The problems are on the capacitors worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A capacitor stores charge on two plates separated by an insulator, with capacitance C = Q/V in farads. These worked answers correct the 20 problems of the AP Physics 2 worksheet on capacitors, from capacitance from charge to doubling the voltage: each solution shows the formula, the substitution and the final answer with its unit.
Check your method as well as your numbers. In AP-style marking, you earn credit for stating what is held constant, writing the right relation, substituting with units and giving a conclusion. Rounding differences in the third significant figure are acceptable.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10
Problem 1 Capacitance from charge
1. C = Q/V = 24 / 6.0 = 4.0 μF.2. Q = CV = 4.0 × 9.0 = 36 μC.
3. One farad is one coulomb per volt: 1 F = 1 C/V.
Problem 2 Prefixes
1. 4.7 × 10⁻⁴ F.2. 2.2 × 10⁻⁸ F.
3. 100 × 10⁻¹² F = 1.0 × 10⁻⁴ μF.
4. 3.3 mF.
Problem 3 Plate capacitor in air
1. C = ε₀A/d = (8.85 × 10⁻¹²)(0.020) / (5.0 × 10⁻⁴) = 3.54 × 10⁻¹⁰ F = 354 pF.2. C is inversely proportional to d, so it halves to 177 pF.
Problem 4 Energy in a capacitor
1. Q = CV = 100 × 10⁻⁶ × 12 = 1.2 × 10⁻³ C.2. U = ½CV² = 0.5 × 100 × 10⁻⁶ × 144 = 7.2 × 10⁻³ J.
Problem 5 Electrons on the plates
1. Q = CV = 2.2 × 10⁻⁶ × 9.0 = 2.0 × 10⁻⁵ C (19.8 μC), negative.2. N = 1.98 × 10⁻⁵ / 1.60 × 10⁻¹⁹ = 1.2 × 10¹⁴ electrons.
Problem 6 Scaling capacitance
1. 2C.2. C/2.
3. 3C.
4. Factors 2 × (1/2) × 3 = 3, so 3C.
Problem 7 Finding the gap
1. d = ε₀A/C.2. d = (8.85 × 10⁻¹²)(0.050) / 1.0 × 10⁻⁹ = 4.4 × 10⁻⁴ m = 0.44 mm.
Problem 8 Charging graph
1. Gradient = 10 / 40 μC = 0.25 V/μC = 2.5 × 10⁵ V/C, which is 1/C, so C = 4.0 μF.2. U = ½QV = 0.5 × 40 × 10⁻⁶ × 10 = 2.0 × 10⁻⁴ J.
3. The area under the graph is the energy stored, the work done in charging.
Problem 9 Three energy formulas
1. V = Q/C = 50 / 10 = 5.0 V.2. U = (50 × 10⁻⁶)² / (2 × 10 × 10⁻⁶) = 1.25 × 10⁻⁴ J.
3. ½CV² = 0.5 × 10 × 10⁻⁶ × 25 = 1.25 × 10⁻⁴ J, which matches.
Problem 10 Dielectric with the battery connected
1. C = 4.0 × 2.0 = 8.0 nF.2. Before: 2.0 × 10⁻⁹ × 9.0 = 18 nC. After: 8.0 × 10⁻⁹ × 9.0 = 72 nC.
3. Before: ½CV² = 8.1 × 10⁻⁸ J. After: 3.2 × 10⁻⁷ J.
4. V and d are fixed, so E = V/d is unchanged overall.
Problem 11 Dielectric in an isolated capacitor
1. Isolated, so Q stays 18 nC.2. V = Q/C = 18 / 8.0 = 2.25 V.
3. U = Q²/2C = (18 × 10⁻⁹)² / (16 × 10⁻⁹) = 2.0 × 10⁻⁸ J, a quarter of the original.
4. The capacitor does work pulling the dielectric in, so energy goes into kinetic energy of the slab and, if it is stopped, thermal energy.
Problem 12 Supercapacitor torch
1. U = ½CV² = 0.5 × 0.50 × 25 = 6.3 J (6.25 J).2. t = U/P = 6.25 / 0.25 = 25 s.
3. At 3.0 V, U = 0.5 × 0.50 × 9 = 2.25 J. Released: 6.25 − 2.25 = 4.0 J.
Problem 13 Air breakdown
1. V = Ed = 3.0 × 10⁶ × 2.0 × 10⁻³ = 6.0 × 10³ V.2. C = (8.85 × 10⁻¹²)(0.040) / (2.0 × 10⁻³) = 1.8 × 10⁻¹⁰ F (177 pF).
3. Q = CV = 1.77 × 10⁻¹⁰ × 6000 = 1.1 × 10⁻⁶ C; U = ½QV = 3.2 × 10⁻³ J.
Problem 14 Inside a dielectric
1. They polarise: the molecules turn so that their positive ends point towards the negative plate and the negative ends towards the positive plate.2. Induced charges on the glass surfaces create a field opposite to the field of the plates, so the net field is smaller.
3. A smaller field means a smaller voltage for the same Q, and C = Q/V is larger by the factor κ.
Problem 15 Measuring a dielectric constant
1. κ = Cfilled / Cair = 190 / 52 = 3.7.2. It resembles paper (κ = 3.7).
3. An air gap between the sheet and the plates, or stray capacitance in the leads, changes the reading.
Problem 16 Pulling the plates apart
1. C halves to 2.5 μF in both.2. A: V = 12 V, Q = 2.5 × 12 = 30 μC, U = ½CV² = 1.8 × 10⁻⁴ J (down from 3.6 × 10⁻⁴ J).
3. B: Q = 60 μC, V = Q/C = 24 V, U = ½QV = 7.2 × 10⁻⁴ J (up).
4. In B you do work pulling opposite charges apart, and the charge cannot leave, so U doubles. In A the battery takes charge back as C falls, so U falls.
Problem 17 Partly filled capacitors
1. C₀ = (8.85 × 10⁻¹²)(0.020) / 1.0 × 10⁻³ = 177 pF.2. Two capacitors in parallel, each with half the area: C = ½C₀ + ½(5C₀) = 3C₀ = 531 pF.
3. Two capacitors in series, each with half the gap: 1/C = 1/(2C₀) + 1/(10C₀) = 6/(10C₀), so C = (5/3)C₀ = 295 pF.
4. Both are larger than C₀, but the side-by-side slab raises C more than the stacked slab with the same volume of dielectric.
Problem 18 Sharing charge
1. Q = CV = 40 μC; U = ½CV² = 0.5 × 4.0 × 10⁻⁶ × 100 = 2.0 × 10⁻⁴ J.2. The charge is conserved and the capacitors are in parallel: V = 40 / (4.0 + 6.0) = 4.0 V.
3. U = ½(10 × 10⁻⁶)(16) = 8.0 × 10⁻⁵ J. Lost: 1.2 × 10⁻⁴ J (60 %).
4. Heating of the connecting wires and switch, a tiny spark, and radiated electromagnetic waves.
Problem 19 Defibrillator capacitor
1. U = ½CV² = 0.5 × 80 × 10⁻⁶ × (3000)² = 3.6 × 10² J.2. Delivered: 0.80 × 360 = 288 J; P = 288 / 5.0 × 10⁻³ = 5.8 × 10⁴ W.
3. Q = CV = 80 × 10⁻⁶ × 3000 = 0.24 C; I = Q/t = 0.24 / 5.0 × 10⁻³ = 48 A.
Problem 20 Doubling the voltage
1. Q = CV, so the second capacitor holds twice the charge.2. U = ½CV², so the second stores four times the energy.
3. E = V/d with the same d, so the field is twice as large.
4. The second one, because its field is twice as close to the breakdown value of the dielectric.
