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Four electric charges +q, +q, -q and –q are placed at the corners of square of side 2L(as shown in the figure). The electric potential at point A, midway between the two charges +q and +q is

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$\frac{1}{4 π ε_{0}} \frac{2 q}{L} (1 + \frac{1}{\sqrt{5}})$

Zero

$\frac{1}{4 π ε_{0}} \frac{2 q}{L} (1 - \frac{1}{\sqrt{5}})$

$\frac{1}{4 π ε_{0}} \frac{2 q}{L} (1 + \sqrt{3})$

answer is D.

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Detailed Solution

By the principal of superposition, we have

$V = \sum \frac{1}{4 π ε_{0}} . \frac{q}{r}$

$V (A) = \frac{2}{4 π ε_{0}} . \frac{q}{L} - \frac{2}{4 π ε_{0}} . \frac{q}{\sqrt{5} L}$

$V (A) = \frac{2 q}{4 π ε_{0} L} [1 - \frac{1}{\sqrt{5}}]$

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