# Capacitance

Any two conductors separated by an insulator or vacuum form a device called capacitor. It is a device to store charge and electric potential energy. To store energy in a capacitor we can simply charge the capacitor; charging a capacitor means transferring electrons from one conductor the other so that one conductor has positive charge and the other has an equal amount of negative charge.

An amount of work is done to transfer the charge from one conductor to the other against the resulting potential difference and that amount of work is stored as the electric potential energy in the capacitor. This is the same as charging a capacitor. To charge a capacitor you can simply connect the conductors of the capacitor to the opposite terminals of a battery. The figure below shows a parallel-plate capacitor whose two conductors are parallel plates separated at a distance in vacuum.

The fact that an amount of work done being stored in a capacitor is related to the electric field between the conductors which follows that the electric field itself is the storehouse of energy. This is one of the most interesting facts you can explore!

The total charge on a capacitor is zero (equal amount of positive and negative charge add to zero net charge) but it is common to represent the magnitude of charge on either conductor as the charge of the capacitor. If we say a capacitor has charge $Q$, we mean that one conductor has charge $+Q$ (higher potential) and other has charge $-Q$(lower potential) assuming $Q$ is positive.

The ratio of the charge on either conductor $Q$ to the potential difference $V_{ab}$ (potential of conductor a with respect to conductor b) between the conductors is constant called capacitance.

$C = \frac{Q}{V_{ab}} \tag{1}$

The capacitance is always a positive quantity; the charge $Q$ and the potential difference $V_{ab}$ are always expressed as positive quantities. The capacitance of a capacitor is constant and depends on the shapes and sizes of the capacitor, the separation of the conductors and an insulator (dielectric) if present between the conductors. We'll talk about this later as well.

The SI unit of capacitance is one farad ($1\text{F}$), that is $1\text{F} = 1\text{C}/\text{V}$. Please don't confuse between columb($\text{C}$) and capacitance($\text{C}$).

The electric field between the conductors is proportional to the charge of the capacitor and the potential difference is also proportional to the charge, so increasing the charge increases the electric field and increases the potential difference but the ratio of the charge to the potential difference remains the same.

The greater amount of capacitance for capacitor means the greater amount of charge for a given potential difference and therefore the capacitors with greater capacitance can store more charge and energy.

In Figure 1 above we have a simple parallel plate capacitor with charge $Q$ (one plate has $+Q$ and another has $-Q$ assuming $Q$ is positive) separated at a distance $d$. Each plate has area $A$. If $\sigma$ (absolute value) is the surface charge density, we have already determined the electric field for this arrangement of charge using Gauss's law in this article which is $E = \sigma / \epsilon_0$. So,

$E = \frac{\sigma }{\epsilon_0}$

We know $\sigma = Q/A$ and,

$E = \frac{Q }{\epsilon_0 A}$

The potential difference $V_{ab}$ for the uniform electric field with separation $d$ is

$V_{ab} = Ed = \frac{Qd}{\epsilon_0 A}$

And finally our capacitance is

$C = \frac{Q}{V_{ab}} = \epsilon_0 \frac{A}{d} \tag{2}$

It follows that the capacitance depends on the area of each plates and the separation distance. For a particular capacitor both $A$ and $d$ are constants and therefore the capacitance is constant for a capacitor. But the capacitance also depends on the material present between the conductors. If there is air instead of vacuum the difference in capacitance is negligible. If there is other material present between the conductors, the capacitance will be largely effected.

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

## Capacitance of 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 to 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 $4\pi {r_b}^2$.

## Capacitance of 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

$E = k\frac{{2\lambda }}{r}$

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 $r_b$.

Electromagnetism