The capacitance of any capacitor is defined as \(C≡\frac{ΔV}{Q}\). In this lesson, we'll be interested in finding the capacitance for what is known as a parallel-plate capacitor. A parallel-plate capacitor is a capacitor whose conductors are two thin plates which are parallel to one another and seperated by an insulator as illustrated in Figure 1. We'll assume that the two conductors are separated by a vacuum.
Let's try to find an expression for the voltage \(ΔV\) across the capacitor and then, after that, we'll substitute it into the equation \(C=\frac{ΔV}{Q}\). The general expression for the voltage between any two points is given by
$$ΔV=q\int_B^A\vec{E}·d\vec{r}.\tag{1}$$
Figure 1: Parallel-plate capacitor. Each conductor is a flat plate with charges \(-Q\) and \(+Q\), areas of \(A\), and separated at a distance of \(d\).
Courtesy of the Department of Physics and Astronomy, Michigan State University
For our purposes we'll let the point \(A\) and \(B\) be the two points illustrated in Figure 1. We showed in a previous lesson (using Guass's law) that the electric field in between two parallel plates is a constant given by \(\vec{E}=\frac{σ}{ε_0}\hat{i}\) where \(σ\) is the magnitude of the charge density on each plate and \(\hat{i}\) is a unit vector pointing from the negatively charged plate to the positively charged plate along the x-axis as illustred in Figure 1. Substituting \(\vec{E}=\frac{σ}{ε_0}\hat{i}\) into Equation (1), we have
$$ΔV=\frac{σq}{ε_0}\int_B^A\hat{i}·d\vec{r}.\tag{2}$$
[Explain why \(\hat{i}·d\vec{r}\) simplifies to \(dr\).]
$$ΔV=\frac{σq}{ε_0}\int_B^Adr=\frac{σqd}{ε_0}.\tag{3}$$
Substituting Equation (3) into \(C=\frac{ΔV}{Q}\), we find that the capacitance of a parallel-plate capacitor (where each conductor is seperated by vacuum) is given by
$$C=\frac{σqd}{qε_0}=$$