The problems are on the capacitor worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A capacitor stores charge on two plates, and its capacitance is C = Q / V in farads. These worked answers correct the 20 problems of the Physics 11-12 worksheet on capacitor, from charge on a capacitor to defibrillator pulse: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the formula, the substitution and the result with its unit. Convert microfarads, nanofarads and kilohms before you calculate, and note that final digits may differ slightly if you round at a different step.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.7.4
Problem 1 Charge on a capacitor
C = 2200 × 10⁻⁶ F. Q = CV = 2200 × 10⁻⁶ × 6.0 = 0.0132 C, so Q = 1.3 × 10⁻² C (13 mC).Problem 2 Prefix conversions
1. C = Q / V = 2.4 × 10⁻⁶ / 12 = 2.0 × 10⁻⁷ F.2. 2.0 × 10⁻⁷ F = 0.20 µF = 200 nF.
Problem 3 Energy in a small capacitor
E = ½CV² = 0.5 × 100 × 10⁻⁶ × 25² = 0.03125 J, so E ≈ 3.1 × 10⁻² J (31 mJ).Problem 4 Time constant
1. τ = RC = 47 × 10³ × 100 × 10⁻⁶ = 4.7 s.2. 5τ = 5 × 4.7 = 24 s (about 23.5 s).
Problem 5 Two capacitors
1. C = 2.0 + 3.0 = 5.0 µF.2. 1/C = 1/2.0 + 1/3.0 = 0.8333 µF⁻¹, so C = 1.2 µF.
Problem 6 Describing a discharge
1. Capacitance is the charge stored per unit potential difference, C = Q / V.2. Both decay exponentially: they start at their maximum and fall quickly at first, then more and more slowly towards zero.
3. After one time constant the voltage is e⁻¹ ≈ 0.37 of its starting value.
Problem 7 Camera flash
1. Q = CV = 150 × 10⁻⁶ × 300 = 0.045 C.2. E = ½CV² = 0.5 × 150 × 10⁻⁶ × 300² = 6.75 J.
3. P = E / t = 6.75 / 1.0 × 10⁻³ = 6750 W ≈ 6.8 × 10³ W.
Problem 8 Memory back-up
1. ΔQ = It = 2.0 × 10⁻³ × 1.5 = 3.0 × 10⁻³ C.2. ΔV = ΔQ / C = 3.0 × 10⁻³ / 1000 × 10⁻⁶ = 3.0 V.
3. V = 9.0 − 3.0 = 6.0 V.
Problem 9 Three in series
1. 1/C = 1/6.0 + 1/3.0 + 1/2.0 = (1 + 2 + 3)/6.0 = 1.0 µF⁻¹, so C = 1.0 µF.2. The same charge is on each: Q = CV = 1.0 × 10⁻⁶ × 12 = 12 µC.
3. V = Q / C gives 2.0 V (6.0 µF), 4.0 V (3.0 µF) and 6.0 V (2.0 µF), which add to 12 V.
Problem 10 Network of capacitors
1. 1/C = 1/6.0 + 1/3.0 = 0.5 µF⁻¹, so C = 2.0 µF.2. The branches are in parallel: C = 4.0 + 2.0 = 6.0 µF.
3. Q = CV = 6.0 × 10⁻⁶ × 9.0 = 54 µC. E = ½CV² = 0.5 × 6.0 × 10⁻⁶ × 9.0² = 243 µJ (2.4 × 10⁻⁴ J).
Problem 11 Discharge calculation
1. τ = RC = 22 × 10³ × 470 × 10⁻⁶ = 10.3 s.2. V = V₀e−t/τ = 12 e−15/10.34 = 12 × 0.2344 = 2.8 V.
3. I₀ = V₀ / R = 12 / 22 × 10³ = 0.55 mA.
Problem 12 Finding τ from a graph
1. V₀ = 8.0 V and 0.37 × 8.0 = 2.96 V ≈ 3.0 V (e⁻¹ × 8.0 = 2.94 V).2. The graph passes through 2.94 V at t = 10 s, so τ = 10 s.
3. C = τ / R = 10 / (10 × 10³) = 1.0 × 10⁻³ F (1000 µF).
Problem 13 Charging a capacitor
1. τ = 4.7 × 10³ × 2200 × 10⁻⁶ = 10.34 s.2. V = V₀(1 − e−t/τ) = 9.0 (1 − e−10/10.34) = 9.0 × 0.6198 = 5.6 V.
3. 6.0 = 9.0 (1 − e−t/τ) gives e−t/τ = 1/3, so t = τ ln 3 = 10.34 × 1.0986 = 11.4 s.
Problem 14 Spot the error
1. The values were not converted to ohms and farads (2.2 kΩ = 2.2 × 10³ Ω and 470 µF = 470 × 10⁻⁶ F), and after one time constant the capacitor is still at 37 % of its voltage, not zero.2. τ = 2.2 × 10³ × 470 × 10⁻⁶ = 1.03 s.
3. V = 0.37 × 6.0 = 2.2 V (exactly 6.0 e⁻¹ = 2.21 V).
Problem 15 Parallel-plate capacitor
1. C = ε₀A / d = (8.85 × 10⁻¹² × 0.050) / (0.50 × 10⁻³) = 8.8 × 10⁻¹⁰ F (0.89 nF).2. Q = CV = 8.85 × 10⁻¹⁰ × 100 = 8.8 × 10⁻⁸ C (89 nC).
3. C is inversely proportional to d, so halving d doubles the capacitance.
Problem 16 Sharing charge
1. Q = CV = 4.0 × 10⁻⁶ × 12 = 48 µC.2. Charge is conserved and the capacitors are now in parallel: V = Q / (C₁ + C₂) = 48 / 6.0 = 8.0 V.
3. Ebefore = ½ × 4.0 × 10⁻⁶ × 12² = 288 µJ and Eafter = ½ × 6.0 × 10⁻⁶ × 8² = 192 µJ. Loss = 96 µJ, transferred as thermal energy in the connecting wires and by radiation as charge moves.
Problem 17 Energy and time constant
1. E₀ = ½CV₀² = 0.5 × 1000 × 10⁻⁶ × 10² = 0.05 J.2. E = ½CV² and V = V₀e−t/τ, so E = ½CV₀²e−2t/τ = E₀e−2t/τ.
3. τ = RC = 1.0 s. 0.25 = e−2t/τ gives 2t/τ = ln 4, so t = τ ln 2 = 0.69 s. Then V = 10 e−0.693 = 5.0 V, half the starting voltage, as expected for a quarter of the energy.
Problem 18 Delay timer design
1. V = V₀(1 − e−t/RC) = 6.0 with V₀ = 9.0 gives e−t/RC = 1/3, so t = RC ln 3.2. R = t / (C ln 3) = 30 / (470 × 10⁻⁶ × 1.0986) = 58 kΩ (5.8 × 10⁴ Ω).
3. t = 56 × 10³ × 470 × 10⁻⁶ × 1.0986 = 28.9 s, about 29 s.
Problem 19 Linearising the decay
1. ln V = 2.303, 1.803, 1.303, 0.803, 0.303.2. V = V₀e−t/τ gives ln V = ln V₀ − t/τ, which has the form y = c + mx with gradient −1/τ. Gradient = (0.303 − 2.303) / 40 = −0.050 s⁻¹.
3. τ = 1 / 0.050 = 20 s. C = τ / R = 20 / (40 × 10³) = 5.0 × 10⁻⁴ F (500 µF).
Problem 20 Defibrillator pulse
1. E = ½CV² = 0.5 × 32 × 10⁻⁶ × (5.0 × 10³)² = 400 J.2. I₀ = V / R = 5000 / 50 = 100 A. τ = RC = 50 × 32 × 10⁻⁶ = 1.6 ms.
3. The energy remaining is E₀e−2t/τ, so the fraction delivered is 1 − e−2 × 2.0/1.6 = 1 − e−2.5 = 0.9179. Energy = 400 × 0.9179 = 367 J ≈ 3.7 × 10² J.
