A spherical capacitor consists of two concentric conducting shells of inner radius $R_1$ and outer r

Question

A spherical capacitor consists of two concentric conducting shells of inner radius $R_1$ and outer radius $R_2$. It is given that $R_2 = e R_1$, where $e$ is Euler's number ($e \approx 2.718$). The gap between the shells is filled with an isotropic dielectric whose relative permittivity varies radially as $K(r) = \alpha/r$, where $\alpha$ is a positive constant and $r$ is the radial distance from the centre. The equivalent capacitance of this capacitor can be expressed as $C = N \pi \epsilon_0 \alpha$. Find the integer value of $N$.

Professor PC Thomas Classes · Answer

Consider an elemental spherical shell of dielectric of radius $r$ and thickness $dr$. Its capacitance is $dC = \frac{K(r) \epsilon_0 (4\pi r^2)}{dr}$. These elemental capacitors are connected in series. Therefore, the total equivalent capacitance is found by integrating their reciprocals: $\frac{1}{C} = \int_{R_1}^{R_2} \frac{dr}{K(r) \epsilon_0 4\pi r^2}$. Substituting $K(r) = \alpha/r$, we get $\frac{1}{C} = \int_{R_1}^{R_2} \frac{dr}{(\alpha/r) \epsilon_0 4\pi r^2} = \int_{R_1}^{R_2} \frac{dr}{4\pi\epsilon_0 \alpha r}$. Evaluating the integral: $\frac{1}{C} = \frac{1}{4\pi\epsilon_0 \alpha} \ln\left(\frac{R_2}{R_1}\right)$. Given $R_2/R_1 = e$, we have $\ln(e) = 1$. Thus, $\frac{1}{C} = \frac{1}{4\pi\epsilon_0 \alpha}$, which gives $C = 4\pi\epsilon_0 \alpha$. Comparing this with $N \pi \epsilon_0 \alpha$, we find $N = 4$.