Question

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

$[k=\frac{1}{4\pi {\epsilon }_0}]$

• A. $2k{\lambda }^{2}l\ln \left(\frac{b}{a}\right)$
• B. $k{\lambda }^{2}l\ln \left(\frac{b}{a}\right)$
• C. $\frac{k{\lambda }^{2}l}{2}\ln \left(\frac{b}{a}\right)$
• D. $\frac{2k{\lambda }^{2}l}{3}\ln \left(\frac{b}{a}\right)$

Answer 1

The correct option is B $k{\lambda }^{2}l\ln \left(\frac{b}{a}\right)$

Let us consider an elemental shell of radius $x$ and width $dx$ as shown in the figure.


The volume of this shell $dV$ can be given as,

$dV=2\pi xldx$

The electric field due to the wire at the shell is,

$E=\frac{2k\lambda }{x}$

The electrostatic field energy stored in the volume of this shell is,

$dU=\frac{1}{2}{\epsilon }_0{E}^{2}dV$

$\Rightarrow dU=\frac{1}{2}{\epsilon }_0{\left(\frac{2k\lambda }{x}\right)}^2\times \left(2\pi xldx\right)$

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

$U=\int dU={\int }_a^b\frac{1}{2}{\epsilon }_0{\left(\frac{2k\lambda }{x}\right)}^2\times \left(2\pi xldx\right)$

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

Since, $k=\frac{1}{4\pi {\epsilon }_0}$

$\Rightarrow U=\frac{k^2{\lambda }^{2}l}{k}{\int }_a^b\frac{1}{x}dx$

$\Rightarrow U=k{\lambda }^{2}l\ln \left(\frac{b}{a}\right)$

Hence, option (b) is the correct answer.