The Capacitor

The capacitor is an electronic device for storing charge. The simplest type is the parallel plate capacitor, illustrated in figure 17.1. This consists of two conducting plates of area $S$ separated by distance $d$. Positive charge $q$ resides on one plate, while negative charge $-q$ resides on the other.

Figure 17.1: Two views of a parallel plate capacitor.

The electric field between the plates is $E = \sigma / \epsilon_0$, where the charge per unit area on the inside of the left plate in figure 17.1 is $\sigma = q/S$. The density on the right plate is just $- \sigma$. All charge is assumed to reside on the inside surfaces and thus contribute to the electric field crossing the gap between the plates.

The above formula for the electric field comes from applying Gauss's law to the sheet of charge on the positive plate. The factor of $1/2$ present in the equation for an isolated sheet of charge is absent here because all of the electric flux exits the Gaussian surface on the right side -- the left side of the Gaussian box is inside the conductor where the electric field is zero, at least in a static situation.

There is no vector potential in this case, so the electric field is related solely to the scalar potential $\phi$. Integrating $E_x = - \partial \phi / \partial x$ across the gap between the conducting plates, we find that the potential difference between the plates is $\Delta \phi = E_x d = qd/( \epsilon_0 S)$, since $E_x$ is known to be constant in this case. This equation indicates that the potential difference $\Delta \phi$ is proportional to the charge $q$ on the left plate of the capacitor in figure 17.1. The constant of proportionality is $d/(\epsilon_0 S)$, and the inverse of this constant is called the capacitance:

$$C = \frac{\epsilon_0 S}{d} ~~~~ \mbox{(parallel plate capacitor)} .$$ (18.1)

The relationship between potential difference, charge, and capacitance is thus

$$\Delta \phi = q/C ~~~ \mbox{or} ~~~ C = q/ \Delta \phi .$$ (18.2)

The equation for the capacitance of the illustrated parallel plates contains just a fundamental constant ($\epsilon_0$) and geometrical factors (area of plates, spacing between them), and represents the amount of charge the parallel plate capacitor can store per unit potential difference between the plates. A word about signs: The higher potential is always on the plate of the capacitor which which has the positive charge.

Note that equation (17.1) is valid only for a parallel plate capacitor. Capacitors come in many different geometries and the formula for the capacitance of a capacitor with a different geometry will differ from this equation. However, equation (17.2) is valid for any capacitor.

Figure 17.2: Parallel plate capacitor with circular plates in a circuit with current $i$ flowing into the left plate and current $i$ flowing out of the right plate. The magnetic field which occurs when the charge on the capacitor is increasing with time is shown at right as vectors tangent to circles. The radially outward vectors represent the vector potential giving rise to this magnetic field in the region where $x > 0$. The vector potential points radially inward for $x < 0$. The $y$ axis is into the page in the left panel while the $x$ axis is out of the page in the right panel.

We now show that a capacitor which is charging or discharging has a magnetic field between the plates. Figure 17.2 shows a parallel plate capacitor with a current $i$ flowing into the left plate and out of the right plate. This is necessarily accompanied by an electric field which is changing with time: $E_x = q/( \epsilon_0 S) = it / ( \epsilon_0 S)$. Such an electric field can be derived from a scalar potential which is a function of time: $\phi = -itx/( \epsilon_0 S)$. However, the Lorentz condition

$$\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\par... ...rtial z} + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0$$ (18.3)

demands that some component of the vector potential $\mbox{\bf A}$ be non-zero under these circumstances, since $\partial \phi / \partial t$ is non-zero.

How much can we infer about the vector potential from the geometry of the capacitor and equation (17.3)? Substituting $\phi = -itx/( \epsilon_0 S)$ into this equation results in

$$\frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\part... ...ac{\partial A_z}{\partial z} = \frac{i x}{\epsilon_0 c^2 S} ,$$ (18.4)

which suggests a number of different possibilities for $\mbox{\bf A}$. For instance, $\mbox{\bf A} = (0, ixy / ( \epsilon_0 c^2 S) , 0)$ and $\mbox{\bf A} = [0, 0, ixz / ( \epsilon_0 c^2 S)]$ both satisfy equation (17.4). However, neither of these trial choices is satisfactory by itself, as they are not consistent with the cylindrical symmetry of the capacitor about the $x$ axis.

A choice of vector potential which is consistent with the shape of the capacitor and which satisfies the Lorentz condition is obtained by combining these two trial solutions:

$$\mbox{\bf A} = [0,ixy/(2 \epsilon_0 c^2 S) , ixz/(2 \epsilon_0 c^2 S)] .$$ (18.5)

This vector potential leads to the magnetic field

$$\mbox{\bf B} = [0,-iz/(2 \epsilon_0 c^2 S),iy/(2 \epsilon_0 c^2 S)] .$$ (18.6)

These fields are illustrated in the right-hand panel of figure 17.2.