Parallel Plate Capacitors

Consider the simplest example of a capacitor: a pair of parallel plates. Let each plate have area $A,$ and let the plates be separated by a distance $d,$ where $d<<A$ (this assumption can be applied to the vast majority of capacitors unless it is explicitly stated otherwise). We will compute the capacitance of this capacitor.

Suppose that one plate is given a charge of $+Q$ and the other plate is given a charge of $-Q.$

Yes, we have $\sigma = \frac{Q}{A}$ as the charge density of the plates.

Now, we must calculate the electric field. This is where the assumption that $d<<A$ helps us, because we may make the simplifying assumption that the plates are effectively infinite in size.

Yes, the electric field strength is $\frac{\sigma}{2\epsilon_0}.$ If you were unsure of how to compute this, a review of Gauss's Law would be helpful.

Now, between the plates, the electric fields produced by each plate add to each other, so the electric field is $E=\frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}.$

Finally, to compute the potential difference between the two plates, we must integrate the electric field over the distance between the plates. Since the electric field is constant, the integration reduces to a simple multiplication: $$V = Ed = \frac{Qd}{\epsilon_0 A}.$$

Now, we use the definition of capacitance to obtain our answer: $$C = \frac{Q}{V}$$ $$C = \frac{\epsilon_0 A}{d}.$$ This is the capacitance of a pair of identical parallel plates.