The questions in this review are original Marlbridge practice questions, not past exam questions, and each combines two or more topics. The rubrics are a suggested Marlbridge rubric, not official scoring. Data: k = 9.0 × 10⁹ N·m²/C²; ε₀ = 8.85 × 10⁻¹² C²/(N·m²); e = 1.60 × 10⁻¹⁹ C; mₑ = 9.11 × 10⁻³¹ kg; mₚ = 1.67 × 10⁻²⁷ kg. Potential is zero very far from all charges. All data are invented.
Big ideas of the unit
- Charge causes the force. Like charges repel and unlike attract: F = k|q₁q₂|/r² along the line between them (Topic 10.1).
- Charge moves, but the total is conserved. Friction, contact and grounding transfer electrons; a nearby charge can separate charge in a neutral object (Topic 10.2).
- A field is force per charge. E = F/q, away from positive and toward negative charges. Fields add as vectors. Inside a conductor in equilibrium, E = 0 (Topic 10.3).
- Potential energy belongs to a system. U_E = kq₁q₂/r for each pair, with signs; slow external work equals ΔU_E (Topic 10.4).
- Potential is energy per charge. V = kq/r adds as a scalar. Conductors in contact reach the same potential (Topic 10.5).
- Field and potential are two views of one thing. E points toward lower V, with size |ΔV/Δr|; equipotentials cross field lines at right angles.
- A capacitor stores separated charge and energy. C = κε₀A/d; between the plates the field is uniform, E = ΔV/d; U_C = ½QΔV (Topic 10.6).
- Energy conservation ties it together. ΔU_E = qΔV and ΔK = −ΔU_E for a charge moving under electric forces alone (Topic 10.7).
Key relationships and methods
| Idea | Relationship or method | Watch out for |
|---|---|---|
| Force (10.1) | F = k|q₁q₂|/r²; equal and opposite on the pair | r² not r; direction from like/unlike |
| Charging (10.2) | Identical conductors share equally; grounding lets charge leave | System total changes only by transfer |
| Field (10.3) | E = F/q; point charge E = k|q|/r²; add components | Adding magnitudes of vectors |
| Conductors (10.3, 10.5) | E = 0 inside; whole conductor at one potential | Insulators can have E ≠ 0 inside |
| Potential energy (10.4) | U_E = kq₁q₂/r per pair; W_ext = ΔU_E | Dropping signs of charges |
| Potential (10.5) | V = Σkq/r; ΔU_E = qΔV; |E| = |ΔV/Δr| | V = 0 does not mean E = 0 |
| Capacitors (10.6) | C = κε₀A/d; Q = CΔV; E = ΔV/d; U_C = ½QΔV | Isolated: Q fixed. Connected: ΔV fixed |
| Energy (10.7) | ΔK = −qΔV | Negative charges speed up toward higher V |
Question 1 (multiple choice · mixed)
Topics 10.5, 10.7. A proton is released from rest on the 80 V equipotential line of a map. Only electric forces act. Which is correct when it reaches the 30 V line?
- (A) It has moved toward higher potential and has speed 9.8 × 10⁴ m/s.
- (B) It has moved toward lower potential and has speed 1.2 × 10⁵ m/s.
- (C) It has moved toward lower potential and has speed 9.8 × 10⁴ m/s.
- (D) It has moved toward lower potential and has speed 4.2 × 10⁶ m/s.
Answer and explanation
Answer: (C). A positive charge is pushed along the field, toward lower potential. ΔK = −qΔV = −(1.60 × 10⁻¹⁹)(30 − 80) = 8.0 × 10⁻¹⁸ J, so v = √(2K/mₚ) = 9.8 × 10⁴ m/s.
- (A) uses the rule for a negative charge.
- (B) uses 80 V instead of the 50 V difference.
- (D) uses the electron mass.
Question 2 (multiple choice · mixed)
Topics 10.1, 10.2, 10.6. A charged capacitor is disconnected from its battery. A slab of plastic is then slid in to fill the gap, and the potential difference between the plates falls. Which explanation is correct?
- (A) The plastic conducts a little charge from one plate to the other, so Q falls and ΔV = Q/C falls.
- (B) The plates’ field shifts the electrons in each plastic molecule toward the positive plate. This induced separation makes a field opposite to the plates’ field, so E and ΔV = Ed fall.
- (C) Free electrons flow through the plastic to the positive plate and cancel part of its charge.
- (D) The plastic pushes the plates farther apart, so the field weakens.
Answer and explanation
Answer: (B). Plastic is an insulator: its charges cannot flow, but they shift within each molecule. Its induced field partly cancels the plates’ field. Q is unchanged, so C = Q/ΔV rises by the factor κ.
- (A) and (C) treat the plastic as a conductor.
- (D) The slab does not move the plates.
Question 3 (constructed response · mixed)
Topics 10.3, 10.4, 10.5. A +2.0 nC charge is fixed at the origin and a −6.0 nC charge is fixed at (0.24 m, 0). Point P is at (0, 0.10 m).
(a) Calculate the magnitude and direction of the net electric field at P. (b) Find the force on an electron placed at P. (c) Calculate the electric potential at P. (d) Calculate the work an external force must do to bring a +1.0 nC charge slowly from very far away to P. Explain the sign. (e) A student adds 1800 N/C and 799 N/C to get 2599 N/C for (a), then adds 180 V and 208 V to get 388 V for (c). Explain each error.
Worked solution and suggested Marlbridge rubric
(a) Distance from the −6.0 nC charge: √(0.24² + 0.10²) = 0.26 m.
| Field at P | Size (N/C) | x (N/C) | y (N/C) |
|---|---|---|---|
| From +2.0 nC (away, +y) | 9.0 × 10⁹ × 2.0 × 10⁻⁹ ÷ 0.10² = 1800 | 0 | +1800 |
| From −6.0 nC (toward it) | 9.0 × 10⁹ × 6.0 × 10⁻⁹ ÷ 0.26² = 799 | 799 × (12/13) = +737 | 799 × (−5/13) = −307 |
| Net | +737 | +1493 |
Size: √(737² + 1493²) = 1.67 × 10³ N/C, at 63.7° above the +x direction.
(b) F = eE = (1.60 × 10⁻¹⁹)(1665) = 2.66 × 10⁻¹⁶ N, opposite to the field (63.7° below the −x direction).
(c) V = 9.0 × 10⁹ [2.0 × 10⁻⁹ ÷ 0.10 + (−6.0 × 10⁻⁹) ÷ 0.26] = 180 V − 207.7 V = −27.7 V.
(d) W = qV_P = (1.0 × 10⁻⁹)(−27.7) = −2.77 × 10⁻⁸ J. Negative because U_E falls: overall the charge is pulled in, and the external force holds it back.
(e) The fields point in different directions, so they add by components. Potentials are scalars, but each keeps the sign of its charge: the −6.0 nC term is −207.7 V.
| Point | What earns it |
|---|---|
| 1 | Correct size and direction of each field at P, using 0.26 m |
| 1 | Components added: |E| = 1.67 × 10³ N/C |
| 1 | Direction 63.7° above +x (or equivalent) |
| 1 | Electron force 2.66 × 10⁻¹⁶ N, opposite to E |
| 1 | V_P = −27.7 V with signed terms |
| 1 | W = −2.77 × 10⁻⁸ J with the sign explained |
| 1 | (e) Vector addition for E; signed scalar addition for V |
Total: 7 points.
Question 4 (constructed response · mixed)
Topics 10.3, 10.5, 10.6, 10.7. Two square plates, 0.15 m × 0.15 m, are 3.0 mm apart in a vacuum and connected to a 450 V supply.
(a) Calculate the capacitance, the charge on each plate and the energy stored. (b) Calculate the field between the plates and the force on a proton there. Show that gravity can be ignored. (c) A proton is released from rest at the positive plate. Find its speed at the negative plate using energy, then check it with a = F/m and constant-acceleration motion. (d) The supply is disconnected. The plates are then pulled apart to 6.0 mm. State what happens to Q, C, ΔV, E and U_C, with values. Where does the extra stored energy come from?
Worked solution and suggested Marlbridge rubric
(a) C = ε₀A/d = (8.85 × 10⁻¹²)(0.0225) ÷ (3.0 × 10⁻³) = 66.4 pF. Q = CΔV = 2.99 × 10⁻⁸ C. U_C = ½C(ΔV)² = 6.72 × 10⁻⁶ J.
(b) E = ΔV/d = 450 ÷ 3.0 × 10⁻³ = 1.50 × 10⁵ V/m, from the positive to the negative plate. F = eE = 2.40 × 10⁻¹⁴ N. The proton’s weight is (1.67 × 10⁻²⁷)(9.8) = 1.6 × 10⁻²⁶ N, about 10¹² times smaller.
(c) ΔK = −qΔV = e(450 V) = 7.20 × 10⁻¹⁷ J, so v = √(2K/mₚ) = 2.94 × 10⁵ m/s. Check: a = F/mₚ = 1.44 × 10¹³ m/s², and v = √(2ad) = √(2 × 1.44 × 10¹³ × 3.0 × 10⁻³) = 2.94 × 10⁵ m/s. Both work because the field is uniform.
(d) Isolated, so Q stays 2.99 × 10⁻⁸ C. C = ε₀A/d halves to 33.2 pF. ΔV = Q/C doubles to 900 V. E = Q/(ε₀A) is unchanged at 1.50 × 10⁵ V/m. U_C = ½QΔV doubles to 1.34 × 10⁻⁵ J. The plates attract, so the external force pulling them apart does positive work: 6.72 × 10⁻⁶ J.
| Point | What earns it |
|---|---|
| 1 | C = 66.4 pF, with A in m² and d in m |
| 1 | Q = 2.99 × 10⁻⁸ C and U_C = 6.72 × 10⁻⁶ J |
| 1 | E = 1.50 × 10⁵ V/m and F = 2.40 × 10⁻¹⁴ N, gravity compared |
| 1 | v = 2.94 × 10⁵ m/s from energy |
| 1 | Matching kinematics check |
| 1 | Q fixed and C halves, so ΔV and U_C double |
| 1 | E unchanged, with a reason |
| 1 | External work against the plates’ attraction supplies the energy |
Total: 8 points. Carry forward an error in C once.
Question 5 (constructed response · mixed)
Topics 10.1, 10.3, 10.4, 10.5, 10.7. Two +3.0 μC charges are fixed at (−0.10 m, 0) and (+0.10 m, 0) on a horizontal table. A small bead of mass 2.0 g and charge +1.0 μC slides on a frictionless insulating rod along the y-axis. It is released from rest at (0, 0.10 m).
(a) Calculate the net electric force on the bead at release, and its acceleration. (b) Calculate the electric potential at the release point and the potential energy of the bead with the two fixed charges. Explain why the potential energy of the fixed pair can be ignored. (c) Calculate the speed of the bead at (0, 0.24 m). (d) Calculate the speed the bead approaches when it is very far away. (e) Explain why you cannot find (c) with v² = 2ay, using a from (a).
Worked solution and suggested Marlbridge rubric
(a) Each fixed charge is √(0.10² + 0.10²) = 0.1414 m away. Each force: (9.0 × 10⁹)(3.0 × 10⁻⁶)(1.0 × 10⁻⁶) ÷ 0.02 = 1.35 N, along the line from that charge. The x-components cancel by symmetry. The y-components add: 2 × 1.35 × (0.10 ÷ 0.1414) = 1.91 N in the +y direction. a = 1.91 ÷ 0.0020 = 955 m/s².
(b) V = 2 × (9.0 × 10⁹)(3.0 × 10⁻⁶) ÷ 0.1414 = 3.82 × 10⁵ V, so U_E = qV = 0.382 J. The fixed charges never move, so their pair energy (0.405 J) never changes.
(c) At y = 0.24 m each distance is 0.26 m: V = 2.08 × 10⁵ V and U_E = 0.208 J. ΔK = −ΔU_E = 0.174 J, so v = √(2 × 0.174 ÷ 0.0020) = 13.2 m/s.
(d) Far away U_E → 0, so K = 0.382 J and v = 19.5 m/s.
(e) The force falls as the bead moves away (to 0.74 N at y = 0.24 m), so the acceleration is not constant. Energy conservation needs only the start and end positions.
| Point | What earns it |
|---|---|
| 1 | Each force 1.35 N; x-components cancel |
| 1 | Net force 1.91 N in +y; a = 955 m/s² |
| 1 | V = 3.82 × 10⁵ V and U_E = 0.382 J, from scalar addition |
| 1 | Fixed pair’s U_E constant, so it does not affect ΔK |
| 1 | v = 13.2 m/s at y = 0.24 m |
| 1 | v = 19.5 m/s far away |
| 1 | Force and acceleration fall with distance, so constant-acceleration equations do not apply |
Total: 7 points.
Question 6 (constructed response · mixed)
Topics 10.2, 10.3, 10.5, 10.7. A solid metal sphere of radius 0.060 m on an insulating stand carries −9.0 nC. Nothing is moving.
(a) Describe where the excess charge is, and state the field inside the metal. Calculate the field just outside the surface and give its direction. (b) Calculate the potential at the surface and at the centre. Justify the value at the centre. (c) An electron starts very far away and moves straight toward the sphere. Calculate the smallest starting speed that lets it reach the surface. (d) The sphere briefly touches an identical neutral metal sphere, and the two are moved far apart. Find the new charge on each sphere and the number of electrons that moved, with their direction. By what factor does the answer to (c) change?
Worked solution and suggested Marlbridge rubric
(a) The excess electrons repel each other onto the outer surface. Inside the metal E = 0. Outside, it acts like a point charge at its centre: E = (9.0 × 10⁹)(9.0 × 10⁻⁹) ÷ 0.060² = 2.25 × 10⁴ N/C, perpendicular to the surface and pointing inward.
(b) V = kQ/R = (9.0 × 10⁹)(−9.0 × 10⁻⁹) ÷ 0.060 = −1350 V at the surface. With E = 0 inside, V does not change with position, so the centre is also at −1350 V.
(c) ΔU_E = qΔV = (−1.60 × 10⁻¹⁹)(−1350 − 0) = +2.16 × 10⁻¹⁶ J. The repelled electron needs K ≥ 2.16 × 10⁻¹⁶ J: v = √(2K/mₑ) = 2.18 × 10⁷ m/s.
(d) Identical spheres share equally: −4.5 nC each. 4.5 × 10⁻⁹ ÷ 1.60 × 10⁻¹⁹ = 2.81 × 10¹⁰ electrons moved from the charged sphere to the neutral one. The total stays −9.0 nC. V at the surface halves to −675 V, so K halves and the minimum speed falls by a factor of 1/√2 = 0.707, to 1.54 × 10⁷ m/s.
| Point | What earns it |
|---|---|
| 1 | Charge on the outer surface and E = 0 inside |
| 1 | E = 2.25 × 10⁴ N/C, perpendicular and inward |
| 1 | V = −1350 V at the surface |
| 1 | Same potential at the centre, because E = 0 inside |
| 1 | ΔU_E = +2.16 × 10⁻¹⁶ J and v = 2.18 × 10⁷ m/s |
| 1 | −4.5 nC each and 2.81 × 10¹⁰ electrons, charged to neutral sphere |
| 1 | Speed factor 0.707 from K ∝ V |
Total: 7 points.
How did you do?
Questions 1–2 are worth 1 point each and Questions 3–6 are worth 29 in total. There is no pass mark: use the checklist that matches the points you lost.
- Coulomb’s law and force components (Q5(a)): Topic 10.1 checklist
- sharing, grounding and polarization (Q2, 6(d)): Topic 10.2 checklist
- adding fields, conductors (Q3(a)–(b), 4(b), 6(a)): Topic 10.3 checklist
- potential energy and external work (Q3(d), 5(b)): Topic 10.4 checklist
- potential and equipotentials (Q1, 3(c), 6(b)): Topic 10.5 checklist
- isolated or connected capacitors (Q2, 4(a), 4(d)): Topic 10.6 checklist
- speeds from energy (Q1, 4(c), 5(c)–(d), 6(c)): Topic 10.7 checklist
For a quicker topic-by-topic check, take the Unit 10 diagnostic.