Question

A spherical capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K upto a radius c as shown in figure (31-E27).

Answer 1

The capacitance of the spherical capacitor is: $C = \frac{4\pi \epsilon_0}{\frac{1}{K}\left(\frac{1}{a} - \frac{1}{c}\right) + \left(\frac{1}{c} - \frac{1}{b}\right)}$ This can also be written in a simplified form as: $C = \frac{4\pi \epsilon_0 Kabc}{b(c-a) + Ka(b-c)}$ or $C = \frac{4\pi \epsilon_0 Kabc}{bc - ab + Kab - Kac}$

Answer 2

The spherical capacitor is divided into two regions:

• Region 1: $a \le r \le c$, filled with dielectric of constant K.
• Region 2: $c \le r \le b$, filled with vacuum ($\epsilon_r = 1$).

Assume charge $+Q$ on the inner shell and $-Q$ on the outer shell. The electric field in Region 1 is $E_1 = \frac{Q}{4\pi K \epsilon_0 r^2}$ and in Region 2 is $E_2 = \frac{Q}{4\pi \epsilon_0 r^2}$.

The potential difference $V$ between the shells is calculated by integrating the electric field from the inner shell (radius $a$) to the outer shell (radius $b$): $V = V_a - V_b = \int_a^b E dr = \int_a^c E_1 dr + \int_c^b E_2 dr$.

Evaluating the integrals: $\int_a^c \frac{Q}{4\pi K \epsilon_0 r^2} dr = \frac{Q}{4\pi K \epsilon_0} \left[-\frac{1}{r}\right]_a^c = \frac{Q}{4\pi K \epsilon_0} \left(\frac{1}{a} - \frac{1}{c}\right)$ $\int_c^b \frac{Q}{4\pi \epsilon_0 r^2} dr = \frac{Q}{4\pi \epsilon_0} \left[-\frac{1}{r}\right]_c^b = \frac{Q}{4\pi \epsilon_0} \left(\frac{1}{c} - \frac{1}{b}\right)$

The total potential difference is the sum of these two parts: $V = \frac{Q}{4\pi K \epsilon_0} \left(\frac{1}{a} - \frac{1}{c}\right) + \frac{Q}{4\pi \epsilon_0} \left(\frac{1}{c} - \frac{1}{b}\right)$ $V = \frac{Q}{4\pi \epsilon_0} \left[ \frac{1}{K}\left(\frac{1}{a} - \frac{1}{c}\right) + \left(\frac{1}{c} - \frac{1}{b}\right) \right]$

The capacitance $C$ is given by $C = \frac{Q}{V}$. $C = \frac{4\pi \epsilon_0}{\frac{1}{K}\left(\frac{1}{a} - \frac{1}{c}\right) + \left(\frac{1}{c} - \frac{1}{b}\right)}$