Find the capacitance of a cylindrical capacitor. The structure of the capacitor is a cylindrical shell inside another cylindrical shell. The two shells become oppositely charged when the capacitor is connected to a power source. The length of the cylinders is $L$, and their radii are $a$ and $b$, with $a<b$.

• The length is $L$
• The inner radius is $a$, and the outer radius is $b$.
• The two cylinders are shells, so all charge will accumulate on the surface.

• Capacitance

• The cylinders are much longer than they are far from one another, i.e., $L >> a, b$.

• We represent capacitance as $$C=\frac{Q}{\Delta V},$$ where $Q$ is the charge on one of the capacitor's conductors (cylinders, in this case), and $\Delta V$ is the potential difference between them.
• We represent the situation below.

In order to find capacitance, we need a charge $Q$ and a potential difference $\Delta V$. But it's impossible to find one without the other. The idea is to assign some charge, give a variable name, and then solve for $\Delta V$. When we compute the capacitance, the assigned charge will cancel out. For now, we say there is charge $Q$ on the inner cylinder, and a charge $-Q$ on the outer cylinder. Since the cylinders are conductors (as they would be in any capacitor), the charge is uniformly distributed on the cylindrical shells.

In order to arrive at potential difference, we will need to go through electric field. Remember from the notes on electric potential that in general we can expression potential difference as $$\Delta V=V_f-V_i=-\int_{r_i}^{r_f} \vec{E}\bullet d\vec{r}$$

We will end up integrating in the radial direction (a convenient choice, as this is how the electric field is directed!) from $a$ to $b$, which will give us the potential difference between the two cylinders.

In order to find the electric field between the cylinders, we will use Gauss' Law. Below, we show a Gaussian surface that is cylindrical and fits inside the capacitor, with a radius $s$, which $a<s<b$.

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