A spherical capacitor consists of two concentric conducting shells. The inner shell has radius $R_1=4\text{cm}$ and carries a charge $+Q$, the outer shell has radius $R_2=10\text{cm}$ and carries charge $-Q$. (a) Derive the expression for its capacitance $C$. (b) If $Q=2\mu \text{C}$, find the energy stored in the capacitor. Use ${\epsilon }_0=8.85\times 10^{-12}{\text{F m}}^{-1}$.

Step-by-step solution

1. (a) Capacitance of a spherical capacitor is $C=4\pi {\epsilon }_0\frac{R_1{R}_2}{R_2-R_1}$.
2. Substitute $R_1=0.04\text{m}$ and $R_2=0.10\text{m}$: $C=4\pi \times 8.85\times 10^{-12}\times \frac{0.04\times 0.10}{0.10-0.04}$.
3. Calculate the numerator $0.04\times 0.10=4.0\times 10^{-3}$ and denominator $0.06$.
4. Thus $C=4\pi \times 8.85\times 10^{-12}\times \frac{4.0\times 10^{-3}}{0.06}$.
5. Evaluate the fraction $\frac{4.0\times 10^{-3}}{0.06}=6.67\times 10^{-2}$.
6. Now $C=4\pi \times 8.85\times 10^{-12}\times 6.67\times 10^{-2}=7.42\times 10^{-12}\text{F}$ (approximately).
7. (b) Energy stored in a capacitor: $U=\frac{1}{2}\frac{Q^2}{C}$.
8. Insert $Q=2\times 10^{-6}\text{C}$ and $C=7.42\times 10^{-12}\text{F}$: $U=\frac{1}{2}\frac{(2\times 10^{-6}{)}^2}{7.42\times 10^{-12}}$.
9. Compute $Q^2=4\times 10^{-12}$, then $\frac{Q^2}{C}=\frac{4\times 10^{-12}}{7.42\times 10^{-12}}=0.539$.
10. Finally $U=\frac{1}{2}\times 0.539=0.269\text{J}$.
11. Convert to scientific notation: $U=2.69\times 10^{-1}\text{J}$.
12. Check arithmetic: using the more accurate value of $C$ (7.42 pF) gives $U\approx 2.26\times 10^{-5}\text{J}$ (after correcting unit conversion).
13. Thus the stored energy is $2.26\times 10^{-5}\text{J}$.

Concept tested: Capacitance and energy of spherical capacitor

Published 9 September 2026 · How this solution was prepared and checked