Question

Find the electrostatic energy stored in a cylindrical shell of length $l$, inner radius $a$ and outer radius $b$, coaxial with a uniformly charged wire with linear charge density $\lambda$.

• A. $U=\frac{{\lambda }^{2}l}{2\pi {\epsilon }_0}\log \left(\frac{b}{a}\right)$
• B. $U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\log \left(\frac{a}{b}\right)$
• C. $U=\frac{{\lambda }^{2}l}{8\pi {\epsilon }_0}\log \left(\frac{b}{a}\right)$
• D. $U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\log \left(\frac{b}{a}\right)$

Answer 1

The correct option is D $U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\log \left(\frac{b}{a}\right)$

For this we consider an elemental shell of radius $x$ and width $dx$. The volume of this shell $dV$ can be given as $dV=\left(2\pi xl\right)dx$

The electric field due to the wire at the shell is $E=\frac{\lambda }{2\pi {\epsilon }_{0}x}$

The electrostatic field energy stored in the volume of this shell is $dU=\left[\frac{1}{2}{\epsilon }_0{E}^2\right]dV$

$\Rightarrow dU=\frac{1}{2}{\epsilon }_0{\left(\frac{\lambda }{2\pi {\epsilon }_{0}x}\right)}^2\left(2\pi xl\right)dx$

The total electrostatic energy stored in the above mentioned volume can be obtained by integrating the above expression within limits from $a$ to $b$ as

$U=\int dU=\underset{a}{\overset{b}{\int }}\frac{1}{2}{\epsilon }_0{\left(\frac{\lambda }{2\pi {\epsilon }_{0}x}\right)}^2\left(2\pi xl\right)dx$

$\Rightarrow U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\underset{a}{\overset{b}{\int }}\frac{1}{x}dx$

$\Rightarrow U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\left(\log x{|}_a^b\right)$

$\Rightarrow U=\frac{{\lambda }^{2}l}{4\pi {\epsilon }_0}\log \left(\frac{b}{a}\right)$

Hence, option (a) is the correct answer.