These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: e = 1.60 × 10⁻¹⁹ C; k = 9.0 × 10⁹ N·m²/C². Metal spheres sit on insulating stands unless the question says otherwise, and “identical” spheres share charge equally on contact. Treat small spheres as point charges. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A neutral plastic rod is rubbed with a neutral cloth. Afterwards the cloth has gained 5.0 × 10¹⁰ electrons from the rod. What is the charge on the rod?
- (A) +8.0 × 10⁻⁹ C
- (B) −8.0 × 10⁻⁹ C
- (C) +3.1 × 10²⁹ C
- (D) 0, because charge is conserved
Answer and explanation
Answer: (A). The rod lost 5.0 × 10¹⁰ electrons, so it is positive: q = (5.0 × 10¹⁰)(1.60 × 10⁻¹⁹ C) = +8.0 × 10⁻⁹ C. The cloth has −8.0 × 10⁻⁹ C, so the rod-plus-cloth system still has zero net charge.
- (B) is the cloth’s charge. The object that loses electrons becomes positive.
- (C) divides by e instead of multiplying.
- (D) applies conservation to the rod alone. Conservation holds for the rod and cloth together; charge moved between them.
Question 2 (multiple choice · core)
A small, neutral piece of aluminium foil hangs from an insulating thread. A positively charged rod is brought near it, without touching. The foil swings toward the rod. Which explanation is correct?
- (A) Electrons in the foil move toward the rod, so the near side is negative and closer to the rod than the positive far side
- (B) Positive charges in the foil move away from the rod, leaving the near side negative
- (C) The rod transfers electrons to the foil through the air, giving the foil a negative charge
- (D) The foil must already have been negatively charged, because neutral objects are not attracted
Answer and explanation
Answer: (A). The rod attracts the foil’s free electrons to the near side. The near side becomes negative and the far side positive, while the foil stays neutral. By Coulomb’s law the nearer, opposite charge feels a stronger force than the farther, like charge, so the net force is toward the rod.
- (B) gives the right charge pattern for the wrong reason. In a metal, the electrons move; the positive nuclei stay put.
- (C) invents a transfer. The rod does not touch the foil, and the question describes induced separation, not charging.
- (D) is a common error. A charged object attracts neutral objects by inducing charge separation.
Question 3 (multiple choice · foundation)
Two identical metal spheres have charges +7.0 nC and −3.0 nC. They are touched together and then separated. What is the charge on each sphere?
- (A) +2.0 nC on each
- (B) +5.0 nC on each
- (C) +4.0 nC on each
- (D) +3.5 nC and −1.5 nC
Answer and explanation
Answer: (A). The total is +7.0 + (−3.0) = +4.0 nC. Identical spheres share it equally: +2.0 nC each.
- (B) adds the sizes of the charges and ignores the minus sign: (7.0 + 3.0)/2.
- (C) gives the whole total to each sphere, so the total after contact would be +8.0 nC, which breaks conservation of charge.
- (D) halves each charge separately. Charge flows until the identical spheres have equal charge, not until each has lost half.
Question 4 (multiple choice · core)
A student wants to give a neutral metal sphere a negative charge without letting a charged rod touch it. Which sequence works?
- (A) Bring a positive rod near; touch the sphere to ground; remove the ground; remove the rod
- (B) Bring a negative rod near; touch the sphere to ground; remove the ground; remove the rod
- (C) Bring a positive rod near; touch the sphere to ground; remove the rod; remove the ground
- (D) Bring a negative rod near; remove the rod; touch the sphere to ground; remove the ground
Answer and explanation
Answer: (A). The positive rod attracts electrons up from the ground into the sphere. Breaking the ground while the rod is still near traps those extra electrons. When the rod is removed, the sphere is negative. Induction gives the charge opposite to the rod’s.
- (B) makes the sphere positive: the negative rod drives electrons out to the ground.
- (C) removes the rod while the ground is still connected, so the extra electrons flow back to Earth and the sphere ends neutral.
- (D) grounds the sphere after the rod has gone, so there is no separation to trap. The sphere ends neutral.
Question 5 (calculation · core)
Two identical small metal spheres, X and Y, are 0.10 m apart. X has charge +12 nC and Y has charge −2.0 nC.
(a) Calculate the electric force between them and state whether it is attractive or repulsive. (b) The spheres are touched together and returned to 0.10 m apart. Find the new charge on each sphere and the new force. (c) How many electrons moved during the contact, and in which direction? (d) A student expects the force to be smaller after contact “because some charge cancelled”. Evaluate this claim.
Worked solution and suggested Marlbridge rubric
- (a) F = k|q_X q_Y|/r² = (9.0 × 10⁹)(12 × 10⁻⁹)(2.0 × 10⁻⁹) ÷ (0.10)² = 2.16 × 10⁻⁵ N, attractive (opposite signs).
- (b) Total = +12 + (−2.0) = +10 nC, so each sphere has +5.0 nC. F = (9.0 × 10⁹)(5.0 × 10⁻⁹)² ÷ (0.10)² = 2.25 × 10⁻⁵ N, now repulsive (same signs).
- (c) Y changed from −2.0 nC to +5.0 nC, a change of +7.0 nC, so Y lost electrons: N = 7.0 × 10⁻⁹ ÷ 1.60 × 10⁻¹⁹ = 4.4 × 10¹⁰ electrons (4.375 × 10¹⁰), moving from Y to X. Check: X changed from +12 to +5.0 nC, a change of −7.0 nC, so X gained the same number.
- (d) The claim is wrong here. Some charge did cancel: the sizes of the two charges add to 14 nC before contact and 10 nC after. But the force depends on the product of the charges: 5.0 × 5.0 = 25 (nC)² after, compared with 12 × 2.0 = 24 (nC)² before. Sharing the charge evenly raised the product, so the force grew slightly, from 2.16 × 10⁻⁵ N to 2.25 × 10⁻⁵ N, and changed from attraction to repulsion.
| Point | What earns it |
|---|---|
| 1 | (a) 2.16 × 10⁻⁵ N (2.2 × 10⁻⁵ N) and attractive |
| 1 | (b) +5.0 nC each, from adding the charges with signs and sharing equally |
| 1 | (b) 2.25 × 10⁻⁵ N (2.3 × 10⁻⁵ N) and repulsive |
| 1 | (c) 4.4 × 10¹⁰ electrons |
| 1 | (c) Direction from Y to X, with a reason (Y became more positive, so it lost electrons) |
| 1 | (d) Rejects the claim by comparing the products of the charges (or the two forces) |
Common error: saying “positive charge moved from X to Y”. The final charges are the same, but in a metal it is electrons that move. Award the direction point only for electrons moving from Y to X.
Question 6 (constructed response · core)
A metal sphere on an insulating stand and a plastic rod each carry a charge of −5.0 nC, spread over their surfaces. A student briefly touches each one at a single point with a grounded wire.
(a) Predict the charge on the metal sphere afterwards. Explain in terms of electron movement and grounding. (b) Predict, qualitatively, what happens to the charge on the plastic rod. Explain why the result differs from (a). (c) Charge is conserved, yet the sphere’s charge changed. Explain how both statements can be true.
Worked solution and suggested Marlbridge rubric
(a) The sphere ends (nearly) neutral. Its extra electrons repel each other and are free to move through the metal, so they flow out through the wire into Earth. Earth is so large that it can take them with no meaningful change in its own charge.
(b) Only the charge at or very near the touched point leaves. The rod is an insulator, so the electrons elsewhere on the rod cannot move through it to reach the wire. The rod stays negative, with almost all of its −5.0 nC.
(c) The system that conserves charge is sphere + Earth. The sphere’s −5.0 nC went into Earth. The sphere’s net charge changed because charge crossed its boundary, which conservation allows.
| Point | What earns it |
|---|---|
| 1 | (a) Sphere ends neutral (or close to it) |
| 1 | (a) Electrons flow from the sphere to ground, because they repel and can move freely in a metal |
| 1 | (b) Rod keeps most of its negative charge |
| 1 | (b) Explanation: charge carriers cannot move easily through an insulator, so only charge near the contact point is removed |
| 1 | (c) Identifies sphere + Earth as the system whose charge is conserved, or says the charge crossed the sphere’s boundary into Earth |
Do not award (a) for “the negative charge is cancelled by positive charge from the ground”: in metals and wires the moving particles are electrons.
Question 7 (constructed response · stretch)
Two identical neutral metal spheres, A and B, on insulating stands are touching each other. A positively charged rod is brought near A, on the side away from B. With the rod held still, the student pulls B away from A. Then the rod is removed.
(a) Draw a diagram of A, B and the rod just before B is pulled away. Show the charge distribution with + and − signs. (b) State the sign of the final charge on each sphere. Compare their magnitudes and justify using conservation of charge. (c) Afterwards B has charge +3.2 nC. State the charge on A. How many electrons moved, and from which sphere to which? (d) The spheres are now placed 0.050 m apart. Calculate the force between them and state its direction. Then predict what happens to the charges if the spheres are touched together again.
Worked solution and suggested Marlbridge rubric
(a) The rod is near A. The two touching spheres act as one conductor. The rod attracts electrons toward it, so the surface of A nearest the rod is negative and the far surface of B is positive. The total charge of A + B is still zero, so equal numbers of − and + signs should be drawn.
(b) A ends negative, B ends positive, with equal magnitudes. A and B started neutral, no charge left or entered the pair (the rod never touched them, and they were on insulating stands), so q_A + q_B = 0.
(c) By conservation, q_A = −q_B = −3.2 nC. B lost electrons and A gained them: N = 3.2 × 10⁻⁹ ÷ 1.60 × 10⁻¹⁹ = 2.0 × 10¹⁰ electrons, which moved from B to A, toward the positive rod.
(d) F = (9.0 × 10⁹)(3.2 × 10⁻⁹)² ÷ (0.050)² = 3.7 × 10⁻⁵ N (3.69 × 10⁻⁵ N), attractive, because the charges are opposite. If they touch again, the total is −3.2 + 3.2 = 0, so each ends neutral.
| Point | What earns it |
|---|---|
| 1 | (a) Negative charge on the side of A nearest the rod, positive on the far side of B, equal amounts |
| 1 | (b) A negative and B positive |
| 1 | (b) Equal magnitudes justified by conservation: the pair started neutral and was isolated |
| 1 | (c) q_A = −3.2 nC, from conservation of charge |
| 1 | (c) 2.0 × 10¹⁰ electrons, from B to A |
| 1 | (d) 3.7 × 10⁻⁵ N, attractive |
| 1 | (d) Both spheres neutral after touching, because the total is zero |
Carry forward a sign error from (b) into (c) and (d) once. Do not award the direction point in (c) for “positive charge moved from A to B”: the moving particles are electrons.
How did you do?
- Q1 or Q5 wrong: re-read “Charge is conserved” and “Charging by contact” in the study guide. Add charges with signs before sharing.
- Q2 wrong: revisit “Induced charge separation” and Figure 1.
- Q3 wrong: redo Worked example 1, writing the total charge first every time.
- Q4 or Q7 wrong: follow Figure 2 step by step, and remember that induction gives the opposite sign.
- Q6 wrong: compare “Grounding” with “Charging by contact” for conductors and insulators.
Then tick off the topic checklist.