The problems are on the capacitance worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Capacitance is the charge stored per unit potential difference, C = Q/ΔV, measured in farads. These worked answers correct the 20 problems of the AP Physics C worksheet on capacitance, from charge on a capacitor to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit in bold. Compare your reasoning about which quantity stays fixed, because choosing the wrong constant is the most common source of errors in dielectric problems.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 10
Problem 1 Charge on a capacitor
1. Q = CΔV = (10 × 10⁻⁶)(6.0) = 6.0 × 10⁻⁵ C (60 μC).2. U = ½CΔV² = ½(10 × 10⁻⁶)(36) = 1.8 × 10⁻⁴ J.
Problem 2 Parallel plate value
1. C = ε₀A/d = (8.85 × 10⁻¹²)(0.0050)/(5.0 × 10⁻⁴) = 8.9 × 10⁻¹¹ F (88.5 pF).2. C ∝ 1/d, so C halves.
Problem 3 Series and parallel
1. 1/C = 1/6 + 1/3 = 1/2, so C = 2.0 μF.2. C = 6.0 + 3.0 = 9.0 μF.
Problem 4 Camera flash
1. Q = CΔV = (470 × 10⁻⁶)(25) = 1.2 × 10⁻² C (11.75 mC).2. U = ½CΔV² = ½(470 × 10⁻⁶)(625) = 0.15 J.
Problem 5 Doubling the gap
1. C = ε₀A/d, so C halves.2. With V fixed by the battery, Q = CV and U = ½CV² both halve; charge flows back to the battery.
Problem 6 Dielectric constant
1. C = κC₀ = 3.5 × 50 = 175 pF.2. Q = CΔV = (175 × 10⁻¹²)(9.0) = 1.6 × 10⁻⁹ C.
Problem 7 Reading a Q–V graph
1. C = slope = 20 μC/10 V = 2.0 μF.2. U = area = ½(20 × 10⁻⁶)(10) = 1.0 × 10⁻⁴ J.
Problem 8 Field between plates
1. σ = Q/A = 2.0 × 10⁻⁹/0.010 = 2.0 × 10⁻⁷ C/m².2. E = σ/ε₀ = 2.0 × 10⁻⁷/8.85 × 10⁻¹² = 2.3 × 10⁴ V/m.
3. ΔV = Ed = 45 V, so C = 2.0 nC/45 V = 44 pF, which matches ε₀A/d.
Problem 9 Energy density
1. u = ½ε₀E² = ½(8.85 × 10⁻¹²)(3.0 × 10⁶)² = 40 J/m³ (39.8).2. Volume = 0.040 × 0.0020 = 8.0 × 10⁻⁵ m³, so U = 39.8 × 8.0 × 10⁻⁵ = 3.2 × 10⁻³ J.
Problem 10 Coaxial cable
1. Take a Gaussian cylinder of radius r and length l: E(2πrl) = λl/ε₀, so E = λ/(2πε₀r).2. C = 2πε₀L/ln(b/a) = (5.56 × 10⁻¹¹)(5.0)/ln 6 = 2.78 × 10⁻¹⁰/1.792 = 1.6 × 10² pF (155 pF).
Problem 11 Concentric spheres
1. E = kQ/r² between the shells, so ΔV = kQ(1/a − 1/b) = kQ(b − a)/(ab) and C = Q/ΔV = ab/[k(b − a)] = 4πε₀ab/(b − a).2. C = (1.112 × 10⁻¹⁰)(0.050 × 0.10)/0.050 = 1.1 × 10⁻¹¹ F (11.1 pF).
3. Let b → ∞: C = 4πε₀R = (1.112 × 10⁻¹⁰)(0.50) = 56 pF.
Problem 12 Dielectric with battery
1. C = κC₀ = 1000 pF.2. Q₀ = (200 pF)(12) = 2.4 nC and Q = (1000 pF)(12) = 12 nC.
3. U₀ = ½(200 × 10⁻¹²)(144) = 14.4 nJ and U = 72 nJ.
Problem 13 Dielectric, no battery
1. The charge Q = 2.4 nC stays constant, because there is nowhere for it to go.2. V = Q/(κC₀) = 2.4 nC/1000 pF = 2.4 V.
3. U = Q²/2C = 2.88 nJ, a drop of 14.4 − 2.88 = 11.5 nJ done as work pulling the slab in.
Problem 14 Capacitor network
1. C₂ + C₃ = 6.0 μF, in series with 4.0 μF: C = (4.0 × 6.0)/10 = 2.4 μF.2. Q = CV = (2.4)(12) = 28.8 μC on C₁ (the series charge).
3. V₁ = 28.8/4.0 = 7.2 V, so the pair has 12 − 7.2 = 4.8 V.
4. Q₂ = (2.0)(4.8) = 9.6 μC and Q₃ = (4.0)(4.8) = 19.2 μC.
Problem 15 Sharing charge
1. Q = (6.0)(10) = 60 μC is conserved, so V = 60/(6.0 + 3.0) = 6.7 V.2. U₀ = ½(6.0 × 10⁻⁶)(100) = 3.0 × 10⁻⁴ J and U = ½(9.0 × 10⁻⁶)(6.67)² = 2.0 × 10⁻⁴ J, so one third is lost.
3. It is dissipated as heat in the connecting wires, and partly radiated, during the transient current.
Problem 16 Two dielectric layers
1. C₁ = κ₁ε₀A/d₁ = 2(1.77 × 10⁻¹³)/10⁻³ = 354 pF and C₂ = 708 pF; in series C = (354 × 708)/1062 = 2.4 × 10² pF (236 pF).2. Q = CΔV = (2.36 × 10⁻¹⁰)(100) = 2.4 × 10⁻⁸ C.
Problem 17 Half-filled plates
1. Both halves are connected to the same two plates, so each has the full voltage; charges add, which is the rule for parallel capacitors.2. C = ε₀(A/2)/d + κε₀(A/2)/d = (8.85 × 10⁻¹²)(0.020)(1 + 4)/(2 × 0.0020) = 2.2 × 10² pF (221 pF).
Problem 18 Work to charge
1. W = ∫₀^Q (q/C) dq = Q²/2C.2. U = (40 × 10⁻⁶)²/(2 × 8.0 × 10⁻⁶) = 1.0 × 10⁻⁴ J.
3. U(60 μC) = 2.25 × 10⁻⁴ J, so the extra work is 1.25 × 10⁻⁴ J.
Problem 19 Energy by integration
1. U = ∫ ½ε₀E² 4πr² dr with E = kQ/r² = (Q²/8πε₀)∫ from a to b of dr/r² = (Q²/8πε₀)(1/a − 1/b).2. U = (4.0 × 10⁻¹⁸/2.224 × 10⁻¹⁰)(10) = 1.8 × 10⁻⁷ J.
3. Q²/2C = 4.0 × 10⁻¹⁸/(2 × 1.112 × 10⁻¹¹) = 1.8 × 10⁻⁷ J, which agrees.
Problem 20 Spot the error
1. Q stays fixed when the capacitor is isolated, so U = Q²/2C falls as C rises.2. U₀ = ½(100 × 10⁻¹²)(2500) = 1.25 × 10⁻⁷ J and U = U₀/2.5 = 5.0 × 10⁻⁸ J.
3. V = 50/2.5 = 20 V.
