Questions & explanations
1. A cube has 12 identical capacitors of capacitance C on each edge. What is the equivalent capacitance between two opposite corners (body diagonal)?
- 7C/12
- 5C/6
- C/2
- 6C/5
Answer: 6C/5
Using symmetry, the three vertices adjacent to the start corner are equipotential, and the three adjacent to the end corner are equipotential. Merging these gives three capacitors in parallel from start to first node (3C), six in parallel between the two nodes (6C), and three in parallel from second node to end (3C). These three groups are in series: 1/C_eq = 1/(3C) + 1/(6C) + 1/(3C) = 5/(6C), so C_eq = 6C/5.
2. A 4 μF capacitor charged to 12 V is connected to an uncharged 6 μF capacitor. Find the energy lost during redistribution.
- 144 μJ
- 288 μJ
- 432 μJ
- 172.8 μJ
Answer: 172.8 μJ
Initial charge Q = C1 V1 = 4×12 = 48 μC. Common voltage V = Q/(C1+C2) = 48/(4+6)=4.8 V. Initial energy Ui = ½ C1 V1² = 0.5×4×144 = 288 μJ. Final energy Uf = ½ (C1+C2) V² = 0.5×10×23.04 = 115.2 μJ. Energy lost = Ui - Uf = 288 - 115.2 = 172.8 μJ. Alternatively, ΔU = ½ (C1 C2/(C1+C2)) V1² = 0.5×(24/10)×144 = 0.5×2.4×144 = 172.8 μJ.
3. A square network has four identical 4 μF capacitors on each side and a 4 μF capacitor across one diagonal. What is the equivalent capacitance between the two corners that are not connected by the diagonal?
- 5C/6
- 6C
- C/6
- 6C/5
Answer: 6C/5
Using symmetry, the two corners not connected by the diagonal are equipotential. The network reduces to two parallel branches: one with a 4 μF capacitor, and the other with a series combination of two 4 μF capacitors (2 μF) in series with a parallel combination of two 4 μF capacitors (8 μF). The equivalent capacitance is 6C/5.
4. A parallel plate capacitor with plate area 0.01 m² and separation 1 mm has a dielectric with K(x) = 4(1+500x) where x is in m from one plate. Its capacitance is closest to:
- 0.44 nF
- 0.22 nF
- 0.88 nF
- 0.11 nF
Answer: 0.44 nF
The capacitance is found by integrating 1/C = ∫₀^d dx/(K(x)ε₀A). With A=0.01 m², d=0.001 m, ε₀=8.85×10⁻¹², K(x)=4(1+500x), we get 1/C = (1/(4ε₀×0.01))∫₀^0.001 dx/(1+500x) = (1/(4ε₀×0.01))×(1/500)ln(1+500×0.001) = (1/(4×8.85e-12×0.01))×(1/500)×ln(1.5). This evaluates to about 2.26×10⁹, so C ≈ 0.44×10⁻⁹ F = 0.44 nF.
5. An infinite ladder of capacitors has each rung and each side segment of capacitance C. The equivalent capacitance between the input terminals is:
- C(1+√3)/2
- C(1+√5)/2
- C(√5-1)/2
- C(√3-1)/2
Answer: C(√5-1)/2
The ladder starts with a series capacitor (side segment) before the first rung. Let Ceq be the equivalent. The first side C is in series with the parallel combination of the first rung C and the rest Ceq. So 1/Ceq = 1/C + 1/(C+Ceq). Solving gives Ceq² + C·Ceq - C² = 0, positive root Ceq = C(√5-1)/2.
6. A 2 μF capacitor (C1) and a capacitor C2 (plate area 0.01 m², gap 1 mm, dielectric slab of thickness 0.5 mm, K=4) are in series. This branch is in parallel with a 3 μF capacitor (C3) and a capacitor C4 identical to C2 in series. Find the equivalent capacitance between the terminals.
- 1.14 μF
- 0.283 μF
- 1.42 μF
- 1.56 μF
Answer: 0.283 μF
C2 = ε₀A / [d - t(1-1/K)] = (8.85×10⁻¹²×0.01) / [0.001 - 0.0005×0.75] = 8.85×10⁻¹⁴ / 0.000625 = 1.416×10⁻¹⁰ F = 141.6 pF ≈ 0.142 μF. C1 (2 μF) >> C2, so series branch C1-C2 gives C_s1 ≈ C2 = 0.142 μF. Similarly C_s2 ≈ 0.142 μF. Parallel combination: C_eq = 0.142 + 0.142 = 0.284 μF ≈ 0.283 μF.
7. A parallel plate capacitor (area A, separation d) is connected to a battery of voltage V. A dielectric slab of constant K filling the gap is slowly pulled out. Which graph correctly shows the work done by the external agent as a function of the slab's position x (0 = fully inside, d = fully out)?
- Linear decrease from (1/2)(K-1)ε₀AV²/d at x=0 to 0 at x=d
- Linear increase from 0 at x=0 to (1/2)(K-1)ε₀AV²/d at x=d
- Constant at (1/2)(K-1)ε₀AV²/d for all x
- Parabolic increase from 0 at x=0 to (1/2)(K-1)ε₀AV²/d at x=d
Answer: Linear increase from 0 at x=0 to (1/2)(K-1)ε₀AV²/d at x=d
With battery connected, V constant. Capacitance C(x) = ε₀A/d + (K-1)ε₀A(1-x/d)/d, linear in x. Work done by external agent W_ext = -ΔU - W_battery = -(1/2)(C_f - C_i)V² - (C_f - C_i)V² = -(1/2)(C_f - C_i)V². Since C_f < C_i, W_ext positive and linear in x, from 0 to (1/2)(K-1)ε₀AV²/d.
8. A parallel-plate capacitor with air dielectric (capacitance C) is connected to a battery of voltage V. A dielectric slab of constant K is slowly inserted fully. Find the work done by the external agent pulling the slab.
- ½(K-1)CV²
- ½(K-1)CV²/K
- (K-1)CV²
- 0
Answer: ½(K-1)CV²
With battery connected, V constant. Initial energy U_i = ½CV². Final capacitance C' = KC, so U_f = ½KC V². ΔU = ½(K-1)CV². Battery work W_batt = VΔQ = V(KCV - CV) = (K-1)CV². By energy conservation, W_batt = ΔU + W_ext, so W_ext = W_batt - ΔU = (K-1)CV² - ½(K-1)CV² = ½(K-1)CV².
9. A parallel plate capacitor with air gap has capacitance 10 μF and is charged to 100 V. The battery is disconnected and the plate separation is doubled. What is the final energy stored?
- 0.10 J
- 0.025 J
- 0.05 J
- 0.20 J
Answer: 0.10 J
Initial energy U_i = ½CV² = 0.5 × 10×10⁻⁶ × 100² = 0.05 J. With battery disconnected, charge Q = CV = 1×10⁻³ C remains constant. Doubling d halves capacitance to C_f = 5 μF. Final energy U_f = Q²/(2C_f) = (1×10⁻³)²/(2×5×10⁻⁶) = 0.10 J. Thus final energy is 0.10 J, option a.
10. A 2 μF capacitor charged to 6 V and a 3 μF capacitor charged to 4 V are connected in parallel (like plates together). What is the energy lost during redistribution?
- 0.024 J
- 24 μJ
- 2.4 μJ
- 0.24 J
Answer: 2.4 μJ
Energy lost = initial energy - final energy. Initial energy = ½×2×6² + ½×3×4² = 36+24 = 60 μJ. Common voltage = (2×6+3×4)/(2+3) = 24/5 = 4.8 V. Final energy = ½×5×4.8² = 57.6 μJ. Loss = 60-57.6 = 2.4 μJ. Alternatively, ΔU = ½×(C1C2/(C1+C2))×(V1-V2)² = ½×(6/5)×4 = 2.4 μJ.
11. In the circuit, a 10 V battery and a 5 V battery are connected with two capacitors C₁=2 μF and C₂=3 μF in series. Find the charge on C₁ in steady state.
- 15 μC
- 10 μC
- 30 μC
- 6 μC
Answer: 6 μC
In steady state, capacitors act as open circuits. Apply KVL: +10 V - V₁ - V₂ - 5 V = 0 => V₁ + V₂ = 5 V. Since capacitors are in series, charge Q is same on both: Q = C₁V₁ = C₂V₂. So V₁ = Q/2, V₂ = Q/3. Then Q/2 + Q/3 = 5 => (5Q/6)=5 => Q=6 μC. Thus charge on C₁ is 6 μC.
12. In a five-capacitor Wheatstone bridge, C1 = 2 μF, C2 = 3 μF, C3 = 4 μF, C4 = 6 μF, and C5 = 5 μF is the bridge capacitor. Find the equivalent capacitance between the two input terminals.
- 3.0 μF
- 2.5 μF
- 3.6 μF
- 4.0 μF
Answer: 3.6 μF
Bridge is balanced because C1/C2 = 2/3 = C3/C4 = 4/6. C5 carries no charge and is removed. Equivalent capacitance: (C1 series C2) in parallel with (C3 series C4). C12 = (2×3)/(2+3)=6/5=1.2 μF. C34 = (4×6)/(4+6)=24/10=2.4 μF. Parallel: C_eq = 1.2+2.4 = 3.6 μF.