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 and assumptions for every question: e = 1.60 × 10⁻¹⁹ C; electron mass m_e = 9.11 × 10⁻³¹ kg; proton mass m_p = 1.67 × 10⁻²⁷ kg; k = 8.99 × 10⁹ N·m²/C². Ignore gravity. Any charges that create the field are fixed in place. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A particle with charge −3.0 μC moves from a point where the potential is +20 V to a point where the potential is −40 V. What is the change in the electric potential energy of the system?
- (A) −1.8 × 10⁻⁴ J
- (B) +6.0 × 10⁻⁵ J
- (C) +1.8 × 10⁻⁴ J
- (D) +1.2 × 10⁻⁴ J
Answer and explanation
Answer: (C). ΔV = V_f − V_i = −40 V − (+20 V) = −60 V. Then ΔU_E = qΔV = (−3.0 × 10⁻⁶ C)(−60 V) = +1.8 × 10⁻⁴ J. A negative charge moving to lower potential gains potential energy.
- (A) has the right size but drops the sign of the charge. That is the result for a positive charge.
- (B) adds the two potentials (−40 V + 20 V = −20 V) instead of subtracting them.
- (D) uses only the final potential, qV_f, which is the final potential energy, not the change.
Question 2 (multiple choice · core)
Particle X has charge +q and mass m. Particle Y has charge +2q and mass 8m. Each starts from rest and is accelerated through the same potential difference. Which statement compares Y with X correctly?
- (A) Y has the same kinetic energy and about 0.35 times the speed.
- (B) Y has twice the kinetic energy and half the speed.
- (C) Y has twice the kinetic energy and about 1.4 times the speed.
- (D) Y has twice the kinetic energy and one quarter of the speed.
Answer and explanation
Answer: (B). The kinetic energy gained is |q||ΔV|, so doubling the charge doubles K. Mass does not affect the energy gained. The speed is v = √(2|q||ΔV|/m), so v_Y/v_X = √[(2q/8m) ÷ (q/m)] = √(1/4) = 0.5.
- (A) assumes every particle gains the same energy, giving √(1/8) ≈ 0.35. Equal energy is only true for equal charges.
- (C) gets K right, then forgets that Y’s mass is eight times larger: √2 ≈ 1.4.
- (D) forgets the square root: it uses v ∝ q/m, giving 2/8 = 1/4.
Question 3 (multiple choice · core)
An electron is released from rest next to the negative plate of a pair of parallel plates and moves across the uniform field to the positive plate. Which graph best describes its kinetic energy K against the distance x it has travelled?
- (A) A curve through the origin that rises steeply at first and then flattens (like y = √x).
- (B) A curve through the origin that starts flat and gets steeper (like y = x²).
- (C) A straight line that starts high and falls to zero.
- (D) A straight line through the origin with positive slope.
Answer and explanation
Answer: (D). In a uniform field V changes linearly with x, so ΔU_E = qΔV is proportional to x. With only the electric force acting, K = −ΔU_E, so K is proportional to x: a straight line through the origin.
- (A) is the shape of speed against x, because v ∝ √K ∝ √x.
- (B) is the shape of K against time. The particle covers more distance each second, so K rises faster and faster in time.
- (C) is the shape of the electric potential energy (measured from its final value), not K.
Question 4 (multiple choice · core)
A charged particle moves only under electric forces from point P to point Q. P and Q are both on the 60 V equipotential line. On the way, the particle passes through a region where the potential is 100 V. How does its speed at Q compare with its speed at P?
- (A) It is the same.
- (B) It is larger, because the particle passed through a region of higher potential.
- (C) It is smaller, because the field did work on the particle along the curved path.
- (D) It cannot be found without knowing the sign of the charge.
Answer and explanation
Answer: (A). Between P and Q, ΔV = 60 V − 60 V = 0, so ΔU_E = 0 and ΔK = 0. The particle may speed up or slow down in between, but it returns to the same kinetic energy at Q.
- (B) uses a point in the middle of the path. Only the start and end potentials decide the total change.
- (C) treats the work done by the electric field as path dependent. It is not: the net work from P to Q is −qΔV = 0.
- (D) is tempting because the sign of q decides whether the particle speeds up or slows down on the way. But qΔV = 0 for any q, so the end speeds are equal either way.
Question 5 (calculation · core)
An electron passes through a small hole in plate A, which is at 0 V, moving at 4.0 × 10⁶ m/s straight toward plate B, which is at −30 V.
(a) Calculate the change in electric potential energy of the system as the electron travels from A to B. (b) Calculate the speed of the electron when it reaches plate B. (c) What is the most negative potential that plate B could have and the electron still reach it?
Worked solution
- (a) ΔV = −30 V − 0 V = −30 V. ΔU_E = qΔV = (−1.60 × 10⁻¹⁹ C)(−30 V) = +4.80 × 10⁻¹⁸ J. The negative electron moves to lower potential, so U_E rises.
- (b) Initial kinetic energy: K_A = ½(9.11 × 10⁻³¹ kg)(4.0 × 10⁶ m/s)² = 7.288 × 10⁻¹⁸ J.
- Energy conservation: K_B = K_A − ΔU_E = 7.288 × 10⁻¹⁸ J − 4.80 × 10⁻¹⁸ J = 2.488 × 10⁻¹⁸ J.
- v_B = √(2K_B/m) = √[2(2.488 × 10⁻¹⁸ J) ÷ (9.11 × 10⁻³¹ kg)] = 2.34 × 10⁶ m/s.
- (c) The electron just reaches B if all of K_A becomes U_E: e|ΔV| = K_A, so |ΔV| = 7.288 × 10⁻¹⁸ J ÷ 1.60 × 10⁻¹⁹ C = 45.55 V. Plate B can be at most about 45.6 V below plate A, that is at about −45.6 V (more exactly −45.55 V).
Suggested mark points (4): 1 for ΔU_E = +4.80 × 10⁻¹⁸ J with the correct sign; 1 for K_A = 7.29 × 10⁻¹⁸ J; 1 for v_B = 2.34 × 10⁶ m/s; 1 for −45.6 V (accept “45.6 V lower than A”).
Common error: adding ΔU_E to K_A, which gives 5.15 × 10⁶ m/s. The electron is moving to lower potential, against the direction it would move if released, so it slows down.
Question 6 (constructed response · core)
A small object with charge +q and mass m is held at rest a distance r₀ from a fixed point charge +Q, then released. Only electric forces act.
(a) Derive an expression for the speed v of the object when it is a distance r from the fixed charge (r > r₀). Write it in terms of q, Q, m, r₀, r and physical constants. (b) Hence write the speed the object approaches when it is very far away. (c) A second, identical object is released from rest at r₀/2 instead. By what factor is its speed when very far away larger than that of the first object? Justify your answer. (d) On one set of axes, sketch the electric potential energy U_E and the kinetic energy K of the first object against r, from r₀ to large r. Label each curve.
Worked solution and suggested Marlbridge rubric
(a) Energy conservation for the object–charge system: K_i + U_i = K_f + U_f. With K_i = 0,
0 + kqQ/r₀ = ½mv² + kqQ/r, so v = √[(2kqQ/m)(1/r₀ − 1/r)].
The term in brackets is positive for r > r₀, so v is real, as it should be.
(b) As r → ∞, 1/r → 0, so v_∞ = √(2kqQ/(m r₀)).
(c) v_∞ ∝ 1/√r₀. Halving r₀ doubles the starting potential energy, so it doubles the kinetic energy when far away. The speed is larger by a factor of √2 ≈ 1.41.
(d) U_E starts at kqQ/r₀ at r = r₀ and falls along a 1/r curve toward zero, never reaching it. K starts at zero at r₀ and rises toward kqQ/r₀, flattening out. At every r, U_E + K = kqQ/r₀ (a horizontal line, if drawn). The two curves are mirror images about half that value.
| Point | What earns it |
|---|---|
| 1 | Starts from energy conservation, with K_i = 0 and U_E = kqQ/r at both positions |
| 1 | Correct expression for v in (a), algebraically equivalent forms accepted |
| 1 | v_∞ = √(2kqQ/(m r₀)) in (b), with the reason that U_E → 0 far away |
| 1 | Factor √2 in (c), linked to v ∝ 1/√r₀ or to the doubled starting energy |
| 1 | Sketch: U_E a decreasing curve from kqQ/r₀ toward zero; K an increasing curve from zero that levels off |
| 1 | Sketch shows (or states) that U_E + K is constant at every r |
Accept a (c) answer that compares the energies numerically with chosen values, provided the factor √2 for speed is stated. Do not award the (c) point for “2” (that is the energy ratio, not the speed ratio).
Question 7 (constructed response · stretch)
Three large parallel plates stand side by side. Plate 1 is at +500 V, plate 2 (in the middle, with a small hole) is at 0 V and plate 3 is at +300 V. A proton is released from rest at plate 1.
(a) Calculate the speed of the proton as it passes through the hole in plate 2. (b) Does the proton reach plate 3? If so, calculate its speed there. (c) Sketch the kinetic energy of the proton against position from plate 1 to plate 3. Describe the shape in each gap. (d) A student says: “If plate 3 were moved twice as far from plate 2, keeping all the potentials the same, the proton would arrive at plate 3 with a lower speed, because the field would slow it down over a longer distance.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) From plate 1 to plate 2, ΔV = 0 − 500 V = −500 V, so ΔK = −qΔV = −(1.60 × 10⁻¹⁹ C)(−500 V) = 8.00 × 10⁻¹⁷ J. v₂ = √[2(8.00 × 10⁻¹⁷ J) ÷ (1.67 × 10⁻²⁷ kg)] = 3.10 × 10⁵ m/s.
(b) From plate 1 to plate 3 overall, ΔV = 300 V − 500 V = −200 V. The proton ends at a lower potential than it started, so it still has kinetic energy there: yes, it reaches plate 3. K₃ = (1.60 × 10⁻¹⁹ C)(200 V) = 3.20 × 10⁻¹⁷ J, so v₃ = 1.96 × 10⁵ m/s. (Equivalently, from plate 2 it climbs 300 V and loses 4.80 × 10⁻¹⁷ J of its 8.00 × 10⁻¹⁷ J.)
(c) Gap 1–2: K rises in a straight line from 0 to 8.00 × 10⁻¹⁷ J (uniform field, so V and K are linear in position). Gap 2–3: K falls in a straight line to 3.20 × 10⁻¹⁷ J. The graph is an upside-down “V” shape, peaking at plate 2, with straight segments.
(d) The claim is wrong. The change in K depends only on the potential difference between the start and end points: K₃ = e(500 V − 300 V) whatever the spacing. With a larger gap the field between plates 2 and 3 is weaker (E = ΔV/d is halved), so the retarding force is halved while the distance doubles. The work done by the field, force × distance, is unchanged. The proton still arrives at 1.96 × 10⁵ m/s; it just takes longer.
| Point | What earns it |
|---|---|
| 1 | Uses ΔK = −qΔV with ΔV = −500 V for 1 → 2 |
| 1 | v₂ = 3.10 × 10⁵ m/s |
| 1 | States the proton reaches plate 3 because V₃ < V₁ (or because it has enough K to climb 300 V) |
| 1 | v₃ = 1.96 × 10⁵ m/s |
| 1 | Sketch: linear rise then linear fall, peak at plate 2, final K above zero |
| 1 | Rejects the claim: K₃ depends only on ΔV, not on distance |
| 1 | Supports the rejection: weaker field over a longer distance gives the same work (E halves, d doubles) |
Accept answers in electronvolts for (a) and (b) if converted correctly to a speed. Carry forward an error from (a) into the second method of (b) once.
How did you do?
- Q1 or Q5 wrong: re-read “From potential to potential energy” in the study guide. Write down the sign of q before you start.
- Q2 wrong: revisit “How fast? The speed after a potential difference” and the X and Y comparison.
- Q3 or Q6(d) wrong: look again at “Energy graphs” and Figure 2.
- Q4 or Q7(d) wrong: remember that only the end-point potentials matter, never the path or the distance.
- Q6 incomplete: work through Worked example 2 again, then redo the derivation with symbols only.
Then tick off the topic checklist.