Q. Four equal charges $Q$ are placed at the four corners of a square of each side is $'a'$. Work done in removing a charge $–Q$ from its centre to infinity is
A) 0
B) $\frac{\sqrt{2}Q^{2}}{4\pi\varepsilon_{0}a}$
C) $\frac{\sqrt{2}Q^{2}}{\pi\varepsilon_{0}a}$
D) $\frac{Q^{2}}{2\pi\varepsilon_{0}a}$
Explanation
(c) Potential at centre $O$ of the square
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$V_{0} = 4\left( \begin{array}{r}
\frac{Q}{4\pi_{\ }\varepsilon_{0}(a/\sqrt{2})} \\
\
\end{array} \right)$
Work done in shifting $( - Q)$ charge from centre to infinity
$W = - Q\left( V_{\infty} - V_{0} \right) = QV_{0}$
$= \frac{4\sqrt{2}Q^{2}}{4\pi\varepsilon_{0}a} = \frac{\sqrt{2}Q^{2}}{\pi\varepsilon_{0}a}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.