A cylindrical capacitor consists of two coaxial conducting cylinders of length $l$ and radii $a$ (inner) and $b$ (outer). Its capacitance is:
A$4\pi\varepsilon_0\dfrac{ab}{b-a}$
B$\dfrac{\varepsilon_0 l}{2\pi\ln(b/a)}$
C$\dfrac{2\pi\varepsilon_0 l}{\ln(b/a)}$
D$\dfrac{2\pi\varepsilon_0 l}{b-a}$
Answer & Solution
Correct answer: C. $\dfrac{2\pi\varepsilon_0 l}{\ln(b/a)}$
Linear charge density $\lambda = Q/l$. Field between shells $E(r) = \lambda/(2\pi\varepsilon_0 r)$. $V = \int_a^b E\,dr = \lambda \ln(b/a)/(2\pi\varepsilon_0)$. So $C = Q/V = 2\pi\varepsilon_0 l / \ln(b/a)$.
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