Determining the capacitance for a parallel-plate capacitor was quiet straightforward but the expression for the capacitance we determined for the parallel-plate capacitor in the previous article is not the same for other capacitances of more complex shapes we are about to discuss in this article.

## Spherical Capacitor

Consider two concentric spherical shells separated by vacuum as shown in Figure 1. The inner shell has charge $+Q$ and the outer shell has charge $-Q$. It means the charge on the capacitor is $Q$ (note that it is a common practice to represent the magnitude of charge on either conductor as the charge on the capacitor, otherwise you know the charge of a capacitor is zero). The outer radius of the inner shell is $r_a$ and the inner radius of the outer shell is $r_b$.

The electric field due to the outer shell has no effect on electric field between the shells. We know from the application of Gauss's law that the electric field inside a conducting sphere is zero. It means the electric field between the shells is due inner shell with charge $+Q$. So take any point between the shells at distance $r$ from the centre and the electric field at that point due to the inner shell is

\[E = k\frac{Q}{r^2}\]

The electric field is radially outward and the small potential difference for a very small displacement $d\vec r$ along the radial line is $\vec E \cdot d\vec r$. Now the potential difference between the plates is determined by integrating this small potential difference from $r_a$ to $r_b$. Note that both $\vec E$ and $d\vec r$ are parallel, so

\[\begin{align*} {V_{ab}} &= \int\limits_{{r_a}}^{{r_b}} {\vec E \cdot d\vec {r} } \\ {\rm{or, }}\quad {V_{ab}} &= \int\limits_{{r_a}}^{{r_b}} {Edr = kQ\frac{{{r_b} - {r_a}}}{{{r_a}{r_b}}}} \end{align*}\]

Now the capacitance $C$ is $C = Q/V_{ab}$ and

\[C = \frac{1}{k}\frac{{{r_a}{r_b}}}{{{r_b} - {r_a}}} = {\epsilon_0}\frac{{4\pi {r_a}{r_b}}}{{{r_b} - {r_a}}}\]

In the above expression of the capacitance the quantity $4\pi r_a r_b$ is the geometric mean of areas $4\pi {r_a}^2$ and $4\pi {r_b}^2$.

## Cylindrical Capacitor

Consider two coaxial conducting cylinders separated by vacuum. The inner cylinder has linear charge density $+ \lambda$ and radius $r_a$ and and the outer cylinder has linear charge density $-\lambda$ and inner radius $r_b$.

Similar to the above discussion the electric field due to the outer cylinder has no effect on the electric field between the cylinders and therefore the electric field between the cylinders is only due to the charge in the inner cylinder. And we have determined the electric field of an infinite line of charge in this article using Gauss's law. The same expression also applies to an infinite cylindrical charge distribution; if you enclose the charged conducting cylinder with a Gaussian surface the charge on the cylinder is the same as the charge concentrated along its axis. So the electric field at distance $r$ between the cylinders is

\[E = k\frac{{2\pi }}{r}\]

Now the potential difference of inner cylinder $a$ with respect to the outer cylinder $b$, $V_{ab}$ is obtained by integrating the small potential difference $\vec E \cdot d\vec r$ in a very small displacement $d\vec r$ along the radial line from $r_a$ to $r_b$ (note that the electric field is radially outward so $\vec E$ and $d\vec r$ are parallel):

\[{V_{ab}} = \int\limits_{{r_a}}^{{r_b}} {E \cdot dr} = 2k\lambda \int\limits_{{r_a}}^{{r_b}} {\frac{{dr}}{r}} = 2k\lambda \ln \left( {\frac{{{r_b}}}{{{r_a}}}} \right)\]

If $L$ is the length of the cylinders, $\lambda = Q/L$ and $C = Q/V_{ab}$ is

\[C = \frac{Q}{{{V_{ab}}}} = \frac{Q}{{2k(Q/L)\ln ({r_b}/{r_a})}} = \frac{L}{{2k\ln ({r_b}/{r_a})}}\]

As you can see from the above expression that the capacitance is proportional to the length of the cylinders and also depends on the radii $r_a$ and $r_b$.