The problems are on the gauss's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electric flux is Φ = ∮E · dA, the field passing through a surface. These worked answers correct the 20 problems of the AP Physics C worksheet on gauss's law, from flux of a uniform field to long charged cylinder: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the Gaussian surface or the integral first, then the substitution and the result with its unit in bold. Use k = 8.99 × 10⁹ N m²/C² and ε₀ = 8.85 × 10⁻¹² C²/(N m²). Check signs and directions, not only magnitudes.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 8
Problem 1 Flux of a uniform field
Φ = E A cosθ.1. Φ = 500 × 0.20 × 1 = 1.0 × 10² N m²/C.
2. Φ = 500 × 0.20 × cos 60° = 50 N m²/C.
3. The normal is at 90°, so Φ = 0.
Problem 2 Charge from flux
Q = ε₀Φ = 8.85 × 10⁻¹² × 3.0 × 10³ = 2.7 × 10⁻⁸ C (27 nC), positive because the flux is outward.Problem 3 Flux from a point charge
Φ = q/ε₀ = 4.0 × 10⁻⁹ / 8.85 × 10⁻¹² = 4.5 × 10² N m²/C. Doubling the radius changes nothing, because only the enclosed charge matters.Problem 4 Field of a point charge
E = kq/r² = 8.99 × 10⁹ × 5.0 × 10⁻⁹ / 0.30² = 5.0 × 10² N/C, directed radially away from the positive charge.Problem 5 Net enclosed charge
1. Q_enc = +2.0 nC, so Φ = 2.0 × 10⁻⁹ / 8.85 × 10⁻¹² = 2.3 × 10² N m²/C.2. The outside charge changes E on the surface but adds no net flux.
Problem 6 Charge densities
1. λ = Q/L = 3.0 × 10⁻⁹ / 0.60 = 5.0 × 10⁻⁹ C/m.2. σ = Q/A = 12 × 10⁻⁹ / 0.30 = 4.0 × 10⁻⁸ C/m².
Problem 7 Charged ring on its axis
E = kQx / (x² + R²)^(3/2). x² + R² = 0.0256 + 0.0144 = 0.0400, and 0.0400^(3/2) = 0.00800. E = 8.99 × 10⁹ × 5.0 × 10⁻⁹ × 0.16 / 0.00800 = 9.0 × 10² N/C along the axis, away from the ring. At the centre, E = 0 by symmetry.Problem 8 Long wire
E = λ/(2πε₀r) = 2kλ/r = 2 × 8.99 × 10⁹ × 3.0 × 10⁻⁹ / 0.40 = 1.3 × 10² N/C, radially outward.Problem 9 Large sheet
E = σ/(2ε₀) = 4.0 × 10⁻⁹ / (2 × 8.85 × 10⁻¹²) = 2.3 × 10² N/C.1. At 0.10 m: 2.3 × 10² N/C, away from the sheet.
2. At 1.0 m: the same, since the field of an infinite sheet does not depend on distance.
Problem 10 Thin spherical shell
(a) Inside a uniform shell Q_enc = 0, so E = 0.(b) E = kQ/r² = 8.99 × 10⁹ × 12 × 10⁻⁹ / 0.25 = 4.3 × 10² N/C, directed inward.
Problem 11 Solid insulating sphere
At 0.10 m the enclosed charge is 10 × (0.10/0.15)³ = 2.96 nC, so E = 8.99 × 10⁹ × 2.96 × 10⁻⁹ / 0.010 = 2.7 × 10³ N/C. At 0.30 m, E = kQ/r² = 8.99 × 10⁹ × 10 × 10⁻⁹ / 0.090 = 1.0 × 10³ N/C.Problem 12 Reading an E–r graph
1. The peak is at the surface: R = 0.10 m.2. E = kQ/R², so Q = E R²/k = 5.4 × 10³ × 0.010 / 8.99 × 10⁹ = 6.0 × 10⁻⁹ C.
3. E = kQ/r² = 8.99 × 10⁹ × 6.0 × 10⁻⁹ / 0.040 = 1.3 × 10³ N/C.
Problem 13 Charge at the centre of a cube
Total: Φ = q/ε₀ = 12 × 10⁻⁹ / 8.85 × 10⁻¹² = 1.4 × 10³ N m²/C. By symmetry each of the six faces carries one sixth: 2.3 × 10² N m²/C.Problem 14 Spot the error
1. E in ∮E · dA is the total field at each point, from all charges.2. Only the net flux depends on the enclosed charge alone; an outside charge changes the local field but contributes zero net flux.
Problem 15 Field between charged plates
1. E = σ/ε₀ = 2.0 × 10⁻⁷ / 8.85 × 10⁻¹² = 2.3 × 10⁴ N/C.2. a = eE/m = 1.6 × 10⁻¹⁹ × 2.26 × 10⁴ / 9.11 × 10⁻³¹ = 4.0 × 10¹⁵ m/s².
Problem 16 Maximum field on a ring’s axis
1. dE/dx = kQ[(x² + R²) − 3x²]/(x² + R²)^(5/2) = 0 gives R² = 2x², so x = R/√2 = 0.085 m.2. E_max = kQ(R/√2)/(1.5R²)^(3/2) = 0.385 kQ/R² = 0.385 × 3.12 × 10³ = 1.2 × 10³ N/C.
Problem 17 Field of a finite rod
1. λ = Q/L. Element dq = λ dx at distance x from P gives dE = kλ dx/x². E = ∫ from a to a+L of kλ dx/x² = kλ(1/a − 1/(a + L)) = kQ / (a(a + L)).2. E = 8.99 × 10⁹ × 8.0 × 10⁻⁹ / (0.10 × 0.50) = 1.4 × 10³ N/C, directed away from the rod.
Problem 18 Non-uniform sphere
1. Q = ∫ A r (4πr²) dr from 0 to R = πAR⁴ = π × 3.0 × 10⁻⁶ × 0.0016 = 1.5 × 10⁻⁸ C.2. Q_enc = πAr⁴, so E (4πr²) = πAr⁴/ε₀ and E = Ar²/(4ε₀) = 3.0 × 10⁻⁶ × 0.010 / (4 × 8.85 × 10⁻¹²) = 8.5 × 10² N/C.
3. E = kQ/r² = 8.99 × 10⁹ × 1.51 × 10⁻⁸ / 0.25 = 5.4 × 10² N/C.
Problem 19 Charge inside a conducting shell
1. Inside the metal E = 0, so the inner surface carries −3.0 nC and the outer surface carries +5.0 nC (net +2.0 nC).2. At 0.05 m, E = kq/r² = 8.99 × 10⁹ × 3.0 × 10⁻⁹ / 0.0025 = 1.1 × 10⁴ N/C. At 0.30 m, 5.0 nC is enclosed: E = 8.99 × 10⁹ × 5.0 × 10⁻⁹ / 0.090 = 5.0 × 10² N/C.
3. E = 0 in the metal.
Problem 20 Long charged cylinder
1. Per length L, Q_enc = ρπr²L, so E(2πrL) = ρπr²L/ε₀ and E = ρr/(2ε₀) = 2.0 × 10⁻⁶ × 0.030 / (2 × 8.85 × 10⁻¹²) = 3.4 × 10³ N/C.2. Outside, all charge within R counts: E = ρR²/(2ε₀r) = 2.0 × 10⁻⁶ × 0.0025 / (1.77 × 10⁻¹¹ × 0.10) = 2.8 × 10³ N/C.
