First slide

Electrostatic potential

Question

Four equal charges Q are placed at the four corners of a square of each side ‘a’ . Work done in removing a charge –Q from its center to infinity is

Easy

A

0
B

$\frac{\sqrt{2} Q^{2}}{4 π ε_{0} a}$
C

$\frac{\sqrt{2} Q^{2}}{π ε_{0} a}$
D

$\frac{Q^{2}}{π ε_{0} a}$

Solution

Potential at center O of the square
$V_{o} = V_{1} + V_{2} + V_{3} + V_{4} = 4 V = 4 (\frac{Q}{4 π ε_{0} r}) = 4 (\frac{Q}{4 π ε_{0} (a / \sqrt{2})}) (∵ r = \frac{a}{\sqrt{2}})$
Work done in shifting (-Q) charge from center to infinity
$W = - Q (V_{\infty} - V_{O}) = Q V_{0} = \frac{4 \sqrt{2} Q^{2}}{4 π ε_{0} a} = \frac{\sqrt{2} Q^{2}}{π ε_{0} a} .$

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