = 36.0 nC
✓ 1
Spheres far apart, so each is treated as isolated.
QA = 28.8 nC, QB = 7.2 nC
✓ 1
Part (b)(iii)
V = 8.99 × 109 × 28.8 × 10−9/0.120 = 2.16 × 103 V
✓ 1
Or using B: 8.99 × 109 × 7.2 × 10−9/0.030; ECF from (b)(ii).
Part (b)(iv)
At the surface of a sphere E = kQ/R² = V/R. Both spheres have the same V, so E ∝ 1/R
✓ 1
EB/EA = 0.120/0.030 = 4 (EA = 1.80 × 104 V m−1, EB = 7.19 × 104 V m−1)
✓ 1
Allow ECF from (b)(iii).
Part (b)(v)
A sharp point behaves like a part of the conductor with a very small radius of curvature. The whole conductor is at one potential, so the field there (∝ 1/radius) is very large, and the field lines are most closely packed there
✓ 1
Allow ECF from (b)(iv): the field at the surface is inversely proportional to the radius.
The strong field ionises the air near the point, and the ions carry charge away, so charge leaks most readily from sharp points
✓ 1
Part (c)(i)
Zero
✓ 1
The two surfaces are at the same potential (the connected spheres form one equipotential), so W = qΔVe = 0
✓ 1
Answers: (a)(i) 2.70 × 103 V · (b)(ii) 28.8 nC and 7.2 nC · (b)(iii) 2.16 × 103 V · (b)(iv) ratio 4 · (c)(i) 0 (the remaining parts are explanations — see the table above)
Model sketch
Model sketch: V = 2.70 kV (constant) for r < 0.120 m and ∝ 1/r beyond; E = 0 inside and ∝ 1/r² beyond, 2.25 × 10⁴ V m⁻¹ at the surface
104D-2-26Potential gradient and permittivityPaper 2Hard13 marks
The membrane of a nerve cell is modelled as a layer of lipid 7.0 nm thick with conducting fluid on each side. The graph shows how the electric potential V varies with distance x measured from the outer surface of the membrane into the cell. The fluid outside the cell is at 0 V.
In a medium of permittivity ε, Coulomb's law is F = q1q2/(4πεr²). The permittivity of water is 7.1 × 10−10 C² N−1 m−2 and that of lipid is 1.8 × 10−11 C² N−1 m−2. Body temperature is 310 K.
Potential across the membrane of a nerve cell. The dashed lines mark the surfaces of the membrane.
(a)
The field in the membrane.
(i)
Determine the magnitude and the direction of the electric field strength in the membrane.
(2)
(ii)
State the electric field strength in the fluid inside the cell, with a reason from the graph.
(1)
(iii)
Calculate the electric force on a sodium ion (Na+, charge +e) inside the membrane.
(1)
(b)
An ion crossing the membrane.
(i)
A sodium ion passes through a channel in the membrane from the outside to the inside of the cell. Determine the work done on the ion by the electric field, in eV and in J.
(2)
(ii)
Compare your answer to (b)(i) with the average random kinetic energy of an ion at body temperature.
(2)
(c)
Why ions dissolve in water but not in lipid.
(i)
Calculate the electric force between a sodium ion and a chloride ion (Cl−, charge −e) 0.50 nm apart in water.
(2)
(ii)
By calculating the work needed to separate the two ions from 0.50 nm to infinity in water and in lipid, explain why ions separate readily in water but not in lipid.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
E = −ΔV/Δx: magnitude = 0.070/7.0 × 10−9 = 1.0 × 107 V m−1
✓ 1
Directed from outside to inside (in the +x direction), from higher to lower potential
✓ 1
Part (a)(ii)
Zero: the potential is constant inside the cell, so the potential gradient is zero
✓ 1
Part (a)(iii)
F = qE = 1.60 × 10−19 × 1.0 × 107 = 1.6 × 10−12 N (into the cell)
✓ 1
ECF from (a)(i).
Part (b)(i)
W = qΔVe: the ion falls through 70 mV, so the field does +0.070 eV of work (the electric potential energy decreases by 0.070 eV)
✓ 1
= 0.070 × 1.60 × 10−19 = 1.1 × 10−20 J
✓ 1
Accept 1.12 × 10−20 J. Sign or statement that the field does positive work is required for the first mark.
The work done by the field is about 1.7 times (accept 1.7–1.8) the average thermal kinetic energy, so the same order of magnitude: the field has a large effect on ion movement, but thermal motion is also significant
Work to separate = electric potential energy magnitude q1q2/(4πεr): in water 5.7 × 10−21 J (0.036 eV)
✓ 1
Or F × r in water.
In lipid, 2.3 × 10−19 J (1.4 eV), about 40 times larger because the permittivity is about 40 times smaller
✓ 1
In water the separation energy is comparable to the average thermal energy (6.4 × 10−21 J), so collisions easily pull the ions apart and they stay dissolved. In lipid it is about 35 times the thermal energy, so ion pairs are not separated and ions do not dissolve in the lipid layer
✓ 1
Allow ECF from the thermal energy found in (b)(ii).
Answers: (a)(i) 1.0 × 107 V m−1, inwards · (a)(ii) 0 · (a)(iii) 1.6 × 10−12 N · (b)(i) 0.070 eV = 1.1 × 10−20 J · (b)(ii) 6.4 × 10−21 J · (c)(i) 1.1 × 10−11 N · (c)(ii) 5.7 × 10−21 J; 2.3 × 10−19 J (the remaining parts are explanations — see the table above)
105D-2-39Electrostatic levitation of lunar dustPaper 2Hard18 marks
On the sunlit side of the Moon, ultraviolet radiation from the Sun releases photoelectrons from the surface, which becomes positively charged. An electric field forms in the region just above the surface. The graph shows how the electric potential V varies with height h above the surface, relative to a point far above it.
Mass of the Moon = 7.35 × 1022 kg; radius of the Moon = 1.74 × 106 m.
Graph drawn to scale.
(a)
Charging the surface.
(i)
Most of the photoelectrons are released by ultraviolet photons of wavelength 122 nm. The work function of the surface material is 5.0 eV. Determine, in eV, the maximum kinetic energy of the photoelectrons.
(2)
(ii)
Explain why the potential of the surface does not rise much above about 5 V.
(2)
(b)
Levitating dust.
(i)
Use the graph to determine the electric field strength at h = 0.50 m and state its direction.
(3)
(ii)
Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg−1.
(1)
(iii)
A dust grain of mass 2.0 × 10−15 kg floats at rest at h = 0.50 m. Determine the number of electrons that the grain has lost.
(3)
(iv)
Explain whether the equilibrium of the grain is stable for small vertical displacements.
(2)
(c)
(i)
A second grain of the same material has a larger radius. Its charge is proportional to its surface area. Deduce, without calculation, whether it floats higher or lower than the grain in (b)(iii).
(2)
(d)
(i)
Estimate the work done by the electric field on the grain in (b)(iii) as it rose from the surface to h = 0.50 m.
As the surface becomes positive, the emitted electrons are attracted back: to escape an electron must do work eΔV against the field
✓ 1
When the surface potential reaches about 5.2 V, even the fastest photoelectrons (5.2 eV) cannot escape, so the surface stops losing electrons; this agrees with the 5.0 V at h = 0 on the graph
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
Tangent drawn to the curve at h = 0.50 m
✓ 1
|E| = |gradient| ≈ 3.0 V m−1
✓ 1
Accept 2.6–3.5 V m⁻¹.
Vertically upwards, away from the surface (the potential decreases with height)
✓ 1
Part (b)(ii)
g = GM/R² = 6.67 × 10−11 × 7.35 × 1022/(1.74 × 106)² = 1.62 N kg−1
✓ 1
Full substitution or 3 s.f. required.
Part (b)(iii)
In equilibrium the electric force balances the weight: qE = mg
✓ 1
The grain is treated as remaining at the same height above a locally flat surface.
q = 2.0 × 10−15 × 1.62/3.0 = 1.1 × 10−15 C
✓ 1
Allow ECF from (b)(i) and (b)(ii).
N = q/e ≈ 6.7 × 103
✓ 1
Accept 5.8–7.9 × 10³.
Part (b)(iv)
The field strength (the gradient of the graph) decreases with height
✓ 1
Allow ECF from the shape of the field found in (b)(i).
If the grain rises, qE < mg and the resultant force is downwards; if it falls, qE > mg and the resultant force is upwards: the force is restoring, so the equilibrium is stable
✓ 1
Part (c)(i)
m ∝ r³ and q ∝ r², so q/m ∝ 1/r: the larger grain needs a stronger field, E = mg/q, to float
✓ 1
The field is stronger closer to the surface, so the larger grain floats lower
✓ 1
Part (d)(i)
W = qΔV with ΔV = 5.0 − 3.0 = 2.0 V read from the graph
✓ 1
Accept 1.9–2.1 V.
W = 1.1 × 10−15 × 2.0 = 2.2 × 10−15 J
✓ 1
Allow ECF from (b)(iii). Accept 2.0–2.3 × 10⁻¹⁵ J.
Part (d)(ii)
The charge on the grain stays constant as it rises
✓ 1
Accept: the grain starts from the surface with the same charge.
Answers: (a)(i) 5.2 eV · (b)(i) 3.0 V m−1 upwards · (b)(ii) 1.62 N kg−1 · (b)(iii) 6.7 × 103 · (c)(i) lower · (d)(i) 2.2 × 10−15 J (the remaining parts are explanations — see the table above)
106IBP2H-031-01-13Electric potential energy in the hydrogen atomPaper 2Hard8 marks
The dome of a Van de Graaff generator used in a science museum can be treated as an isolated, positively charged conducting sphere of radius 0.20 m. Air breaks down when the electric field strength reaches 3.0 × 106 V m−1. Mass of a proton = 1.67 × 10−27 kg Elementary charge = 1.60 × 10−19 C
(a)
Show that the maximum potential of the dome is about 6 × 105 V.
(2)
(b)
Calculate the maximum charge on the dome.
(2)
(c)
A proton is released from rest at the surface of the fully charged dome. Calculate its speed when it is very far from the dome.
(2)
(d)
Determine the speed of the proton when it is 0.60 m from the centre of the dome.
(2)
Show mark scheme
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Mark
Notes
Part (a)
At the surface, E = kQ/R2 and V = kQ/R, so V = ER
✓ 1
V = 3.0 × 106 × 0.20 = 6.0 × 105 V
✓ 1
Part (b)
Q = ER2/k = 3.0 × 106 × 0.202/8.99 × 109
✓ 1
Q = 1.3 × 10−5 C
✓ 1
Part (c)
eV = ½mv2, so v = √(2 × 1.60 × 10−19 × 6.0 × 105/1.67 × 10−27) (ECF from (a))
✓ 1
v = 1.1 × 107 m s−1
✓ 1
Part (d)
V at 0.60 m = 6.0 × 105 × 0.20/0.60 = 2.0 × 105 V, so ΔV = 4.0 × 105 V
✓ 1
v = √(2 × 1.60 × 10−19 × 4.0 × 105/1.67 × 10−27) = 8.8 × 106 m s−1 (ECF from (a))
✓ 1
Exam tip: The kinetic energy gained comes from a potential difference. Find V at both points first. Do not use E × distance, because the field is not uniform.
107IBP2H-031-01-14Field and potential of two point chargesPaper 2Hard8 marks
Two point charges are fixed on the x-axis: +3.0 nC at x = 0 and −1.0 nC at x = 0.40 m.
Diagram not to scale
(a)
Determine the position between the charges where the electric potential is zero.
(2)
(b)
Calculate the magnitude of the electric field strength at the point found in (a) and state its direction.
(3)
(c)
Outline why the field is not zero at a point where the potential is zero.
(1)
(d)
Calculate the work done by an external agent to move a +2.0 nC charge slowly from the point in (a) to x = 0.20 m.
(2)
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Notes
Part (a)
k(3.0 × 10−9)/x = k(1.0 × 10−9)/(0.40 − x)
✓ 1
x = 0.30 m
✓ 1
Part (b)
Both fields point in the +x direction there, so their magnitudes add
✓ 1
E = 8.99 × 109(3.0 × 10−9/0.302 + 1.0 × 10−9/0.102) (ECF from (a))
✓ 1
E = 1.2 × 103 V m−1 in the +x direction
✓ 1
Part (c)
Field strength depends on the potential gradient, not on the value of the potential; the gradient is not zero at this point
✓ 1
Part (d)
V at 0.20 m = 8.99 × 109(3.0 × 10−9 − 1.0 × 10−9)/0.20 = 90 V
✓ 1
W = qΔV = 2.0 × 10−9 × (90 − 0) = 1.8 × 10−7 J
✓ 1
Common error: Potentials add as scalars with their signs, but fields add as vectors. Mixing the two methods is the most common source of lost marks here.
108IBP2H-031-01-16Electric potential and field from a graphPaper 2Hard10 marks
A charged conducting sphere of radius 0.050 m is used to test insulation in an electric-vehicle charging cable. The electric potential V is measured at distances r from the centre of the sphere, all outside the sphere: r / m: 0.10, 0.20, 0.30, 0.40, 0.50 V / kV: 9.0, 4.5, 3.0, 2.25, 1.8
(a)
Show that the data are consistent with V ∝ 1/r.
(2)
(b)
Determine the charge on the sphere.
(2)
(c)
Estimate the electric field strength at r = 0.20 m, using the readings at 0.10 m and 0.30 m.
(2)
(d)
Calculate the exact field strength at r = 0.20 m and explain why the estimate in (c) is larger.
(2)
(e)
Calculate the work needed to move a +5.0 nC test charge from r = 0.50 m to r = 0.20 m.
(2)
Show mark scheme
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Notes
Part (a)
Calculates Vr for at least three rows (or uses an equivalent ratio method)
✓ 1
Vr = 900 V m every time, a constant, so V ∝ 1/r
✓ 1
Part (b)
Q = Vr/k = 900/(8.99 × 109)
✓ 1
Q = 1.0 × 10−7 C
✓ 1
Part (c)
E ≈ −ΔV/Δr = (9.0 − 3.0) × 103/0.20
✓ 1
E ≈ 3.0 × 104 V m−1
✓ 1
Part (d)
E = kQ/r2 = V/r = 4500/0.20 = 2.25 × 104 V m−1. Allow ECF from (b) if kQ/r2 is used
✓ 1
The V–r curve is steeper nearer the sphere (it curves like 1/r), so the chord from 0.10 m to 0.30 m is steeper than the tangent at 0.20 m
✓ 1
Part (e)
W = qΔV = 5.0 × 10−9 × (4.5 − 1.8) × 103
✓ 1
W = 1.35 × 10−5 J ≈ 1.4 × 10−5 J. Accept 1.3 × 10−5 J to 1.4 × 10−5 J
✓ 1
Exam tip: Field strength is the magnitude of the potential gradient. For a curved V–r graph, a chord gives only an estimate; the tangent at the point gives the exact value.
109IBP2S-030-01-16Charged conducting spherePaper 2Hard8 marks
In orbit, a spacecraft can become charged by the surrounding plasma. Model the spacecraft as a conducting sphere of radius 1.5 m. Its surface potential is −2.0 kV relative to the distant plasma. Take k = 8.99×109 N m2 C−2 and e = 1.60×10−19 C.
(a)
Calculate the charge on the spacecraft.
(2)
(b)
Determine the magnitude and direction of the electric field strength just outside the surface of the spacecraft.
(2)
(c)
Outline why the field lines outside the spacecraft are radial.
(1)
(d)
A singly charged positive oxygen ion of mass 2.66×10−26 kg starts essentially at rest far from the spacecraft. Determine the speed at which it strikes the surface.
(3)
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Notes
Part (a)
Q = Vr/k = (−2000 × 1.5) / 8.99×109 .
✓ 1
Q = −3.3×10−7 C .
✓ 1
Part (b)
E = kQ/r2 = |V|/r = 2000/1.5 = 1.3×103 V m−1 (N C−1) . Allow error carried forward from the charge calculated earlier.
✓ 1
The field is directed radially inwards, towards the centre, because the charge is negative .
✓ 1
Part (c)
Either of the following earns the mark: the charge spreads uniformly over the spherical conducting surface, so by symmetry the field outside is the same as that of a point charge at the centre; OR the conducting surface is an equipotential, so the field lines meet it at right angles, which on a sphere means they are radial .
✓ 1
Part (d)
The kinetic energy gained equals the loss in electric potential energy: eΔV = 1.60×10−19 × 2000 = 3.2×10−16 J .
✓ 1
Set ½mv2 = 3.2×10−16 J, so v2 = 2 × 3.2×10−16 / 2.66×10−26 = 2.4×1010 m2 s−2 .
✓ 1
v = 1.55×105 m s−1 (accept 1.6×105 m s−1) .
✓ 1
Exam tip: For a charge accelerated through a potential difference, use energy conservation (qΔV = ½mv²). Kinematics equations do not apply here, because the field is not uniform.
110IBP2H-031-00-12Charged sphere in a uniform fieldPaper 2Hard8 marks
In an electron-beam lithography machine, the beam is steered by a pair of parallel deflection plates. The plates are 5.0 cm long and 2.0 cm apart, with a potential difference of 150 V across them. An electron enters midway between the plates, moving parallel to them at 2.0 × 107 m s−1. Gravitational effects on the electron are negligible. Take e = 1.60 × 10−19 C and me = 9.11 × 10−31 kg.
Diagram not to scale
(a)
Calculate the electric field strength between the plates.
(1)
(b)
Calculate the acceleration of the electron between the plates.
(2)
(c)
Determine the deflection of the electron, perpendicular to the plates, as it leaves the plates. Hence deduce whether the electron leaves the plates without striking either of them.
(3)
(d)
Explain why the path of the electron between the plates is parabolic.
(2)
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Notes
Part (a)
E = V/d = 150/0.020 = 7.5 × 103 V m−1
✓ 1
Part (b)
F = eE = 1.60 × 10−19 × 7.5 × 103 = 1.2 × 10−15 N
✓ 1
a = F/m = 1.2 × 10−15/9.11 × 10−31 = 1.3 × 1015 m s−2 (ECF from (a))
✓ 1
Part (c)
Time between the plates t = 0.050/(2.0 × 107) = 2.5 × 10−9 s
✓ 1
y = ½at2 = 0.5 × 1.32 × 1015 × (2.5 × 10−9)2 = 4.1 × 10−3 m (ECF from (b))
✓ 1
The electron starts 1.0 cm from each plate, and 4.1 mm < 1.0 cm, so it leaves without striking a plate (ECF)
✓ 1
Part (d)
There is no force parallel to the plates, so the velocity parallel to the plates is constant and x ∝ t
✓ 1
There is a constant force, and so a constant acceleration, perpendicular to the plates, so y ∝ t2 and therefore y ∝ x2
✓ 1
Exam tip: Treat the two directions separately. The time comes from the motion parallel to the plates, and that time is then used for the deflection. Compare the deflection with half the plate separation, not the full separation.
111IBP2H-031-00-14Charged conducting spherePaper 2Hard9 marks
A small satellite in Earth's shadow becomes charged by the surrounding plasma. Model it as an isolated conducting sphere of radius 1.5 m whose surface potential reaches −2.0 kV. Take e = 1.60 × 10−19 C and me = 9.11 × 10−31 kg.
(a)
Calculate the charge on the satellite.
(2)
(b)
An electron leaves the surface with negligible speed and is repelled by the satellite. Calculate its speed when it is very far from the satellite.
(2)
(c)
Determine the speed of the electron when it is 1.5 m from the surface of the satellite.
(3)
(d)
The satellite carries a thin, pointed antenna. Explain why electrical discharge is most likely to start at the tip of the antenna.
112IBP2H-031-00-16Charged conducting spherePaper 2Hard9 marks
In a demonstration of how lightning conductors work, two isolated conducting spheres of radii 2.0 cm and 6.0 cm are placed far apart. The small sphere carries a charge of 40 nC and the large sphere is uncharged. The spheres are then joined by a long, thin conducting wire. Assume the wire carries negligible charge.
Spheres far apart; not to scale
(a)
Explain why charge flows along the wire, and state when the flow stops.
(2)
(b)
Show that the final charge on the small sphere is 10 nC.
(2)
(c)
Calculate the final potential of the spheres.
(2)
(d)
Determine the ratio \(\frac{\text{surface field strength of small sphere}}{\text{surface field strength of large sphere}}\). Relate your answer to the pointed shape of a lightning conductor.
(3)
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Part (a)
Initially the small sphere is at a higher potential than the large sphere, so there is a potential difference along the wire
✓ 1
Charge flows until both spheres are at the same potential
✓ 1
Part (b)
Equal potentials: kQ1/0.020 = kQ2/0.060, so Q2 = 3Q1
✓ 1
Charge is conserved: Q1 + 3Q1 = 40 nC, so Q1 = 10 nC (and Q2 = 30 nC)
✓ 1
Part (c)
V = 8.99 × 109 × 10 × 10−9/0.020
✓ 1
V = 4.5 × 103 V
✓ 1
Part (d)
Esmall = V/r = 2.2 × 105 V m−1 and Elarge = 7.5 × 104 V m−1 (ECF from (c))
✓ 1
Ratio = 3, the inverse ratio of the radii
✓ 1
At the same potential, the field is strongest where the radius of curvature is smallest, so a sharp point produces a very strong field that ionises the air
✓ 1
Exam tip: Connected conductors share a potential, not a charge. Set the potentials equal and use conservation of charge.
113IBP2S-030-03-15Charged conducting spherePaper 2Hard11 marks
An engineer models the effect of rounded and pointed conductors on high-voltage equipment. Conducting sphere A, of radius 0.30 m, is charged to a potential of 90 kV. It is then connected by a long thin wire to an uncharged conducting sphere B, of radius 0.10 m. The spheres are far apart, and any charge on the wire can be ignored.
Spheres far apart; not to scale
(a)
Calculate the initial charge on A.
(2)
(b)
Explain why charge flows along the wire and why the flow stops.
(2)
(c)
Determine the final potential of the spheres.
(3)
(d)
Calculate the electric field strength at the surface of each sphere.
(2)
(e)
Outline, with reference to your answer to (d), why charge leaks away more readily from pointed parts of a conductor.
There is a potential difference between A (90 kV) and B (0 V), so electrons flow along the wire from B to A (equivalently, conventional current flows from A to B) ; the flow stops when both spheres are at the same potential .
✓ 2
Part (c)
Charge is conserved: QA + QB = 3.0×10−6 C ; using Q = VR/k for each sphere, V(RA + RB)/k = 3.0×10−6 C, or equivalently V = 90 kV × 0.30/0.40 ; V = 6.75×104 V (67.5 kV); accept 6.7×104 V to 6.8×104 V, including 6.74×104 V from the rounded charge . ECF from (a).
✓ 3
Part (d)
EA = V/RA = 6.75×104/0.30 = 2.25×105 V m−1 ; EB = V/RB = 6.75×104/0.10 = 6.75×105 V m−1 . Accept values consistent with the answer to (c), e.g. 2.2×105 and 6.7×105 V m−1. ECF from (c).
✓ 2
Part (e)
At the same potential E = V/R, so the smaller sphere B has the larger surface field, and a sharp point (very small radius of curvature) has a much larger field still ; at a sharp point the field can exceed the breakdown field strength of air (about 3×106 V m−1), ionising the air so that charge leaks away .
✓ 2
Exam tip: Connected conductors share the same potential; they do not share the charge equally. Write the equal-potential condition first, then apply charge conservation. For spheres, V = 90 kV × RA/(RA + RB) avoids rounding errors.
114IBP2H-031-02-11Electric potential energy in the hydrogen atomPaper 2Hard10 marks
In a medical X-ray tube, electrons are released with negligible speed from a heated cathode. They are accelerated through a potential difference of 80 kV towards a tungsten target. The tube current is 5.0 mA. e = 1.60×10−19 C, me = 9.11×10−31 kg, c = 3.00×108 m s−1.
(a)
Define electric potential at a point.
(2)
(b)
Calculate the kinetic energy, in J, gained by one electron.
(1)
(c)
Calculate the speed of an electron predicted by Newtonian mechanics, and comment on your answer.
(3)
(d)
Determine the number of electrons reaching the target per second and the power delivered to the target.
(2)
(e)
The target is connected by a conductor back to the high-voltage supply. Explain, with reference to conservation of charge, why this connection is needed.
(2)
Show mark scheme
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Notes
Part (a)
The work done per unit charge
✓ 1
in bringing a small positive test charge from infinity to that point
✓ 1
Part (b)
Ek = eV = 1.60×10−19 × 80×103 = 1.28×10−14 J
✓ 1
Part (c)
v = √(2Ek/me) = √(2 × 1.28×10−14 / 9.11×10−31)
✓ 1
v = 1.7×108 m s−1 (allow error carried forward from the kinetic energy calculated in the previous part)
✓ 1
This is about 0.56c, a significant fraction of the speed of light, so the Newtonian value is not reliable and a relativistic treatment is needed; the true speed is lower
✓ 1
Part (d)
N = I/e = 5.0×10−3 / 1.60×10−19 = 3.1×1016 s−1
✓ 1
P = IV = 5.0×10−3 × 80×103 = 400 W (accept P = N × Ek = 3.125×1016 × 1.28×10−14 = 400 W)
✓ 1
Part (e)
Without the connection, arriving electrons would accumulate on the target. The target would become increasingly negative and repel further electrons, so the beam would stop
✓ 1
With the connection, electrons return through the circuit to the supply. Charge is neither created nor destroyed, so the current is the same at every point in the circuit
✓ 1
Common error: Convert kV to V before multiplying by e: 80 kV = 8.0×10⁴ V. A slip in the power of ten here carries through to every later answer. Always compare a calculated speed with c.
115IBP2H-031-02-13Magnetic field patternsPaper 2Hard9 marks
The two heavy cables, X and Y, connecting the battery of an electric bus run parallel and close together. During start-up each cable carries a current of 150 A, in opposite directions. In a plane perpendicular to the cables, use coordinates in which X passes through the origin (0, 0) and Y passes through (3.0 cm, 0). The current in X flows out of the plane towards the reader, and the current in Y flows into the plane. Point P is at (0, 4.0 cm). μ0 = 4π×10−7 T m A−1.
Plane perpendicular to the cables; not to scale
(a)
Outline the pattern of the magnetic field lines around cable X alone, in the plane described.
(2)
(b)
Calculate the magnitudes of the magnetic fields at P due to X and due to Y.
(2)
(c)
Determine the magnitude of the resultant magnetic field at P.
(3)
(d)
Explain why the magnetic field 2 m from the pair of cables is much smaller than the field 2 m from cable X alone.
(2)
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Notes
Part (a)
Concentric circles centred on X, with spacing that increases with distance from X
✓ 1
Directed anticlockwise as seen by the reader (right-hand grip rule, current towards the reader)
✓ 1
Part (b)
BX = (2×10−7×150)/0.040 = 7.5×10−4 T
✓ 1
YP = 5.0 cm, so BY = (2×10−7×150)/0.050 = 6.0×10−4 T
✓ 1
Part (c)
BX is along −x; BY is perpendicular to YP, with components 0.8×6.0×10−4 = 4.8×10−4 T along +x and 0.6×6.0×10−4 = 3.6×10−4 T along +y
✓ 1
Resultant components: −2.7×10−4 T (x) and 3.6×10−4 T (y)
✓ 1
B = √(2.72 + 3.62)×10−4 = 4.5×10−4 T; ECF from (b)
✓ 1
Part (d)
At large distances both cables are almost the same distance away, so their fields have almost equal magnitude
✓ 1
Because the currents are opposite, the fields are almost opposite in direction and nearly cancel
✓ 1
Exam tip: Magnetic fields from separate wires are vectors. Resolve each one into components before adding, and never add the magnitudes directly unless the fields are parallel.
116IBP2H-031-02-16Electric potential energy of a system of chargesPaper 2Hard8 marks
In the core of the Sun, two protons can fuse only if they come within about 1.0×10−15 m of each other. Consider two protons that start far apart and approach each other head-on with equal speeds. k = 8.99×109 N m2 C−2, e = 1.60×10−19 C, kB = 1.38×10−23 J K−1.
(a)
Show that the electric potential energy of the two protons at a separation of 1.0×10−15 m is about 2×10−13 J.
(2)
(b)
Explain why each proton needs an initial kinetic energy equal to half the electric potential energy calculated in the previous part in order to reach this separation.
(2)
(c)
Estimate the temperature at which the mean kinetic energy of a proton equals the kinetic energy each proton needs, as found in the previous part.
(2)
(d)
The temperature of the Sun's core is about 1.5×107 K. Comment on this in light of the temperature you estimated in the previous part.
(2)
Show mark scheme
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Notes
Part (a)
Ep = ke2/r = 8.99×109×(1.60×10−19)2/1.0×10−15
✓ 1
Ep = 2.3×10−13 J ≈ 2×10−13 J
✓ 1
Part (b)
The total momentum is zero (equal masses, equal and opposite velocities), so at closest approach both protons are momentarily at rest and all the kinetic energy has been converted to electric potential energy
✓ 1
By energy conservation, with initial potential energy ≈ 0 when far apart, 2Ek = Ep, so each proton needs Ek = Ep/2 ≈ 1.15×10−13 J
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Part (c)
(3/2)kBT = 1.15×10−13 J
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T = 2×1.15×10−13/(3×1.38×10−23) ≈ 5.6×109 K; allow error carried forward from earlier values
✓ 1
Part (d)
The core temperature is several hundred times (≈ 370 times) too low for the mean proton energy to overcome the electrostatic repulsion
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Fusion still occurs because some protons are in the high-energy tail of the speed distribution and/or quantum tunnelling lets them pass through the Coulomb barrier
✓ 1
Exam tip: In a head-on collision between identical particles with equal and opposite momenta, the potential energy at closest approach comes from the kinetic energy of both particles.
117IBP2S-030-00-16Field and potential of two point chargesPaper 2Hard9 marks
An electrostatic dust-removal system for a Mars rover's solar panel is modelled with two fixed point charges. Each charge is +4.0 nC, and they are 6.0 cm apart. A charged dust grain carrying +2.0 nC sits on the perpendicular bisector of the line joining them, 4.0 cm from that line. Ignore gravity and any other charges. (k = 8.99 × 109 N m2 C−2)
Diagram not to scale
(a)
Explain why the resultant electric force on the grain is directed along the perpendicular bisector.
(2)
(b)
Calculate the magnitude of the resultant electric force on the grain.
(3)
(c)
Calculate the electric potential at the position of the grain due to the two fixed charges.
(2)
(d)
Determine the minimum work needed to bring the grain from a very large distance to this position.
(2)
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Notes
Part (a)
Both repulsive forces have equal magnitude, because the charges are equal and each is the same distance from the grain
✓ 1
By symmetry, their components parallel to the line joining the charges are equal and opposite and cancel. Only the components along the bisector remain, pointing away from the line of charges
✓ 1
Part (b)
Distance from each charge to the grain: r = √(3.02 + 4.02) = 5.0 cm
✓ 1
Each force: F = kQq/r2 = 8.99 × 109 × 4.0 × 10−9 × 2.0 × 10−9/(0.050)2 = 2.9 × 10−5 N
V = 2kQ/r, adding the two contributions as scalars
✓ 1
V = 2 × 8.99 × 109 × 4.0 × 10−9/0.050 = 1.4 × 103 V
✓ 1
Part (d)
W = qΔV, taking V = 0 at infinity
✓ 1
W = 2.0 × 10−9 × 1.44 × 103 = 2.9 × 10−6 J. Allow error carried forward from the potential calculated in the previous part
✓ 1
Common error: Forces must be resolved into components before they are added, but potentials are scalars and simply add. Do not apply the 4/5 factor to the potential.
118IBP2S-030-05-16Electric potential energy in the hydrogen atomPaper 2Hard8 marks
A small spherical space probe has radius 0.80 m and mass 400 kg. Sunlight causes photoemission of electrons from its surface. As a result, the probe charges to an electric potential of +5.0 V, measured relative to a point at infinity. Treat the probe as an isolated conducting sphere and ignore the surrounding plasma.
(a)
Calculate the charge on the probe.
(2)
(b)
Calculate the number of electrons the probe has lost.
(1)
(c)
A photoelectron leaves the surface with kinetic energy 3.0 eV. Explain whether it escapes to infinity.
(2)
(d)
Determine the ratio of the magnitude of the electric potential energy to the magnitude of the gravitational potential energy for an electron at the probe's surface.
(3)
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Notes
Part (a)
V = kQ/r, so Q = Vr/k = 5.0 × 0.80 / 8.99×109
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Q = 4.4×10−10 C
✓ 1
Part (b)
n = 4.45×10−10 / 1.60×10−19 = 2.8×109, ECF from (a)
✓ 1
Part (c)
To reach infinity, the electron must do work eΔV = 5.0 eV against the electric field, i.e. it needs at least 5.0 eV of kinetic energy
✓ 1
3.0 eV < 5.0 eV, so it does not escape: it stops and returns to the probe
Common error: Potential (V) and potential energy (J) are different quantities. Multiply V by the charge e to obtain the electron's electric potential energy.
119IBP3H-D2-20r8-15Field and potential of two point chargesPaper 2Hard10 marks
Two fixed point charges, +9.0 nC at x = 0 and −4.0 nC at x = 0.50 m, lie on the x-axis. Use k = 8.99 × 109 N m2 C−2 and e = 1.60 × 10−19 C.
Diagram not to scale
(a)
Determine the x-coordinate of the point on the axis, outside the two charges, where the resultant electric field is zero.
(4)
(b)
Calculate the electric potential at this point.
(3)
(c)
Calculate the work done by an external agent in bringing a proton from infinity to this point. Use the unrounded value of the potential.
(2)
(d)
Explain why the potential is not zero although the field is zero at this point.
(1)
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Part (a)
Zero field must lie beyond the smaller charge (−4.0 nC), at distance d from it
✓ 1
9.0/(d + 0.50)2 = 4.0/d2, so 3/(d + 0.50) = 2/d
✓ 1
d = 1.0 m
✓ 1
x = 1.5 m
✓ 1
Part (b)
V = k(Q1/r1 + Q2/r2) with r1 = 1.5 m, r2 = 1.0 m
✓ 1
Sum of potentials: 8.99 × 109 × (6.0 − 4.0) × 10−9
✓ 1
V = 18.0 V (17.98 V)
✓ 1
Part (c)
W = eV
✓ 1
W = 2.88 × 10−18 J (0.00000000000000000288 J)
✓ 1
Part (d)
Field strength is the gradient of potential, not its value; the potential is at a stationary value (zero gradient) but not zero
✓ 1
Common error: Zero field does not mean zero potential: the field is the negative gradient of the potential, and the potential can have any value where that gradient vanishes.
120IBP3H-D2-20r8-16Electric potential energy of a system of chargesPaper 2Hard11 marks
Three identical small spheres, each of charge +5.0 nC, are held at the vertices of an equilateral triangle of side 0.20 m. Use k = 8.99 × 109 N m2 C−2.
Diagram not to scale
(a)
State the magnitude of the electric field at the centre of the triangle and give a reason.
(2)
(b)
Calculate the electric potential at the centre of the triangle. The centre is 0.20/√3 m from each vertex.
(3)
(c)
Calculate the total electric potential energy of the system.
(3)
(d)
The spheres each have mass 1.0 × 10−6 kg. They are released from rest and move apart symmetrically. Calculate the speed of each sphere when they are very far apart. Use the unrounded value of U.
(3)
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Notes
Part (a)
Zero
✓ 1
The three equal fields have equal magnitude at 120° to each other, so they sum to zero by symmetry
✓ 1
Part (b)
Distance from each vertex = 0.1155 m
✓ 1
V = 3kq/r (scalar sum of three equal potentials)
✓ 1
V = 1.17 × 103 V (1170 V)
✓ 1
Part (c)
Three pairs, each with energy kq2/a
✓ 1
U = 3 × 8.99 × 109 × (5.0 × 10−9)2/0.20
✓ 1
U = 3.37 × 10−6 J (0.00000337 J)
✓ 1
Part (d)
Total kinetic energy at large separation equals the initial potential energy: 3 × ½mv2 = U
✓ 1
v2 = 2U/(3m) = 2.2475 m2 s−2
✓ 1
v = 1.50 m s−1
✓ 1
Exam tip: At the centre of a symmetric arrangement the field vectors cancel but the potentials, being scalars, add; keep the two ideas separate.
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