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. All data are invented for practice. Data for every question: e = 1.60 × 10⁻¹⁹ C, 1/(4πε₀) = 8.99 × 10⁹ N·m²/C² and 1 nC = 10⁻⁹ C. Stands and threads are insulating unless a question says otherwise. A scientific calculator is assumed; round only at the end.
Question 1 (multiple choice · foundation)
A rod and a cloth are both neutral. After the rod is rubbed with the cloth, the rod’s charge is −6.4 nC. Which statement is correct?
- (A) The cloth’s charge is +6.4 nC, and about 4.0 × 10¹⁰ electrons moved from the cloth to the rod.
- (B) The cloth’s charge is −6.4 nC, because both objects were rubbed in the same way.
- (C) The cloth’s charge is +6.4 nC, and about 4.0 × 10¹⁰ protons moved from the rod to the cloth.
- (D) The cloth is still neutral, because rubbing created the charge on the rod.
Answer and explanation
Answer: (A). The rod and cloth together form a system with zero net charge. Nothing else is involved, so the cloth must carry +6.4 nC. The rod became negative by gaining electrons: N = (6.4 × 10⁻⁹ C) ÷ (1.60 × 10⁻¹⁹ C) = 4.0 × 10¹⁰.
- (B) breaks conservation of charge: the total would change from 0 to −12.8 nC.
- (C) has the right charges but the wrong carrier. Protons are bound in nuclei; in solids, electrons move.
- (D) treats charge as created. Charge is only moved, so the cloth must carry the opposite charge.
Question 2 (multiple choice · core)
Two identical metal spheres carry +7.0 nC and −3.0 nC. They are touched together and then separated. What is the final 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 charge is (+7.0) + (−3.0) = +4.0 nC. It is conserved, and identical conductors share it equally: +2.0 nC each.
- (B) averages the sizes, (7.0 + 3.0) ÷ 2, ignoring the negative sign.
- (C) gives each sphere the whole total, so the system would now hold +8.0 nC. That breaks conservation.
- (D) halves each charge separately. The total would then be +2.0 nC, and the spheres would not end with equal charges.
Question 3 (multiple choice · core)
A small, light, neutral ball of aluminium foil hangs from an insulating thread. A positively charged rod is brought near it but does not touch it. What happens?
- (A) The ball moves towards the rod, because electrons gather on the near side of the ball.
- (B) The ball moves away from the rod, because the rod induces a positive charge on the ball.
- (C) Nothing happens, because the ball has no net charge.
- (D) The ball moves towards the rod, because electrons jump through the air from the rod onto the ball.
Answer and explanation
Answer: (A). The foil is a conductor. The rod attracts its free electrons to the near side and leaves the far side positive. The attracted (negative) charge is closer to the rod than the repelled (positive) charge. Because the force falls as 1/r², the attraction wins, and the ball moves towards the rod.
- (B) Induction separates charge but does not change the ball’s net charge, which stays zero. A neutral object is not repelled.
- (C) Zero net charge does not mean zero force. The charge distribution changes.
- (D) The rod is positive, so it has a shortage of electrons; it would not give electrons to the ball. No contact happens, so the ball’s net charge does not change.
Question 4 (calculation · core)
Three identical metal spheres start with charges A = +9.0 nC, B = −3.0 nC and C = 0. In order: A touches B; then B touches C; then A touches C. Each pair is separated before the next contact.
(a) Find the final charge on each sphere. (b) Show that charge is conserved. (c) In the first contact, which way did electrons move, and how many?
Worked solution
- (a) A touches B: total +6.0 nC, so A = B = +3.0 nC.
- B touches C: total +3.0 + 0 = +3.0 nC, so B = C = +1.5 nC.
- A touches C: total +3.0 + 1.5 = +4.5 nC, so A = C = +2.25 nC.
- Final: A = +2.25 nC, B = +1.5 nC, C = +2.25 nC.
- (b) Start: 9.0 − 3.0 + 0 = +6.0 nC. End: 2.25 + 1.5 + 2.25 = +6.0 nC. The totals match.
- (c) A changed from +9.0 nC to +3.0 nC, a change of −6.0 nC, so A gained electrons; they came from B. N = (6.0 × 10⁻⁹) ÷ (1.60 × 10⁻¹⁹) = 3.75 × 10¹⁰ ≈ 3.8 × 10¹⁰ electrons, from B to A.
Suggested mark points (4): 1 for adding charges before sharing at each step; 1 for all three final charges; 1 for the conservation check; 1 for the electron number with direction B → A.
Common error: saying positive charge moved from A to B. In metals, electrons move; A became less positive by gaining electrons.
Question 5 (calculation · core)
A neutral metal-coated ball is near a rod with charge +4.0 nC. Treat the rod as a point charge. The ball’s induced charges can be modelled as −0.50 nC at 0.040 m from the rod and +0.50 nC at 0.060 m from the rod, on the same line.
(a) Calculate the force from the rod on each induced charge. (b) Find the net force on the ball and state its direction. (c) Explain why a negatively charged rod would also attract the ball.
Worked solution
- (a) Near charge (attracted): F = kQq/r² = (8.99 × 10⁹)(4.0 × 10⁻⁹)(0.50 × 10⁻⁹) ÷ (0.040)² = 1.1 × 10⁻⁵ N towards the rod.
- Far charge (repelled): F = (8.99 × 10⁹)(4.0 × 10⁻⁹)(0.50 × 10⁻⁹) ÷ (0.060)² = 5.0 × 10⁻⁶ N away from the rod.
- (b) Net force = 1.124 × 10⁻⁵ − 0.499 × 10⁻⁵ = 6.2 × 10⁻⁶ N towards the rod. The forces differ by the factor (0.060/0.040)² = 2.25 because the charges are at different distances.
- (c) A negative rod repels electrons to the far side, so the near side becomes positive. The charge of opposite sign to the rod is again the closer one, so attraction again outweighs repulsion.
Suggested mark points (4): 1 for each correct force with direction (2 in total); 1 for the net force 6.2 × 10⁻⁶ N towards the rod; 1 for explaining (c) with the “opposite charge is closer” argument.
Common error: adding the two forces. They act in opposite directions, so you subtract.
Question 6 (constructed response · core)
A student has a positively charged glass rod, a neutral metal sphere on an insulating stand and a wire that can connect the sphere to a water pipe that is grounded.
(a) Draw (or describe in words) the charge distribution on the sphere when the rod is held near it, without the wire. (b) Describe the steps that will give the sphere a net charge without the rod touching it. For each step, say how electrons move. (c) State the sign of the sphere’s final charge and explain how conservation of charge applies to the process. (d) The student repeats the experiment, but removes the rod before disconnecting the wire. Predict the sphere’s final charge and explain.
Worked solution and suggested Marlbridge rubric
(a) Negative charge on the side nearer the rod, an equal positive charge on the far side. Net charge zero.
(b) (1) Hold the rod near the sphere. (2) Connect the wire from the sphere to the grounded pipe while the rod stays near: the positive rod attracts electrons, which flow up from Earth into the sphere. (3) Disconnect the wire while the rod is still near, so the extra electrons cannot leave. (4) Remove the rod; the extra electrons spread over the sphere.
(c) The final charge is negative, opposite to the rod. Charge is conserved because the electrons the sphere gained came from Earth; Earth’s charge changes by the same amount with the opposite sign. The rod’s charge does not change, because it never touches the sphere.
(d) The sphere ends neutral. With the rod gone, nothing holds the extra electrons, and the wire is still connected, so they flow back to Earth.
| Point | What earns it |
|---|---|
| 1 | (a) Negative near side, positive far side, with the net charge stated as zero |
| 1 | (b) Ground connected while the rod is near, with electrons flowing from Earth to the sphere |
| 1 | (b) Ground removed before the rod |
| 1 | (c) Negative final charge, with charge gained from Earth (conservation) |
| 1 | (d) Neutral, with the reason that electrons return to Earth through the wire |
Accept a correctly labelled diagram for (a) and (b) instead of words. Do not award the (b) points if the rod touches the sphere.
Question 7 (constructed response · stretch)
A student claims: “When a rod is rubbed with a cloth, the charge gained by the rod is equal and opposite to the charge gained by the cloth.” She rubs a rod with a cloth four times, starting with both neutral each time, and measures each charge with a charge sensor. Each reading has an uncertainty of ±0.2 nC. Her data (invented for this question) are below.
| Trial | Rod charge (nC) | Cloth charge (nC) |
|---|---|---|
| 1 | −5.1 | +4.9 |
| 2 | −3.8 | +3.9 |
| 3 | −6.2 | +6.1 |
| 4 | −4.4 | +2.6 |
(a) For each trial, calculate the sum of the two charges. (b) Does the data from trials 1 to 3 support the claim? Justify your answer using the uncertainty. (c) In trial 4, the student held the cloth in her bare hand. Use conservation of charge to explain the result. (d) Suggest one change to the procedure that would test your explanation in (c).
Worked solution and suggested Marlbridge rubric
(a) Sums: trial 1, −0.2 nC; trial 2, +0.1 nC; trial 3, −0.1 nC; trial 4, −1.8 nC.
(b) Yes. Each reading may be off by up to 0.2 nC, so a sum of two readings may be off by up to 0.4 nC even if the true sum is zero. All three sums (−0.2, +0.1 and −0.1 nC) are within ±0.4 nC of zero. The data are consistent with equal and opposite charges.
(c) The sum, −1.8 nC, is far outside ±0.4 nC, so the rod-and-cloth system did not keep zero net charge. Charge must have crossed the system boundary. The cloth (+2.6 nC) is less positive than the rod’s −4.4 nC suggests, so about 1.8 nC of negative charge, roughly 1.1 × 10¹⁰ electrons, must have reached the cloth from outside. The student’s hand is a conducting path to her body, a large system that can supply electrons. Charge is still conserved for the larger system of rod, cloth and student.
(d) For example: hold the cloth with an insulating handle or glove and repeat trial 4. If the sum returns to within ±0.4 nC of zero, the hand was the leak. (Repeating several trials each way would make the comparison stronger.)
| Point | What earns it |
|---|---|
| 1 | All four sums correct |
| 1 | Combines the uncertainties to ±0.4 nC (or argues that each sum is within the reading uncertainty) |
| 1 | Concludes that trials 1–3 support the claim, using the comparison |
| 1 | Trial 4: identifies that charge crossed the boundary of the rod-and-cloth system |
| 1 | Electrons flowed from the hand/body to the cloth, making it less positive; charge conserved for the larger system |
| 1 | A sensible test that isolates the cloth from the hand, with a stated expected result |
Accept “about 1.8 nC moved” without the electron count. Do not accept “charge was destroyed” or “the sensor is wrong” without evidence for (c).
How did you do?
- Q1 or Q2 wrong: re-read “Charge is conserved” and Worked example 1 in the study guide.
- Q3 or Q5 wrong: revisit “Induced charge separation (polarization)”.
- Q4 wrong: add the charges before sharing, one contact at a time, as in Worked example 1.
- Q6 incomplete: work through Figure 1 and Worked example 2 again, with a positive rod instead of a negative one.
- Q7 incomplete: practise comparing a result with its uncertainty, then choosing the system and its boundary.
Then tick off the topic checklist.