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: ε₀ = 8.85 × 10⁻¹² C²/(N·m²) and 1/(4πε₀) = 8.99 × 10⁹ N·m²/C². “Infinite” lines and sheets model objects much larger than the distances involved. A scientific calculator is assumed; round only at the end.
Question 1 (multiple choice · foundation)
A point charge +q sits at the exact centre of a closed cube. What is the electric flux through one face of the cube?
- (A) q/(6ε₀)
- (B) q/ε₀
- (C) q/(4πε₀)
- (D) 0
Answer and explanation
Answer: (A). By Gauss’s law the flux through the whole closed cube is q_enc/ε₀ = q/ε₀. The charge is at the centre, so by symmetry each of the six identical faces receives the same share: q/(6ε₀).
- (B) is the flux through the whole cube, not one face.
- (C) mixes up ε₀ with the Coulomb constant: it is what you get by thinking “flux = kq”. Its units are those of flux, but the value is wrong: the total flux is q/ε₀ = 4πkq, and one face receives a sixth of that, q/(6ε₀).
- (D) is the result for a charge outside the cube, where every field line that enters also leaves.
Question 2 (multiple choice · core)
A long straight wire carries a uniform charge per unit length λ = 3.0 × 10⁻⁸ C/m. What is the magnitude of the electric field 0.20 m from the wire, far from its ends?
- (A) 27 N/C
- (B) 2.7 × 10³ N/C
- (C) 6.7 × 10³ N/C
- (D) 1.3 × 10⁴ N/C
Answer and explanation
Answer: (B). A coaxial cylinder of radius r and length L gives E(2πrL) = λL/ε₀, so E = λ/(2πε₀r) = (3.0 × 10⁻⁸) ÷ (2π × 8.85 × 10⁻¹² × 0.20) = 2.70 × 10³ N/C.
- (A) puts r = 20 (centimetres) into an SI formula without converting to metres.
- (C) treats the wire as a point charge λ and uses λ/(4πε₀r²). That ignores the cylindrical symmetry, and the units do not even come out as N/C.
- (D) uses λ/(2πε₀r²), squaring r by habit from the point-charge formula. The curved area of the cylinder is 2πrL, which is first order in r.
Question 3 (multiple choice · core)
A sphere of radius R has charge density ρ = ρ₀ r/R. What is the ratio E(R/2) / E(R)?
- (A) 1/2
- (B) 1/4
- (C) 1/16
- (D) 4
Answer and explanation
Answer: (B). Integrating gives q_enc(r) = πρ₀r⁴/R, so E = q_enc/(4πε₀r²) = ρ₀r²/(4ε₀R). E ∝ r², so halving r gives (1/2)² = 1/4. Alternatively: q_enc(R/2) is 1/16 of Q, and the Gaussian area is 1/4 as large, so E is (1/16) ÷ (1/4) = 1/4 of E(R).
- (A) uses E ∝ r, which is only true for a uniform density.
- (C) is the enclosed-charge fraction. It forgets to divide by the smaller Gaussian area.
- (D) uses the outside rule E ∝ 1/r² at a point inside the sphere.
Question 4 (calculation · core)
A thin spherical shell of radius 0.25 m carries +6.0 × 10⁻⁹ C spread uniformly over its surface. A point charge of −2.0 × 10⁻⁹ C is fixed at its centre. Find the magnitude and direction of E at (a) r = 0.10 m and (b) r = 0.40 m from the centre.
Worked solution
Use concentric spherical Gaussian surfaces; by symmetry E is radial and constant on each.
- (a) r = 0.10 m is inside the shell, so q_enc = −2.0 × 10⁻⁹ C only. E = |q_enc|/(4πε₀r²) = (8.99 × 10⁹)(2.0 × 10⁻⁹) ÷ (0.10)² = 1.8 × 10³ N/C, pointing inward (towards the centre), because q_enc is negative.
- (b) r = 0.40 m is outside, so q_enc = +6.0 × 10⁻⁹ − 2.0 × 10⁻⁹ = +4.0 × 10⁻⁹ C. E = (8.99 × 10⁹)(4.0 × 10⁻⁹) ÷ (0.40)² = 2.2 × 10² N/C (225 N/C), pointing outward.
Suggested mark points (3): 1 for using only the central charge inside the shell; 1 for the net enclosed charge +4.0 × 10⁻⁹ C outside; 1 for both magnitudes with correct directions.
Common error: including the shell’s charge at r = 0.10 m. The shell lies outside that Gaussian surface, so it adds nothing to the flux, and by symmetry nothing to E there.
Question 5 (calculation · core)
A large, thin, flat insulating sheet carries a uniform surface charge density σ = 5.0 × 10⁻⁹ C/m². (a) Use a Gaussian pillbox to derive the field near the sheet. (b) Calculate E at 0.010 m and at 0.50 m from the sheet.
Worked solution
- (a) By symmetry E is perpendicular to the sheet and points away from it on both sides. Take a cylinder (pillbox) with end area A that straddles the sheet. The curved side is parallel to E, so its flux is zero. Each end has flux EA. Enclosed charge = σA.
- Gauss’s law: 2EA = σA/ε₀, so E = σ/(2ε₀).
- (b) E = (5.0 × 10⁻⁹ C/m²) ÷ (2 × 8.85 × 10⁻¹² C²/(N·m²)) = 2.8 × 10² N/C (282 N/C) at both distances. The field of an infinite sheet does not depend on distance.
Suggested mark points (3): 1 for a pillbox with zero flux through the curved side; 1 for 2EA = σA/ε₀; 1 for 282 N/C at both points with a statement that E does not depend on distance.
Common error: writing EA = σA/ε₀ (only one end), which gives 565 N/C. That would be the field just outside a conductor, where the field inside is zero (Unit 10), not for a thin isolated sheet.
Question 6 (constructed response · core)
A very long solid insulating cylinder of radius R has a uniform volume charge density ρ > 0.
(a) Describe a Gaussian surface you would use and explain why it makes the integral simple. (b) Derive E(r) for r < R and for r > R. (c) Show that your two expressions agree at r = R. (d) Sketch E against r from r = 0 to r = 3R, labelling the value at r = R.
Worked solution and suggested Marlbridge rubric
(a) A coaxial cylinder of radius r and length L. By symmetry E points radially outward with the same size at every point on the curved side, so ∮E·dA there is E(2πrL). E is parallel to the flat ends, so they contribute zero flux.
(b) Inside: q_enc = ρπr²L, so E(2πrL) = ρπr²L/ε₀ and E = ρr/(2ε₀). Outside: q_enc = ρπR²L, so E(2πrL) = ρπR²L/ε₀ and E = ρR²/(2ε₀r).
(c) At r = R, inside gives ρR/(2ε₀) and outside gives ρR²/(2ε₀R) = ρR/(2ε₀). They agree, so E is continuous.
(d) A straight line from (0, 0) rising to ρR/(2ε₀) at r = R, then a 1/r curve falling to half that value at 2R and one third at 3R. The curve outside falls more slowly than the 1/r² curve of a sphere.
| Point | What earns it |
|---|---|
| 1 | Coaxial cylinder, with a reason: E is perpendicular and constant on the curved side and the ends give zero flux |
| 1 | Correct inside result ρr/(2ε₀), using enclosed charge ρπr²L |
| 1 | Correct outside result ρR²/(2ε₀r), using all the charge within radius R |
| 1 | Continuity check at r = R shown, not just stated |
| 1 | Sketch: linear inside, 1/r outside, peak value ρR/(2ε₀) labelled at r = R |
Accept the outside field written as λ/(2πε₀r) with λ = ρπR², the charge per unit length, which is equivalent.
Question 7 (constructed response · stretch)
A sphere of radius R has charge density ρ(r) = ρ₀(1 − r/R) for r ≤ R, where ρ₀ > 0. There is no charge outside the sphere.
(a) Show that the total charge is Q = πρ₀R³/3. (b) Derive E(r) for r ≤ R. (c) Show that your result matches the field outside the sphere at r = R. (d) Find the distance from the centre at which E is greatest, and the maximum value of E.
Worked solution and suggested Marlbridge rubric
(a) q_enc(r) = ∫₀ʳ ρ₀(1 − r′/R) 4πr′² dr′ = 4πρ₀[r³/3 − r⁴/(4R)]. At r = R: Q = 4πρ₀R³(1/3 − 1/4) = 4πρ₀R³/12 = πρ₀R³/3.
(b) Spherical Gaussian surface, radius r: E(4πr²) = q_enc/ε₀, so E = (ρ₀/ε₀)[r/3 − r²/(4R)] for r ≤ R.
(c) Inside formula at r = R: (ρ₀/ε₀)(R/3 − R/4) = ρ₀R/(12ε₀). Outside, E = Q/(4πε₀r²); at r = R this is (πρ₀R³/3)/(4πε₀R²) = ρ₀R/(12ε₀). They match.
(d) dE/dr = (ρ₀/ε₀)[1/3 − r/(2R)] = 0 gives r = 2R/3. The second derivative, −ρ₀/(2ε₀R), is negative, so this is a maximum. E_max = (ρ₀/ε₀)[2R/9 − (4R²/9)/(4R)] = (ρ₀/ε₀)(2R/9 − R/9) = ρ₀R/(9ε₀). This is larger than the surface value ρ₀R/(12ε₀): because the density is highest at the centre, the field peaks inside the sphere.
| Point | What earns it |
|---|---|
| 1 | Sets up q_enc as ∫ρ 4πr′² dr′ with correct limits |
| 1 | Evaluates the integral and gets Q = πρ₀R³/3 |
| 1 | Uses a spherical Gaussian surface with q_enc(r), not Q, to get E inside |
| 1 | Continuity check at r = R against Q/(4πε₀R²) |
| 1 | Sets dE/dr = 0 and finds r = 2R/3 |
| 1 | E_max = ρ₀R/(9ε₀), with a comment that it exceeds the surface value |
Accept a justification of the maximum from the shape of E(r) (a downward parabola in r) instead of the second derivative. Carry forward an error in q_enc into (b)–(d) once.
How did you do?
- Q1 wrong: re-read “From flux to Gauss’s law” in the study guide.
- Q2, Q5 or Q6 wrong: revisit “Choosing a Gaussian surface” and “Standard results”, especially the line and sheet derivations.
- Q3, Q4 wrong: check which charge is enclosed; see Worked example 1.
- Q7 incomplete: work through Worked example 2 again, step by step.
Then tick off the topic checklist.