What this is for. This diagnostic shows which Unit 10 topics to revisit, with at least one short question per topic. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your result is not a predicted score.
How to take it. Work without notes for about 30 minutes, and write each answer before opening the explanation. A scientific calculator is assumed. Data for every question: k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C², ε₀ = 8.85 × 10⁻¹² C²/(N·m²) and e = 1.60 × 10⁻¹⁹ C. Take V = 0 far away and at ground. All conductors are in electrostatic equilibrium, and all data are invented for practice.
Question 1 (multiple choice · 10.1)
A thin, hollow metal sphere of radius 0.20 m carries +4.0 nC and is far from other charges. Point P is inside the empty sphere, 0.10 m from its centre. Which row gives the field and the potential at P?
- (A) E = 0 and V = 0
- (B) E = 3.6 × 10³ N/C and V = 360 V
- (C) E = 0 and V = 180 V
- (D) E = 0 and V = 360 V
Answer and explanation
Answer: (C). The empty cavity of a conductor has no field. E = 0 means V is constant, so V at P equals the sphere’s potential: kQ/R = (8.99 × 10⁹)(4.0 × 10⁻⁹) ÷ 0.20 = 180 V.
- (A) Zero field means constant V, not zero V.
- (B) uses the point-charge formulas at r = 0.10 m. That model works only outside the sphere.
- (D) has the right field but uses kQ/r with r = 0.10 m.
If you missed this: read “Four results for a conductor in equilibrium” in the 10.1 study guide.
Question 2 (multiple choice · 10.1)
A hollow metal object of irregular shape carries a net charge of +5.0 nC. A −2.0 nC point charge hangs inside its cavity on an insulating thread. A +3.0 nC charge is then brought close to the outside of the object. Which statement is correct?
- (A) The cavity wall carries no charge and the outer surface carries all +5.0 nC.
- (B) The cavity wall carries +2.0 nC and the outer surface +7.0 nC.
- (C) The cavity wall carries less than +2.0 nC, because the outside charge pulls some of it to the outer surface.
- (D) The cavity wall carries +2.0 nC and the outer surface +3.0 nC, spread unevenly; the field in the cavity is not changed by the outside charge.
Answer and explanation
Answer: (D). A Gaussian surface inside the metal encloses no net charge, so the wall carries +2.0 nC. The rest, 5.0 − 2.0 = +3.0 nC, sits on the outer surface. The outside charge rearranges only the outer charge; the metal shields the cavity.
- (A) forgets the induced wall charge. Without it, there would be a field in the metal.
- (B) adds 2.0 nC instead of subtracting it; the total must stay +5.0 nC.
- (C) Gauss’s law in the metal fixes the wall charge at +2.0 nC whatever happens outside.
If you missed this: read “Cavities and electrostatic shielding” in the 10.1 study guide.
Question 3 (multiple choice · 10.2)
Metal sphere A, of radius R, is charged until its potential is V₀. It is then joined by a long thin wire to a distant, neutral metal sphere B of radius 2R. What is the final potential of the two spheres?
- (A) V₀/2
- (B) V₀/3
- (C) V₀
- (D) 2V₀/3
Answer and explanation
Answer: (B). The spheres end at equal potential, so q ∝ R: A keeps one third of the charge and B takes two thirds. A’s potential is k(Q/3)/R = V₀/3. Check with B: k(2Q/3)/(2R) = V₀/3.
- (A) shares the charge equally. B would then be at only V₀/4, so charge would still flow.
- (C) Charge left A, so its potential fell.
- (D) divides B’s charge, 2Q/3, by A’s radius.
If you missed this: read “Two spheres joined by a long wire” in the 10.2 study guide.
Question 4 (multiple choice · 10.2)
A neutral metal sphere stands on an insulating base. A positively charged rod is held near it, and the sphere is connected to ground. Next, the rod is taken away. Only then is the ground wire removed. What is the final charge on the sphere?
- (A) Zero
- (B) Negative, because electrons came up from ground while the rod was near
- (C) Positive, because the rod’s charge was shared with the sphere
- (D) Negative, but smaller than the charge induced while the rod was near
Answer and explanation
Answer: (A). While the rod is near, electrons flow up from ground. But when the rod is removed with the ground still connected, the sphere must return to V = 0 with no charge nearby, so Q = 0: the electrons flow back.
- (B) needs the other order: break the ground connection first, then remove the rod.
- (C) The rod never touched the sphere.
- (D) Once the rod has gone, nothing holds induced charge on a grounded sphere.
If you missed this: read “Charging by grounding” in the 10.2 study guide.
Question 5 (short answer · 10.1 and 10.2)
An isolated metal sphere of radius 0.15 m is far from other charges. A field probe just outside its surface reads 2.4 × 10⁴ N/C, pointing towards the sphere.
(a) Find the surface charge density and the total charge on the sphere. (b) Find the potential of the sphere. (c) The sphere is connected to ground. State the final charge, and find the number of electrons that move and their direction.
Answer and explanation
(a) Just outside a conductor, E = σ/ε₀. The field points inward, so σ is negative: σ = −ε₀E = −(8.85 × 10⁻¹²)(2.4 × 10⁴) = −2.1 × 10⁻⁷ C/m² (−2.12 × 10⁻⁷). Q = σ(4πR²) = (−2.124 × 10⁻⁷)(4π × 0.15²) = −6.0 × 10⁻⁸ C (−60 nC).
(b) V = kQ/R = (8.99 × 10⁹)(−6.0 × 10⁻⁸) ÷ 0.15 = −3.6 × 10³ V. (Check: |V| = ER.)
(c) Grounding with nothing nearby makes V = 0, so Q = 0. The excess electrons leave: (6.0 × 10⁻⁸) ÷ (1.60 × 10⁻¹⁹) = 3.8 × 10¹¹ electrons, from the sphere to ground.
Check yourself: 1 mark each for σ with its sign, Q, V, and the electron count with direction (4 in total). Using σ/(2ε₀) doubles Q: that is the thin-sheet formula, not the conductor result.
If you missed this: for (a)–(b), read the 10.1 study guide; for (c), “Ground: the ideal reference” in the 10.2 study guide.
Question 6 (multiple choice · 10.3)
A coaxial cable 10 m long has an inner wire of radius 0.50 mm and an outer conducting tube of radius 2.0 mm. Its capacitance is C. A second cable of the same length has radii 1.0 mm and 4.0 mm. What is its capacitance?
- (A) 2C
- (B) 4C
- (C) C/2
- (D) C
Answer and explanation
Answer: (D). C = 2πε₀L/ln(b/a). Both cables have b/a = 4 and the same L, so the same capacitance (about 401 pF).
- (A) assumes C grows with size, like an isolated sphere’s; but the gap grows too.
- (B) assumes area grows as the square of the radius. For a cylinder of fixed length, area grows only as r.
- (C) looks only at the gap, which doubles, and forgets the larger surfaces.
If you missed this: read “A method for any capacitor” and Worked example 2 in the 10.3 study guide.
Question 7 (short answer · 10.3)
A parallel-plate capacitor in air has plates of area 0.040 m², 2.0 mm apart. A probe shows that the field between the plates is 1.5 × 10⁴ V/m.
(a) Find the charge density on each plate and the charge Q on each plate. (b) Find the potential difference and the capacitance, and check that C = Q/ΔV. (c) Find the stored energy. (d) The charge on the plates is doubled without changing the geometry. By what factor does the stored energy change? Sketch the shape of a graph of U against Q.
Answer and explanation
(a) Between the plates E = σ/ε₀, so σ = (8.85 × 10⁻¹²)(1.5 × 10⁴) = 1.33 × 10⁻⁷ C/m², positive on one plate and negative on the other. Q = σA = 5.3 × 10⁻⁹ C (5.31 nC).
(b) ΔV = Ed = (1.5 × 10⁴)(2.0 × 10⁻³) = 30 V. C = ε₀A/d = (8.85 × 10⁻¹²)(0.040) ÷ (2.0 × 10⁻³) = 1.77 × 10⁻¹⁰ F (177 pF). Check: Q/ΔV = 5.31 × 10⁻⁹ ÷ 30 = 1.77 × 10⁻¹⁰ F.
(c) U = ½QΔV = ½(5.31 × 10⁻⁹)(30) = 8.0 × 10⁻⁸ J.
(d) C is fixed, so U = Q²/(2C): doubling Q makes U four times larger. The U–Q graph is a parabola through the origin, curving upward.
Check yourself: 1 mark each for σ and Q, ΔV and C, U, and the factor 4 with a parabola (4 in total).
If you missed this: read “The parallel-plate capacitor” and “Energy stored in a capacitor” in the 10.3 study guide.
Question 8 (multiple choice · 10.4)
An air-filled parallel-plate capacitor stays connected to a battery. The free charge density on its positive plate is σ₀. A slab with κ = 4 is then slid in to fill the gap. What is the size of the bound charge density on the face of the slab next to the positive plate?
- (A) 3σ₀/4
- (B) 3σ₀
- (C) 4σ₀
- (D) σ₀/4
Answer and explanation
Answer: (B). ΔV is fixed and C rises by κ, so the free charge rises to 4σ₀. Then σᵢ = σ(1 − 1/κ) = 4σ₀ × ¾ = 3σ₀. Check: the net field, (4σ₀ − 3σ₀)/ε₀, is unchanged, as it must be with ΔV and d fixed.
- (A) uses the old free charge σ₀. That is right for an isolated capacitor, not a connected one.
- (C) is the new free charge on the plate, not the bound charge on the slab.
- (D) is the part of the free charge left uncancelled, σ/κ, in the isolated case.
If you missed this: read “The induced field and the dielectric constant” and “Isolated or connected” in the 10.4 study guide.
Question 9 (short answer · 10.3 and 10.4)
A spherical capacitor has an inner sphere of radius 0.050 m and a thin outer shell of radius 0.060 m. The gap is completely filled with oil of dielectric constant 2.5. It is connected to a 200 V supply.
(a) Find the capacitance. (b) Find the charge on the inner sphere. (c) Find the field in the oil just outside the inner sphere. (d) Find the total bound charge on the oil’s inner face, with its sign.
Answer and explanation
(a) Air-filled: C₀ = 4πε₀ab/(b − a) = (0.050 × 0.060) ÷ [(8.99 × 10⁹)(0.010)] = 3.34 × 10⁻¹¹ F. Filling the gap multiplies this by κ: C = 8.3 × 10⁻¹¹ F (83 pF).
(b) Q = CΔV = (8.34 × 10⁻¹¹)(200) = 1.7 × 10⁻⁸ C (16.7 nC).
(c) In the oil, E = kQ/(κr²). At r = a: E = (8.99 × 10⁹)(1.67 × 10⁻⁸) ÷ (2.5 × 0.050²) = 2.4 × 10⁴ V/m, pointing outward. Leaving out κ gives 6.0 × 10⁴ V/m.
(d) The bound charge cancels the fraction (1 − 1/κ) of the free charge: −Q(1 − 1/2.5) = −1.0 × 10⁻⁸ C. It is negative because it faces the positive inner sphere.
Check yourself: 1 mark each for C with the factor κ, Q, E with κ, and the bound charge with its sign (4 in total).
If you missed this: for (a)–(b), read the 10.3 study guide; for (c)–(d), “Capacitance with a dielectric” in the 10.4 study guide.
Your next step
Count a short answer as missed if you lost more than one mark.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 10.1 Electrostatics with conductors | 1, 2, 5(a)–(b) | 10.1 study guide |
| 10.2 Redistribution of charge between conductors | 3, 4, 5(c) | 10.2 study guide |
| 10.3 Capacitors | 6, 7, 9(a)–(b) | 10.3 study guide |
| 10.4 Dielectrics | 8, 9(c)–(d) | 10.4 study guide |
How to use your result
- Missed nothing? Go straight to the mixed unit review, which joins the topics together.
- Missed one or two topics? Read each of those study guides, then do the matching practice sets before the review.
- Missed three or more topics? Work through 10.1 to 10.4 in order.
- Guessed correctly? Treat it as missed.