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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