Cylindrical Capacitors

Suppose that a capacitor is built with two long cylinders of length $L,$ one with radius $R_1$ and the other with radius $R_2,$ where $R_2\gt R_1.$ We wish to calculate the capacitance of this capacitor.

Suppose that the inner cylinder has charge $Q$ and the outer cylinder has charge $-Q.$ The linear charge density is $\lambda = \frac{Q}{L}.$ We must calculate the potential difference between the two cylinders. To do so, we must first calculate the electric field at radius $r$ (where $R_1\lt r\lt R_2$) using Gauss's Law. We choose a coaxial cylinder of radius $r$ and length $x$ as our Gaussian surface: $$2\pi rx\times E(r) = \frac{\lambda x}{\epsilon_0}$$ $$E(r) = \frac{\lambda}{2\pi\epsilon_0 r}$$ We can now calculate the magnitude of the electric potential $V$ between the two cylinders: $$|V| = \int_{R_1}^{R_2}E(r)dr$$ $$|V| = \int_{R_1}^{R_2}\frac{\lambda}{2\pi\epsilon_0 r}dr$$ $$|V| = \frac{\lambda}{2\pi\epsilon_0}\ln\frac{R_2}{R_1}$$ Now, we calculate the capacitance: $$C = \frac{Q}{|V|}$$ $$C = \frac{2\pi L\epsilon_0}{\ln\frac{R_2}{R_1}}$$