Question

A thin spherical conducting shell is uniformly charged. The electric potential energy of shell will be :
$($Charge on shell is $q_0$ & radius is $R)$

• A. $\frac{q_0^2}{4\pi {ϵ}_{0}R}$
• B. $\frac{q_0^2}{8\pi {ϵ}_{0}R}$
• C. $\frac{q_0^2}{2{ϵ}_{0}R}$
• D. $\frac{q_0^2}{4\pi {ϵ}_{0}R}$

Answer 1

The correct option is B $\frac{q_0^2}{8\pi {ϵ}_{0}R}$

Let that charge be $q$ on the shell at some instant.

Now a further charge $dq$ has been added to the shell from infinity. So potential at the surface of the shell at this instant is

$V=\frac{q}{4\pi {ϵ}_{0}R}$

Now potential energy associated with small charge $dq$ brought from infinity is

$du=Vdq$

Therefore, total electric Potential energy $U$ of shell is,

$U=\int du$

$\Rightarrow U=\int Vdq={\int }_0^{q_0}\frac{q}{4\pi {ϵ}_{0}R}dq$

$\Rightarrow U=\frac{1}{4\pi {ϵ}_{0}R}{\int }_0^{q_0}qdq$

$\Rightarrow U=\frac{q_0^2}{8\pi {ϵ}_{0}R}$

Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question?}\\ \text{Tip:The limit has been put from}q=0\\ \text{to (q=q\_0) while evaluating integral,}\\ \text{because we assume that all the charge}\\ \text{has been assembled to shell from infinity.}\end{array}$