Q. A square of side $a$has charge $\mathcal{Q}$at its centre and charge q at one of the corners. The work required to be done in moving the charge q from the corner to the diagonally opposite corner is
A) Zero
B) $\frac{\mathcal{Q}q}{4\pi\varepsilon_{0}a}$
C) $\frac{\mathcal{Q}q\sqrt{2}}{4\pi\varepsilon_{0}a}$
D) $\frac{\mathcal{Q}q}{2\pi\varepsilon_{0}a}$
Explanation
(a) The potential due to charge $q$ at distance $r$ is given by
$V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$
If $W$ be the work done in moving the charge from $A\ to\ B$ then the potential difference $(V)$ is given by
$V_{A} - V_{B} = \frac{W}{q}$
Both work (W) and charge (q) are scalar quantities hence potential difference $\left( V_{A} - V_{B} \right)$will also be a scalar quantity.
Here,
$V_{A} = V_{B} = \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}}{a/\sqrt{2}}$
Since, $\mathcal{Q}$is same for both,
$V_{A} - V_{B} = 0$
$W = 0$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.