A spherical capacitor consists of two concentric conducting shells. The inner shell has radius $R_1=3,\text{cm}$, the outer shell has radius $R_2=12,\text{cm}$, and the space between them is filled with a dielectric of relative permittivity $\kappa =2$. (a) Find the capacitance of the capacitor. (b) If the capacitor is charged to a potential difference of $150,\text{V}$, 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. $R_1=0.03,\text{m}$, $R_2=0.12,\text{m}$
2. Use the formula for a spherical capacitor with a dielectric. $C=4pivarepsilo{n}_{0}kappadfrac{R}_1{R}_2{R}_2-R_1$
3. Insert the values. $C=4\pi \times 8.85\times 10^{-12}\times 2\times \frac{0.03\times 0.12}{0.12-0.03}$
4. Compute the numerator $0.03\times 0.12=0.0036$
5. Compute the denominator $0.12-0.03=0.09$
6. Calculate $C=4\pi \times 8.85\times 10^{-12}\times 2\times \frac{0.0036}{0.09}$
7. Simplify $\frac{0.0036}{0.09}=0.04$
8. Hence $C=4\pi \times 8.85\times 10^{-12}\times 2\times 0.04$
9. Calculate $4\pi \times 0.04\approx 0.502$
10. Thus $C=0.502\times 8.85\times 10^{-12}\times 2\approx 8.0\times 10^{-13},\text{F}$
11. Use $U=\frac{1}{2}C{V}^2$ to find energy. $U=\frac{1}{2}\times 8.0\times 10^{-13}\times (150{)}^2$
12. Compute $(150{)}^2=22500$
13. Compute $8.0\times 10^{-13}\times 22500=1.8\times 10^{-8}$
14. Half of that gives $U=9.0\times 10^{-9},\text{J}$

Concept tested: Capacitance and energy of spherical capacitor

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked