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. This set is for the calculus-based course. Data for every question: k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C², e = 1.60 × 10⁻¹⁹ C and 1 nC = 10⁻⁹ C. All charges are at rest. A scientific calculator is assumed; round only at the end.
Question 1 (multiple choice · foundation)
At a point P the electric field is 4.0 × 10³ N/C directed east. An electron is placed at P. What is the electric force on the electron?
- (A) 6.4 × 10⁻¹⁶ N, east
- (B) 6.4 × 10⁻¹⁶ N, west
- (C) 2.5 × 10²² N, west
- (D) 6.4 × 10⁻¹⁶ N, but its direction cannot be found without knowing where the source charges are
Answer and explanation
Answer: (B). F = qE, so |F| = eE = (1.60 × 10⁻¹⁹ C)(4.0 × 10³ N/C) = 6.4 × 10⁻¹⁶ N. The electron’s charge is negative, so the force is opposite to E: west.
- (A) has the right size but uses the direction of E. That is only correct for a positive charge.
- (C) divides E by e instead of multiplying. Its unit would be N/C², not N.
- (D) is wrong: E already gives the direction. You do not need the sources once you know E at the point.
Question 2 (multiple choice · core)
A point charge Q produces a field of magnitude E₀ at a point 0.20 m away. A small test charge q at that point feels a force F₀. Now Q is doubled, and a test charge 2q is placed 0.40 m from Q. Which row gives the new field magnitude at the new point and the new force on the test charge?
- (A) E₀/2 and F₀
- (B) E₀ and 2F₀
- (C) E₀/2 and F₀/2
- (D) E₀/4 and F₀/2
Answer and explanation
Answer: (A). E = kQ/r². Doubling Q doubles E; doubling r divides E by 4. So E = 2/4 × E₀ = E₀/2. The force is F = (2q)(E₀/2) = qE₀ = F₀.
- (B) treats E as ∝ 1/r. In fact E ∝ 1/r².
- (C) has the correct field but forgets that the force is proportional to the test charge, which doubled.
- (D) forgets that Q was doubled. Its force, 2q × E₀/4 = F₀/2, follows correctly from that wrong field.
Question 3 (multiple choice · core)
Two spheres of the same radius R each carry a net charge +Q and are far apart. Sphere A is solid metal in electrostatic equilibrium. Sphere B is a solid insulator with its charge spread uniformly through its volume. How do the field magnitudes compare at a distance R/2 from each centre and at a distance 2R from each centre?
- (A) At R/2: zero for A, non-zero for B. At 2R: equal.
- (B) At R/2: zero for both. At 2R: equal.
- (C) At R/2: zero for A, non-zero for B. At 2R: larger for A, because its charge is on the surface, closer to the point.
- (D) At R/2: non-zero for A, zero for B. At 2R: equal.
Answer and explanation
Answer: (A). In equilibrium, A’s excess charge sits on its surface and E = 0 inside the metal. B’s charge is spread through its volume, so the field at R/2 is not zero. Outside, both distributions are spherically symmetric with net charge +Q, so each acts like a point charge at its centre: both fields are kQ/(4R²).
- (B) applies the conductor result to an insulator. An insulator can have a field inside.
- (C) gets the inside right, but outside only the net charge and the symmetry matter.
- (D) reverses the two materials.
Question 4 (calculation · core)
Two charges, each +6.0 nC, are fixed at (−0.20 m, 0) and (+0.20 m, 0). Point P is at (0, 0.15 m).
(a) Find the magnitude and direction of the net electric field at P. (b) Find the force on an electron placed at P. (c) The charge at (+0.20 m, 0) is replaced by −6.0 nC. Find the new net field at P.
Worked solution
- Each charge is √(0.20² + 0.15²) = 0.25 m from P. Each field has magnitude E = (8.99 × 10⁹)(6.0 × 10⁻⁹) ÷ (0.25)² = 863 N/C.
- (a) Each field points away from its charge. By symmetry the x-components cancel. The y-components are each 863 × (0.15/0.25) = 517.8 N/C and add: E = 1.04 × 10³ N/C (1036 N/C) in the +y direction.
- (b) F = eE = (1.60 × 10⁻¹⁹)(1036) = 1.66 × 10⁻¹⁶ N in the −y direction, opposite to E because the electron is negative.
- (c) The right-hand field now points towards its charge, so the y-components cancel and the x-components add: each is 863 × (0.20/0.25) = 690.4 N/C. E = 1.38 × 10³ N/C (1381 N/C) in the +x direction.
Suggested mark points (4): 1 for 0.25 m and 863 N/C from each charge; 1 for cancelling x-components and 1036 N/C in +y; 1 for the force with its −y direction; 1 for 1381 N/C in +x in (c).
Common error: adding 863 + 863 = 1726 N/C in (a).
Question 5 (calculation · core)
A charge q_A = +8.0 nC is fixed at x = 0 and a charge q_B = −2.0 nC is fixed at x = 0.30 m.
(a) Find the net field at the midpoint, x = 0.15 m. (b) Find the position on the x-axis, other than at infinity, where the net field is zero. (c) Explain why the field cannot be zero anywhere between the charges, or to the left of q_A.
Worked solution
- (a) At the midpoint, E_A = (8.99 × 10⁹)(8.0 × 10⁻⁹) ÷ (0.15)² = 3196 N/C in +x (away from q_A). E_B = (8.99 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.15)² = 799 N/C, also in +x (towards q_B). Net: 4.0 × 10³ N/C (3996 N/C) in the +x direction.
- (b) The fields are opposite only outside the pair, and the point must be nearer the smaller charge: to the right of q_B. Set the magnitudes equal: k(8.0 nC)/x² = k(2.0 nC)/(x − 0.30)². Taking square roots: x/(x − 0.30) = 2, so x = 0.60 m.
- Check: each field there has magnitude 200 N/C (199.8 N/C), in opposite directions.
- (c) Between the charges, E_A and E_B both point right, so they cannot cancel. To the left of q_A, they are opposite, but every point there is closer to the larger charge, so E_A always wins.
Suggested mark points (4): 1 for both midpoint fields in +x and their sum; 1 for placing the zero point right of q_B, with a reason; 1 for x = 0.60 m; 1 for explaining both excluded regions.
Common error: solving the magnitude equation for 0 < x < 0.30 m, which gives x = 0.20 m. The magnitudes are equal there (1798 N/C each), but both fields point in +x, so E = 3.6 × 10³ N/C, not zero.
Question 6 (constructed response · core)
A student wants to find the charge Q on a small metal sphere (radius 0.050 m) on an insulating stand. They have a probe (a tiny conducting ball on an insulating rod), a force sensor that holds the probe, a metre rule and a meter that measures the probe’s charge.
(a) Describe a procedure the student could use. Include what to measure, how to keep the probe acting as a test charge, and one way to reduce error. (b) With a probe charge of q₀ = +2.0 nC, the student records:
| r (m) | 0.10 | 0.15 | 0.20 | 0.25 | 0.30 |
|---|---|---|---|---|---|
| F (N) | 5.45 × 10⁻⁵ | 2.36 × 10⁻⁵ | 1.36 × 10⁻⁵ | 8.5 × 10⁻⁶ | 6.1 × 10⁻⁶ |
Calculate E at each distance. Then state which quantities to plot to get a straight line, and plot the graph with labelled axes, units and a suitable scale. (c) Use the slope of your best-fit line to find Q. (d) The student repeats the reading at r = 0.10 m with q₀ = +1.0 nC and finds F = 2.74 × 10⁻⁵ N. What does this tell you?
Worked solution and suggested Marlbridge rubric
(a) At several distances along one line, measure r from the centre of the sphere to the probe, and the force F on the probe. Measure q₀ and calculate E = F/q₀. Keep q₀ much smaller than Q so the probe does not move charge on the metal sphere; check by repeating a reading with a smaller q₀. Keep other objects away, repeat readings and average them, and re-measure q₀ in case it leaks away.
(b) E = F/q₀ gives 27 250, 11 800, 6800, 4250 and 3050 N/C. Since E = kQ/r², plot E against 1/r² (100, 44.4, 25.0, 16.0 and 11.1 m⁻²) to get a straight line through the origin. The points lie close to such a line.
(c) The best-fit slope is about 271 N·m²/C (for example, the line passes near 27 100 N/C at 100 m⁻²). Slope = kQ, so Q = 271 ÷ (8.99 × 10⁹) = 3.0 × 10⁻⁸ C (30 nC). Measuring r from the centre works because, outside a spherically symmetric sphere, the field is that of a point charge at the centre.
(d) E = (2.74 × 10⁻⁵ N) ÷ (1.0 × 10⁻⁹ C) = 2.74 × 10⁴ N/C. With 2.0 nC it was 2.725 × 10⁴ N/C, a difference of about 0.5%, which is within measurement error. Halving q₀ halved F but did not change E, so the 2.0 nC probe was small enough to act as a test charge.
| Point | What earns it |
|---|---|
| 1 | Measures F and r (from the centre) at several distances, and uses E = F/q₀ |
| 1 | A valid way to keep the probe a test charge (small q₀, checked by changing q₀) and one valid error-reduction step |
| 1 | Correct E values and the choice of E against 1/r² (or F against 1/r²), with a reason |
| 1 | Graph: labelled axes with units, sensible scale, points plotted, straight best-fit line |
| 1 | Slope from the line (not one data point) and Q ≈ 3.0 × 10⁻⁸ C |
| 1 | (d): F/q₀ unchanged, so the probe does not disturb the source |
Accept F against 1/r² (slope kQq₀ ≈ 5.4 × 10⁻⁷ N·m²). Accept Q from 2.9 × 10⁻⁸ C to 3.1 × 10⁻⁸ C.
Question 7 (constructed response · stretch)
Three point charges lie on the y-axis: +q at (0, a), +q at (0, −a) and −2q at the origin, where q > 0.
(a) Explain, using symmetry, why the net field at any point (x, 0) on the positive x-axis has no y-component. (b) Derive an expression for the net field E_x at (x, 0), with x > 0, in terms of k, q, a and x. (c) Evaluate E_x at x = a, x = 2a and x = 10a, as multiples of kq/a², and state the direction of the field. (d) For a single point charge, E(10a)/E(2a) would be (2/10)² = 0.04. Use your values from (c) to compare, and explain the difference physically.
Worked solution and suggested Marlbridge rubric
(a) The two +q charges are mirror images in the x-axis, equally far from (x, 0). Their fields are equal in size, one tilted up and one down, so the y-components cancel. The −2q charge is on the x-axis, so its field there is along the axis.
(b) Each +q is √(x² + a²) away, so each field has magnitude kq/(x² + a²). Its x-component is that times x/√(x² + a²). The −2q field points towards the origin, in −x, with size 2kq/x². So:
E_x = 2kqx/(x² + a²)^(3/2) − 2kq/x²
(c) In units of kq/a²: at x = a, E_x = 2/2^(3/2) − 2 = −1.29; at x = 2a, E_x = 4/5^(3/2) − 1/2 = −0.142; at x = 10a, E_x = 20/101^(3/2) − 0.02 = −2.96 × 10⁻⁴. All are negative, so the field points in −x, towards the origin.
(d) E(10a)/E(2a) = 2.96 × 10⁻⁴ ÷ 0.142 = 0.0021, about twenty times smaller than 0.04. The net charge is +q + q − 2q = 0, so far away the fields of the +q pair and the −2q charge nearly cancel. The small difference left falls off much faster than 1/x² (for x ≫ a, as 1/x⁴).
| Point | What earns it |
|---|---|
| 1 | Symmetry argument for cancelling y-components, with the field of −2q along the axis |
| 1 | Correct x-component of each +q field, kqx/(x² + a²)^(3/2) |
| 1 | Correct −2kq/x² term with the right sign, giving the full expression |
| 1 | All three values correct, with the direction (towards the origin) |
| 1 | Ratio about 0.002 compared with 0.04 |
| 1 | Physical reason: zero net charge, so the far field falls faster than a point charge’s |
Carry forward an error in (b) once. The 1/x⁴ statement is background, not needed for credit.
How did you do?
- Q1 or Q2 wrong: re-read “What an electric field is” and “The field of a point charge” in the study guide.
- Q3 wrong: revisit “Fields of charged conductors and insulators”.
- Q4, Q5 or Q7 wrong: work through Worked example 2 again and the five-step superposition method.
- Q6 incomplete: see “Measuring a field” and Worked example 3.
Then tick off the topic checklist.