Physics Bootcamp

Two concetric metal spherical shells make up a spherical capacitor.

The capacitance of a spherical capacitor with radii \(R_1 \lt R_2\) of shells without anything between the plates is

\begin{equation} C = 4\pi\epsilon_0\, \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)^{-1}.\label{eq-spherical-capacitor-capacitance}\tag{34.3.1} \end{equation}

We have seen before that if we have a material of dielectric constant \(\epsilon_r\) filling the space between plates, the capacitance in (34.3.1) will increase by a factor of the dielectric constant.

To prove the formula given in Eq. (34.3.1), we place positive \(+Q\) on the inner shell and \(-Q\) on the outer shell. We will find potential difference \(V\) and then get \(C\) from \(Q/V\text{.}\) To find the potential between the plates, we integrate electric field from negative plate to positive plate. Therefore, we first find electric field between the plates.

Using Gauss's law for a spherical surface with radius \(r\) between plates, we get

\begin{equation*} E = \begin{cases} 0 \amp r \lt R_1\\ \dfrac{1}{4\pi\epsilon_0}\, \dfrac{Q}{r^2} \amp R_1 \lt r \lt R_2\\ 0 \amp r \gt R_2 \end{cases} \end{equation*}

With zero of potential at \(r=\infty\text{,}\) potential difference can be shown by integrating \(-\vec E\cdot d\vec r = -E dr\) from \(r=R_2\) to \(r=R_1\text{.}\) If you are working through this book with a Calculus background you should do that integral. For others, I will just give the result.

From this we get the capacitance by finding \(V/Q\text{.}\)

It is interesting to note that you can get capacitance of a single spherical conductor from this formula by taking the radius of the outer shell to infinity, \(R_2\rightarrow \infty\text{.}\) Since we will have only one sphere, let us denote its radius by \(R\text{.}\)