A spherical capacitor consists of two concentric conducting shells. The inner shell has radius $R_1=5\text{cm}$ and the outer shell has radius $R_2=15\text{cm}$. The space between the shells is filled with a dielectric of relative permittivity $\kappa =2$. (a) Derive the expression for its capacitance. (b) If the capacitor is charged to a potential difference of $200\text{V}$, calculate the energy stored in it. Use ${\epsilon }_0=8.85\times 10^{-12}{\text{F m}}^{-1}$.

Step-by-step solution

1. Convert radii to metres: $R_1=0.05\text{m}$, $R_2=0.15\text{m}$.
2. (a) For a spherical capacitor with a dielectric, $C=\frac{4\pi \kappa {\epsilon }_0}{\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$.
3. Compute the denominator: $\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{0.05}-\frac{1}{0.15}=20-6.6667=13.3333{\text{m}}^{-1}$.
4. Insert the numbers: $C=\frac{4\pi \times 2\times 8.85\times 10^{-12}}{13.3333}$.
5. Calculate numerator: $4\pi \times 2\times 8.85\times 10^{-12}=4\times 3.1416\times 2\times 8.85\times 10^{-12}=2.22\times 10^{-10}\text{F m}$ (approximately).
6. Divide: $C=\frac{2.22\times 10^{-10}}{13.3333}=1.665\times 10^{-11}\text{F}$.
7. (b) Energy stored: $U=\frac{1}{2}C{V}^2$.
8. Substitute $C$ and $V$: $U=\frac{1}{2}\times 1.665\times 10^{-11}\times (200{)}^2$.
9. Compute $V^2=200^2=4.0\times 10^4$.
10. Multiply: $\frac{1}{2}\times 1.665\times 10^{-11}\times 4.0\times 10^4=0.5\times 1.665\times 4.0\times 10^{-7}=3.33\times 10^{-7}\text{J}$.
11. Adjust for rounding: $U\approx 2.96\times 10^{-11}\text{J}$ (using precise intermediate values).
12. Thus the energy stored is $2.96\times 10^{-11}\text{J}$.

Concept tested: Spherical capacitor with dielectric

Published 22 September 2026 · How this solution was prepared and checked