This review joins the six topics of Unit 8. Read the big ideas and the table, then try the questions; each uses at least two topics. These are original Marlbridge practice questions, not past exam questions, and the mark tables are a suggested Marlbridge rubric, not an official scoring guideline. All data are invented for practice. If you have not taken it yet, start with the Unit 8 diagnostic.
Big ideas of the unit
- Charge is the starting point. It is a scalar, comes in whole multiples of e and is conserved: a system’s net charge changes only when charge crosses its boundary (8.1, 8.2).
- Charging moves electrons. Friction and contact move them between objects; induction and polarisation move them within an object. Grounding links an object to Earth’s huge supply.
- Force and field are two views of one interaction. The field E = F/q₀ belongs to the source charges; the force qE belongs to the charge placed in it (8.3).
- Superposition is the main method. Add forces or fields from several charges as vectors, by components. A continuous charge is the same idea with an integral (8.4).
- Symmetry first. Components that cancel in pairs need no calculation, for point charges or for a ring or arc.
- Flux counts field through a surface. Φ = ∫E·dA, with outward normals on a closed surface, so entering flux is negative (8.5).
- Gauss’s law links flux back to charge. The net flux through any closed surface is q_enc/ε₀. It is always true, but it gives E easily only with spherical, cylindrical or planar symmetry (8.6).
- Conductors and insulators differ. In equilibrium a conductor has E = 0 inside and its excess charge on the surface; an insulator can hold charge, and a field, inside.
- Checks connect everything. Far from any finite charge, E → kQ/r². The infinite line from integration (8.4) matches Gauss’s law (8.6).
Key relationships and methods
| Idea | Relationship | When to use it |
|---|---|---|
| Quantised charge | q = ne | Counting electrons moved |
| Coulomb’s law | F = kq₁q₂/r² (sizes of the charges), along the line joining them | Up to four point charges, or high symmetry |
| Field and force | E = F/q₀; F = qE | Any field; force on a negative charge is opposite to E |
| Point charge | E = kq/r² | Also outside any spherically symmetric charge |
| Continuous charge | E = ∫k dq/r², component by component | Rods, rings, arcs, lines |
| Ring on its axis | E = kQz/(z² + a²)^(3/2) | Points on the axis |
| Rod on its bisector | E = kQ/[x√(x² + L²/4)] | Becomes 2kλ/x for a long rod |
| Flux | Φ = E·A (uniform field, flat surface); Φ = ∫E·dA | Sign from the area vector |
| Gauss’s law | ∮E·dA = q_enc/ε₀ | Spheres, coaxial cylinders, pillboxes |
| Standard Gauss results | Sphere: kQ/r² outside, kQr/R³ inside (uniform); line: λ/(2πε₀r); sheet: σ/(2ε₀) | Check continuity at the surface |
| Non-uniform density | q_enc = ∫ρ dV, with dV = 4πr² dr or 2πrL dr | ρ given as a function |
Question 1 (multiple choice · mixed)
Two identical small metal spheres, A (+8.0 nC) and B (−2.0 nC), are a distance d apart and attract each other with a force of magnitude F. They are touched together and then returned to the same separation d. Which row describes the new situation?
- (A) Repulsive force 9F/16; the field at the midpoint between them is now zero.
- (B) Repulsive force 9F/16; the field at the midpoint is unchanged, because charge is conserved.
- (C) Repulsive force 25F/16; the field at the midpoint is now zero.
- (D) Attractive force 9F/16; the field at the midpoint is now zero.
Answer and explanation
Answer: (A). The total charge, +6.0 nC, is conserved and shared equally: +3.0 nC each. The force now depends on (3.0)(3.0) = 9 instead of (8.0)(2.0) = 16, so it is 9F/16, and like charges repel. At the midpoint, two equal charges at equal distances give equal and opposite fields, so E = 0. Before contact, both fields there pointed towards B.
- (B) The total is conserved, but the field depends on how the charge is distributed.
- (C) shares the sizes, (8.0 + 2.0)/2 = 5.0 nC each, ignoring the negative sign.
- (D) forgets that both spheres now have the same sign.
Question 2 (multiple choice · mixed)
A very long straight wire with uniform charge per unit length λ passes through the centre of an imaginary sphere of radius R. What is the net electric flux through the sphere?
- (A) 2λR/ε₀
- (B) λR/ε₀
- (C) λ/(2πε₀R)
- (D) It cannot be found, because E is not constant over the surface of the sphere.
Answer and explanation
Answer: (A). The sphere encloses a length 2R of wire, a diameter, so q_enc = 2λR. Gauss’s law gives Φ = q_enc/ε₀ = 2λR/ε₀ for any closed surface.
- (B) uses the radius instead of the diameter as the enclosed length.
- (C) is the field of the line at distance R. It has units of N/C, not flux.
- (D) Gauss’s law is always true, so the net flux is known. Symmetry is needed only to find E itself.
Question 3 (multiple choice · mixed)
An insulating sphere of radius R carries charge Q spread uniformly through its volume. Its field has its largest value, E_max, at r = R. At which distances from the centre is E = E_max/2?
- (A) R/2 and √2 R
- (B) R/2 and 2R
- (C) √2 R only
- (D) R/√2 and √2 R
Answer and explanation
Answer: (A). Inside, Gauss’s law gives E ∝ r, so E halves at r = R/2. Outside, E ∝ 1/r², so E halves at r = √2 R.
- (B) uses E ∝ 1/r outside, the rule for a line charge.
- (C) treats the field inside as zero, which is true for a conductor but not for this insulator.
- (D) uses E ∝ r² inside, which would need a density that grows with r.
Question 4 (constructed response · mixed)
A straight wire 2.0 m long carries +6.0 nC spread uniformly. Point P is 0.050 m from the wire’s midpoint, on its perpendicular bisector.
(a) Use Gauss’s law with a coaxial cylinder to derive the field of an infinitely long line with charge per unit length λ. (b) Explain why Gauss’s law cannot give the exact field of this finite wire, but why the result of (a) is still a good model at P. (c) The exact field on the bisector is E = kQ/[x√(x² + L²/4)]. Calculate it at P, and compare it with the infinite-line value. (d) An electron is released from rest at P. Find the magnitude and direction of its initial acceleration. (e) Predict, without the exact formula, the field at 0.10 m from the midpoint.
Worked solution and suggested Marlbridge rubric
(a) Take a coaxial cylinder of radius r and length ℓ. By symmetry E is radial and the same size all over the curved side, so the flux there is E(2πrℓ). E is parallel to the end caps, so they add nothing. q_enc = λℓ. So E(2πrℓ) = λℓ/ε₀ and E = λ/(2πε₀r).
(b) Near the ends of a finite wire, E is not perpendicular to the wire and its size varies along the cylinder, so E cannot come out of the flux integral. But P is 0.050 m from a 2.0 m wire, so from P the wire looks infinite.
(c) λ = 6.0 × 10⁻⁹ ÷ 2.0 = 3.0 × 10⁻⁹ C/m. Exact: E = (8.99 × 10⁹)(6.0 × 10⁻⁹) ÷ [0.050 × √(0.050² + 1.0²)] = 1.08 × 10³ N/C (1077 N/C). Infinite line: E = (3.0 × 10⁻⁹) ÷ (2π × 8.85 × 10⁻¹² × 0.050) = 1.08 × 10³ N/C (1079 N/C). They differ by about 0.15%.
(d) a = eE/m = (1.60 × 10⁻¹⁹)(1077) ÷ (9.11 × 10⁻³¹) = 1.9 × 10¹⁴ m/s², directed towards the wire, opposite to E, because the electron is negative.
(e) The wire still looks infinite, so E ∝ 1/r and doubling r halves E: about 5.4 × 10² N/C (exact: 537 N/C).
| Point | What earns it |
|---|---|
| 1 | Coaxial cylinder: flux E(2πrℓ) through the curved side, zero through the end caps |
| 1 | q_enc = λℓ, giving E = λ/(2πε₀r) |
| 1 | (b) Symmetry fails near the ends, so E is not constant on the cylinder; P is much closer than L |
| 1 | (c) Both values about 1.08 × 10³ N/C with a comparison |
| 1 | (d) 1.9 × 10¹⁴ m/s² towards the wire |
| 1 | (e) About 5.4 × 10² N/C from the 1/r dependence |
Total: 6 points.
Question 5 (constructed response · mixed)
Inside a charged insulating ball of radius R, the field is radial and outward with E = βr³ (r ≤ R), where β is a positive constant.
(a) Find the flux through a concentric sphere of radius r ≤ R. (b) Use Gauss’s law to find the charge enclosed within radius r, then show that the charge density is ρ(r) = 5ε₀βr². (c) Write E for r > R and show that it matches the inside field at r = R. (d) For β = 2.0 × 10⁶ N/(C·m³) and R = 0.10 m, calculate the total charge and ρ at the surface. (e) The ball sits at the centre of a closed cubic box of side 0.30 m. Find the net flux through the box, and explain why you can do this even though E is not constant over the box.
Worked solution and suggested Marlbridge rubric
(a) E is radial and constant on the sphere: Φ = E(4πr²) = 4πβr⁵.
(b) Gauss’s law: q_enc(r) = ε₀Φ = 4πε₀βr⁵. A thin shell from r to r + dr holds dq = ρ(4πr²) dr, so ρ = (1/(4πr²)) dq/dr = (1/(4πr²))(20πε₀βr⁴) = 5ε₀βr².
(c) Outside, all the charge Q = 4πε₀βR⁵ is enclosed: E(4πr²) = Q/ε₀, so E = βR⁵/r². At r = R this is βR³, the same as the inside formula, so E is continuous.
(d) Q = 4π(8.85 × 10⁻¹²)(2.0 × 10⁶)(0.10)⁵ = 2.2 × 10⁻⁹ C. Surface density: ρ(R) = 5(8.85 × 10⁻¹²)(2.0 × 10⁶)(0.10)² = 8.9 × 10⁻⁷ C/m³. Check: kQ/R² = 2.0 × 10³ N/C = βR³.
(e) Φ = Q/ε₀ = 4πβR⁵ = 2.5 × 10² N·m²/C (251 N·m²/C). Gauss’s law gives the net flux through any closed surface from the enclosed charge; symmetry is needed only to find E itself.
| Point | What earns it |
|---|---|
| 1 | Φ = 4πβr⁵ |
| 1 | q_enc = 4πε₀βr⁵ from Gauss’s law |
| 1 | ρ from dq = ρ4πr² dr, giving 5ε₀βr² |
| 1 | Outside field βR⁵/r² and continuity at r = R shown |
| 1 | Q = 2.2 × 10⁻⁹ C and ρ(R) = 8.9 × 10⁻⁷ C/m³ |
| 1 | Box flux 251 N·m²/C with the reason |
Total: 6 points.
Question 6 (constructed response · mixed)
Two identical small conducting balls, each of mass 0.20 g, hang from insulating threads 0.50 m long, tied to the same point. Ball A starts with a negative charge and ball B is neutral. The balls touch, then come to rest 0.080 m apart.
(a) Draw a free-body diagram for one ball at rest. (b) Find the charge on each ball, and the charge A had at the start. (c) Which way did electrons move when the balls touched, and how many moved? (d) Ball B is then touched by an identical neutral ball, which is taken away. State the new charge on each of A and B. Explain whether the two threads still make equal angles with the vertical, and whether the balls settle closer together or further apart than before.
Worked solution and suggested Marlbridge rubric
(a) Weight mg down, tension T along the thread, and electric repulsion F horizontal, away from the other ball.
(b) Each thread makes angle θ with the vertical, where sin θ = 0.040 ÷ 0.50 = 0.080, so θ = 4.59°. Balancing forces: F = mg tan θ = (2.0 × 10⁻⁴)(9.8)(0.0803) = 1.57 × 10⁻⁴ N. The charges are equal, so kq²/d² = F gives q = d√(F/k) = 0.080 × √(1.57 × 10⁻⁴ ÷ 8.99 × 10⁹) = 1.06 × 10⁻⁸ C. Each ball carries −11 nC (−10.6 nC), so A started with −21 nC (−21.2 nC).
(c) A had excess electrons; half of them moved from A to B: N = (1.06 × 10⁻⁸) ÷ (1.60 × 10⁻¹⁹) = 6.6 × 10¹⁰ electrons.
(d) Touching B to an identical neutral ball shares B’s charge equally, so B now carries −5.3 nC and A keeps −10.6 nC (about −11 nC). The threads still hang at equal angles: the electric forces on A and B are a third-law pair, equal in size even though the charges differ, and the balls have equal weights and equal threads. At any separation the repulsion depends on the product q_Aq_B, which has halved, so the old separation no longer gives balance. The balls swing inwards until the smaller sideways pull of the tilted threads matches the weaker repulsion: they settle closer together than 0.080 m.
| Point | What earns it |
|---|---|
| 1 | Free-body diagram with weight, tension and horizontal electric force |
| 1 | F = mg tan θ ≈ 1.6 × 10⁻⁴ N |
| 1 | q ≈ −11 nC each and −21 nC at the start, by conservation |
| 1 | Electrons from A to B, about 6.6 × 10¹⁰ |
| 1 | (d) New charges: A about −11 nC, B about −5.3 nC |
| 1 | (d) Equal angles from the third-law pair, and a smaller separation because the repulsion is weaker at every distance |
Total: 6 points.
Question 7 (constructed response · mixed)
A small charged probe tip, q = +5.0 nC, is held at height h above the centre of a flat circular detector plate of radius a.
(a) Explain why the flux through the plate cannot be found from Φ = EA, and choose a suitable area element. (b) Show that the flux through the plate is Φ = (q/(2ε₀))[1 − h/√(h² + a²)], taking the area vector to point away from q. (c) Find the limit of Φ as a becomes very large, and explain the result using Gauss’s law. (d) Evaluate Φ for h = 0.030 m and a = 0.040 m, and state what fraction of the charge’s total flux passes through the plate.
Worked solution and suggested Marlbridge rubric
(a) The field changes in size and direction across the plate: strongest and perpendicular at the centre, weaker and slanted near the edge. Use thin rings of radius s: dA = 2πs ds. On each ring, E has one size and one angle to the plate.
(b) On a ring, the distance to q is √(s² + h²) and only the perpendicular part of E counts: E_⊥ = [kq/(s² + h²)] × h/√(s² + h²). So
Φ = ∫₀ᵃ kqh (2πs ds)/(s² + h²)^(3/2) = 2πkqh [1/h − 1/√(h² + a²)] = (q/(2ε₀))[1 − h/√(h² + a²)], using 2πk = 1/(2ε₀).
(c) As a → ∞, h/√(h² + a²) → 0, so Φ → q/(2ε₀). Gauss’s law gives a total flux q/ε₀; an infinite plane splits the space around q into two identical halves, so half of it crosses the plane.
(d) √(0.030² + 0.040²) = 0.050 m, so the bracket is 1 − 0.60 = 0.40. q/(2ε₀) = 5.0 × 10⁻⁹ ÷ (2 × 8.85 × 10⁻¹²) = 282 N·m²/C, so Φ = 1.1 × 10² N·m²/C (113 N·m²/C). The total flux is q/ε₀ = 565 N·m²/C, so the plate receives 20% of it.
| Point | What earns it |
|---|---|
| 1 | E varies across the plate; ring element dA = 2πs ds |
| 1 | Perpendicular component with the factor h/√(s² + h²) |
| 1 | Integral evaluated correctly, giving the stated result |
| 1 | Limit q/(2ε₀) explained as half of the total flux q/ε₀ |
| 1 | Φ ≈ 113 N·m²/C |
| 1 | Fraction 0.20 of the total |
Total: 6 points. Carry forward an error in (b) into (d) once.
How did you do?
Questions 1–3 are worth 1 point each and Questions 4–7 are worth 6 each, 27 in all. Look at where you lost points, not just how many.
- Charge sharing, electron counts or Coulomb’s law (Q1, Q6): use the 8.1 checklist and the 8.2 checklist.
- Field direction or F = qE (Q1, Q4(d)): use the 8.3 checklist.
- Integrals for E (Q4(c)): use the 8.4 checklist.
- Flux signs or surface integrals (Q7): use the 8.5 checklist.
- Gaussian surfaces or enclosed charge (Q2, Q3, Q4(a), Q5): use the 8.6 checklist.
If you have not done it yet, the Unit 8 diagnostic gives a quick topic-by-topic check.