Question

A point charge $q$ is surrounded by eight fixed identical point charges at the same distance $r$ as shown in figure. How much work will be done by external force to remove point charge $q$ from centre $O$ to infinity very slowly?

• A. $\frac{-2q^2}{\pi {ϵ}_{0}r}$
• B. $\frac{+2q^2}{\pi {ϵ}_{0}r}$
• C. $0$
• D. $\frac{-4q^2}{\pi {ϵ}_{0}r}$

Answer 1

The correct option is A $\frac{-2q^2}{\pi {ϵ}_{0}r}$

From the figure, we can see that net electrostatic force on charge at centre $O$ is not zero and has a tendency to repel it out of the system. So, to not make it accelerate $($or to make it move slowly$)$, we have to apply external force of equal magnitude in the opposite direction of electrostatic force
$($also opposite to displacement$)$. So, work done by external force will be negative.

Also, we know, if all the charges are kept at equal distance $\left(r\right)$ from the point where $V$ is to be evaluated, then we can write, $V=\frac{k{q}_{total}}{r}$
So, potential at centre, $V_O=\frac{q_{total}}{4\pi {ϵ}_{o}r}$
$\Rightarrow V_O=\frac{8q}{4\pi {ϵ}_{o}r}=\frac{2q}{\pi {ϵ}_{o}r}$
Now, $V_O=\frac{-W_{\text{ext}(O\to \infty )}}{q_o}$
$\Rightarrow W_{\text{ext}(O\to \infty )}=-q_{o}V=-\frac{2q^2}{\pi {ϵ}_{o}r}$

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: From the given problem it is evident that}\\ \text{charge at 'O' will have natural tendency to move}\\ \text{to infinity due to repulsion from other charges.}\\ \text{Hence,}{W}_{\text{elec}}\to +\text{ve and}{W}_{\text{ext}}\to -ve.\end{array}$