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 it 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 a 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 $$\Delta V=V_f-V_i=-\int_{r_i}^{r_f} \vec{E}\bullet \text{d}\vec{r}$$

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 $a<s<b$.

We have done a similar problem involving Gauss's Law before. In order to find the electric flux in terms of the electric field, we write $$\Phi=\int\vec{E}\bullet \text{d}\vec{A}$$

The integral is over the entire Gaussian surface, upon which $\vec{E}\bullet \text{d}\vec{A}$ takes on different values. On the top and bottom of the Gaussian surface, the electric field is directed radially, whereas the area-vectors point up and down, respectively. Both cases yield $\vec{E}\bullet \text{d}\vec{A}=0$, so we can safely ignore this part of the integral. On the wall of the Gaussian surface, the electric field and the area-vectors are both directed radially, so their dot product simplifies to $$\vec{E}\bullet \text{d}\vec{A} = E(s)\text{d}A$$

We write $E(s)$ because the electric field magnitude depends only on the distance from the central vertical axis. We can rotate the entire system about this axis and it does not change, so $E$ cannot depend on $\phi$. We can shift the entire system along this axis, and it does not change (using the “$L$ is very large” assumption), so $E$ cannot depend on $z$. all that remains is $r$. And since all along the Gaussian surface, $r=s$, we write $E(s)$.

At this point, we can apply Gauss' Law. We equate flux to the charge enclosed divided by $\epsilon_0$: \begin{align*} \frac{Q_{enclosed}}{\epsilon_0} &= \int\vec{E}\bullet \text{d}\vec{A} \\ &= \int E(s)\text{d}A \\ &= E(s) \int \text{d}A \\ &= E(s) A_{wall} \\ &= E(s) (2\pi sh) \end{align*}

The last thing we need is $Q_{enclosed}$. This is simply the fraction of $Q$ that the Gaussian surface encloses. Since the height of the Gaussian cylinder is $h$, we have $Q_{enclosed}=\frac{h}{L}Q$. We can now write the magnitude of the electric field at a radius $s$ from the central vertical axis (given that $a<s<b$). $$E(s)=\frac{Q_{enclosed}}{\epsilon_0}\frac{1}{2\pi sh} = \frac{Q}{2\pi\epsilon_0 L s}$$

This is the magnitude of the electric field between the conducting cylinders. We can now find the potential difference between the plates. Notice below that we simplify the dot product in the same way we simplified the flux through the wall of the Gaussian surface. \begin{align*} |\Delta V| &= \int_{r_i}^{r_f} \vec{E}\bullet \text{d}\vec{r} \\ &= \int_a^b E(s)\text{d}s \\ &= \frac{Q}{2\pi\epsilon_0 L} \int_a^b \frac{1}{s}\text{d}s \\ &= \left. \frac{Q}{2\pi\epsilon_0 L} \log (s)\right|_{s=a}^{s=b} \\ &= \frac{Q}{2\pi\epsilon_0 L}\log\left(\frac{b}{a}\right) \end{align*}

Now, the capacitance: $$C=\frac{Q}{\Delta V} = \frac{2\pi\epsilon_0 L}{\log\left(\frac{b}{a}\right)}$$