1D · Capacitors & Energy — C Proofs, Numericals + Solved PYQs

Prove that the energy stored in a capacitor of capacitance $C$ is $\tfrac{1}{2}CV^2$, where $V$ is the potential.

Recall

Steps

1. Start uncharged ($q = 0$). Move a small charge $dq$ against the voltage $v = q/C$: $dW = (q/C)\,dq$.
2. Integrate $q = 0 \to Q$: $U = \dfrac{1}{C}\!\int_0^Q q\,dq = \dfrac{Q^2}{2C}$ (reason: $\int q\,dq = q^2/2$, evaluated gives $Q^2/2$).
3. Substitute $Q = CV$: $U = \dfrac{(CV)^2}{2C} = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV$.

Answer

Exam tip

Show that the capacitance of a parallel-plate capacitor is $C = \varepsilon_0 A/d$. What changes if a slab of dielectric constant $K$ is introduced?

Recall

Steps

1. Surface density $\sigma = Q/A$. Field between plates: $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$ (reason: Gauss's law on a pillbox, uniform field, edges ignored).
2. Voltage across the gap: $V = Ed = Q d/(\varepsilon_0 A)$.
3. Capacitance: $C = Q/V = Q/(Q d/\varepsilon_0 A) = \varepsilon_0 A/d$ — Eq. (2).
4. With dielectric slab: $E$ drops to $E/K$ at the same $Q$, so $V$ drops to $V/K$, and $C \to K C$ — Eq. (3).

Answer

Exam tip

Find the energy stored in a capacitor of capacitance 2 pF with potential 1 kV.

Recall

Steps

1. Convert to SI: $C = 2$ pF $= 2\times 10^{-12}$ F and $V = 1$ kV $= 10^3$ V.
2. Write the formula: $U = \tfrac{1}{2}CV^2$.
3. Substitute: $U = 0.5\times 2\times 10^{-12}\times (10^3)^2 = 10^{-12}\times 10^6 = 10^{-6}$ J.
4. Convert: $10^{-6}$ J $= 1\ \mu$J. The small energy is reasonable — pF is a tiny capacitance.

Answer

Exam tip

An air-filled parallel-plate capacitor has capacitance $C$. What will it be if immersed half in oil of dielectric constant 1.6?

Recall

Steps

1. State the assumption: plates dipped vertically so the lower half is in oil and the upper half is in air — two regions side by side → parallel combination. (If oil filled the bottom half of the gap depth-wise, it would be series — but the vertical-dip reading is the standard one.)
2. Each region has area $A/2$: $C_{\text{air}} = \varepsilon_0 (A/2)/d = C/2$ and $C_{\text{oil}} = K\varepsilon_0 (A/2)/d = KC/2$.
3. Parallel sum: $C' = C/2 + KC/2 = C(1+K)/2 = C(1+1.6)/2 = C(2.6)/2 = 1.3\,C$.
4. Limit check: $K = 1$ (pure air) gives $C' = C$. ✓

Answer

Exam tip

Capacitance of a spherical capacitor: two concentric metallic spheres, outer earthed. (2018: diameters 20 cm and 30 cm.)

Recall

Fig 1.2: inner sphere of radius $a$ holds $+Q$, outer sphere of radius $b$ is earthed. Between them $E = Q/4\pi\varepsilon_0 r^2$; integrating from $a$ to $b$ gives $C = 4\pi\varepsilon_0 ab/(b-a)$.

Steps

1. Let inner radius be $a$, outer radius be $b$. Charge $+Q$ sits on the inner sphere; $-Q$ is induced on the inside of the outer sphere (reason: the outer sphere is earthed, so its outer surface carries no net charge).
2. Potential difference: $V = \int_a^b E\,dr = \dfrac{Q}{4\pi\varepsilon_0}\!\int_a^b\!\dfrac{dr}{r^2} = \dfrac{Q}{4\pi\varepsilon_0}\!\left(\dfrac{1}{a} - \dfrac{1}{b}\right) = \dfrac{Q(b-a)}{4\pi\varepsilon_0 ab}$.
3. Capacitance: $C = Q/V = \dfrac{4\pi\varepsilon_0 ab}{b-a}$. Limit: $b \to \infty$ gives $4\pi\varepsilon_0 a$ (isolated sphere). ✓
4. 2018 numbers: diameters 20 cm and 30 cm → radii $a = 0.10$ m, $b = 0.15$ m. $b - a = 0.05$ m. Substitute: $C = 4\pi\times 8.85\times 10^{-12}\times (0.10\times 0.15)/0.05 = 33.3$ pF.

Answer

Exam tip

Find the capacitance per unit length of a cylindrical capacitor, the outer cylinder being earthed.

Recall

Steps

1. Inner radius $a$, outer radius $b$, outer cylinder earthed. A coaxial Gaussian cylinder gives $E = \lambda/2\pi\varepsilon_0 r$ between the conductors (reason: cylindrical symmetry, flux only through the curved face).
2. Potential difference: $V = \int_a^b E\,dr = \dfrac{\lambda}{2\pi\varepsilon_0}\!\int_a^b\!\dfrac{dr}{r} = \dfrac{\lambda}{2\pi\varepsilon_0}\ln(b/a)$ (reason: $\int dr/r = \ln r$, evaluated gives $\ln b - \ln a = \ln(b/a)$).
3. Charge on length $l$ is $\lambda l$. Total capacitance: $C = \dfrac{\lambda l}{V} = \dfrac{2\pi\varepsilon_0 l}{\ln(b/a)}$.
4. Per unit length: $C/l = 2\pi\varepsilon_0/\ln(b/a)$. Unit: F/m. Limit: thin gap ($b \approx a$) gives a large $C$, as expected.

Answer

Exam tip

Two capacitors $C_1, C_2$ charged to potentials $V_1, V_2$ are connected together. Calculate the loss of energy on sharing of charges.

Recall

Steps

1. Initial energy: $U_i = \tfrac{1}{2}C_1 V_1^2 + \tfrac{1}{2}C_2 V_2^2$.
2. After connection, total charge $Q = C_1 V_1 + C_2 V_2$ spreads over $C_1 + C_2$ (reason: charge conservation, same polarity). Common voltage: $V = (C_1 V_1 + C_2 V_2)/(C_1 + C_2)$.
3. Final energy: $U_f = \tfrac{1}{2}(C_1 + C_2)V^2 = \tfrac{1}{2}\dfrac{(C_1 V_1 + C_2 V_2)^2}{C_1 + C_2}$.
4. Difference: $\Delta U = U_i - U_f = \tfrac{1}{2}C_1 V_1^2 + \tfrac{1}{2}C_2 V_2^2 - \tfrac{1}{2}\dfrac{(C_1 V_1 + C_2 V_2)^2}{C_1 + C_2}$. Expand and simplify: $\Delta U = \dfrac{C_1 C_2\,(V_1 - V_2)^2}{2(C_1 + C_2)}$.
5. Limit check: $V_1 = V_2 \Rightarrow \Delta U = 0$. ✓ The loss is non-negative and zero only when the voltages already match.

Answer

Exam tip