Question

A thin conducting pipe of radius $r_1$ and length ll is nested inside a thin cylindrical shell with the same length and radius $r_2$. This cylindrical capacitor is connected to the terminals of a battery with voltage, $V$. The charge applied to the curved surface of the inner pipe is $+Q$.
Derive an expression for the capacitance of this system in terms of the given variables and fundamental constants.

• A. $\frac{8\pi l{ϵ}_0}{ln\left(\frac{r_2}{r_1}\right)}$
• B. $\frac{4\pi l{ϵ}_0}{ln\left(\frac{r_1}{r_2}\right)}$
• C. $\frac{2\pi l{ϵ}_0}{ln\left(\frac{r_2}{r_1}\right)}$
• D. $\frac{2\pi l{r}_2{ϵ}_0}{(r_2-r_1{)}^2}$
• E. $C=\frac{2\pi l{r}_2{ϵ}_0}{(r_2-r_1)}$

Answer 1

Answer: (C)

$\frac{2\pi l{ϵ}_0}{ln\left(\frac{r_2}{r_1}\right)}$

The electric field at a distance $r$ from the center of coaxial cylinders using Gauss' Law=$E=\frac{Q}{2\pi {ϵ}_{0}rl}$

Hence potential difference between the plates

$=V={\int }_{r_1}^{r_2}\frac{Q}{2\pi {ϵ}_{0}rl}dr$

$⟹Q=\left(\frac{2\pi {ϵ}_{0}l}{ln\left(\frac{r_2}{r_1}\right)}\right)V$

$⟹C=\frac{2\pi l{ϵ}_0}{ln\left(\frac{r_2}{r_1}\right)}$