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Suppose that a capacitor is built with two long cylinders of length L,L, one with radius R1R_1 and the other with radius R2,R_2, where R2>R1.R_2\gt R_1. We wish to calculate the capacitance of this capacitor.

Suppose that the inner cylinder has charge QQ and the outer cylinder has charge −Q.-Q. The linear charge density is λ=QL.\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 rr (where R1<r<R2R_1\lt r\lt R_2) using Gauss's Law. We choose a coaxial cylinder of radius rr and length xx as our Gaussian surface: 2πrx×E(r)=λxϵ02\pi rx\times E(r) = \frac{\lambda x}{\epsilon_0} E(r)=λ2πϵ0rE(r) = \frac{\lambda}{2\pi\epsilon_0 r} We can now calculate the magnitude of the electric potential VV between the two cylinders: ∣V∣=∫R1R2E(r)dr|V| = \int_{R_1}^{R_2}E(r)dr ∣V∣=∫R1R2λ2πϵ0rdr|V| = \int_{R_1}^{R_2}\frac{\lambda}{2\pi\epsilon_0 r}dr ∣V∣=λ2πϵ0ln⁡R2R1|V| = \frac{\lambda}{2\pi\epsilon_0}\ln\frac{R_2}{R_1} Now, we calculate the capacitance: C=Q∣V∣C = \frac{Q}{|V|} C=2πLϵ0ln⁡R2R1C = \frac{2\pi L\epsilon_0}{\ln\frac{R_2}{R_1}}

Type a concept, like “slope” or “photosynthesis”.

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