The capacitor

In our analysis of two parallel sheets of equal amounts of opposite charge (Two parallel sheets of charge ) we learned that the field was effectively constant and restricted to the region between the two sheets.

Indeed, thinking more broadly about this result, it is not surprising.  Matter is made up of positive and negative charges and since they attract each other strongly, mostly they are near each other and cancel each other out. As a result, we typically don't see anything but neutral matter. If we imagine starting from a single sheet with positive and negative charges, we need to do work to pull the positive and negative charge sheets apart. In the pulling apart process, we have to do work on the charges and therefore build up electrical potential energy. We can therefore think of our two-sheet configuration as a way to store electrical energy as a separation of charge. Let's see how this works.

The capacitance of two parallel sheets

Let's suppose we have two sheets of equal and opposite charge that are quite close together. Suppose each sheet has an area $A$, they have opposite charges, $Q$ and $-Q$ (we assume $Q$ is positive), that are spread uniformly, and they are a (small) distance apart, $d$. With this system, we can model the sheets as infinite and conclude that, except near the edges, the field in between the plates will be 

$$E = 4\pi k_C \sigma = 4\pi k_C \frac{Q}{A}$$

Since we are interested in the energy, we want to re-express this relationship in terms of the potential difference between the two plates.

If we draw an x-axis perpendicular to the plates with its origin at the blue (positive) plate and the positive direction towards the red one, then the E field (x component) will be a positive constant between the plates and 0 outside. The graph will look something like the graph shown at the right. (We are ignoring the discrete nature of the charges. This will only show up if we get really close to a single charge.)

From this, we can infer the potential difference between the sheets in one of two ways.

First, we know that the potential energy difference can be calculated from the work needed to move the sheets apart, which is simply force times distance moved.  This is the application of the work-energy theorem for the electric force and PE. Now all we have to do is divide out the test charge on both sides of the equation. Dividing out the test charge from the electric potential energy gives us the potential $V$. To divide out the test charge from the work we simply have to divide the force by the test charge and end up with the electric field.

So, with the charge of the test charge divided out the work-energy theorem becomes a relation between a potential and a field:

$$\Delta V = =\int^f_i{\overrightarrow{E}\cdot \overrightarrow{dr}}$$

where i refers to the starting (initial) point and f to the ending (final) point. Since $E$  is a constant between the plates, if we start at 0 and integrate along the x direction the integral becomes simple:

$$\Delta V = =\int^f_i{{E dx}}= -Ed$$

Alternatively, we can use that the E field is the gradient of the potential. Since we only have an x component, we have

$$E_x = -\frac{\partial V}{\partial x} = -\frac{dV}{dx} = -\frac{\Delta V}{\Delta x}$$

$$\Delta V = - E \Delta x = -Ed$$

This gives the same result — not a surprise since the derivative is the opposite of the integral and they are representing two aspects of the same relationship between the field and the potential.

As a result, we can replace the E field, $E$, by $ΔV/d$ to get:

$$\frac{\Delta V}{d} = 4\pi k_C \frac{Q}{A}$$

We've ignored the sign here and are just representing a relation between magnitudes. That's because there is both a $Q$ and $-Q$ (for a total charge equal to 0) and the sign of $ΔV$ depends on which side we take as initial and which as final. So the sign really depends on how we choose to describe it. But the magnitude relationship holds no matter what choices we make.

This relation tells us that the charge separation on the sheets is associated with a potential difference. Expressing $Q$ as something times the potential difference, $\Delta V$, allows us to identify the capacitance of this system. 

$$Q = C\Delta V$$

$$C = \frac{A}{4 \pi k_C d}$$

This is the simplest capacitor (since the E field is constant inside) and gives up a way to create a region of space with a constant E field.

Note that if we use the definition $k_C = \frac{1}{4\pi \epsilon_0}$, the capacitance of two parallel plates of area $A$ and separation $d$ becomes

$$C = \epsilon_0 A/d$$.

Workout: The capacitor

Joe Redish 2/20/12 &Wolfgang Losert 2/22/13