**A long cylindrical conductor has a radius $r_{a}$ and a linear charge density $+ \lambda$. It is surrounded by a coaxial cylindrical conducting shell with inner radius $r_{b}$ and linear charge density $- \lambda$. Calculate the capacitance per unit length for this capacitor, assuming that there is vacuum in the space between cylinders.**

First we calculate the electric field between the two cylinders by using Gauss’s Law: (Note that $\lambda = \frac{q}{L}$)

$$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E &= \frac{q}{2 \pi r \epsilon_{0} L} \\ &= \frac{\lambda}{2 \pi r \epsilon_{0}} \end{aligned}$$

We will then perform integration on E to get V.

$$\begin{aligned} – \frac{dV}{dr} &= E \\ – \frac{dV}{dr} &= \frac{\lambda}{2 \pi \epsilon_{0} r} \\ \int\limits_{V_{a}}^{V_{b}} dV &= \, – \int\limits_{r_{a}}^{r_{b}} \frac{\lambda}{2 \pi \epsilon_{0} r} \, dr’ \\ V_{b}-V_{a} &= \, – \frac{\lambda}{2 \pi \epsilon_{0}} \left[ \text{ln} \, r \right]_{r_{a}}^{r_{b}} \\ – (V_{a}-V_{b}) &= \, – \frac{\lambda}{2 \pi \epsilon_{0}} \, \text{ln} \, \left( \frac{r_{b}}{r_{a}} \right) \\ V_{ab} &= \frac{\lambda}{2 \pi \epsilon_{0}} \, \text{ln} \frac{r_{b}}{r_{a}} \end{aligned}$$

Hence,

$$\begin{aligned} C &= \frac{Q}{V_{ab}} \\ &= \frac{2 \pi \epsilon_{0} L}{\text{ln} \, \frac{r_{b}}{r_{a}}} \end{aligned}$$

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