Three equal charges (q) are kept at three corners of equilateral triangle of side ‘d’. The electric potential at the centre of the triangle is

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$\frac{1}{4 {πε}_{0}} \frac{3 \sqrt{3} q}{d}$

$\frac{1}{4 {πε}_{0}} \frac{\sqrt{3} q}{d}$

$\frac{1}{4 {πε}_{0}} \frac{3 q}{d}$

$\frac{1}{4 {πε}_{0}} \frac{4 \sqrt{2} q}{d}$

answer is A.

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Detailed Solution

$\begin{array}{l} V_{net} = V_{1} + V_{2} + V_{3}; V = \frac{1}{4 {πε}_{0}} \times \frac{3 q}{(\frac{d}{\sqrt{3}})} \\ V = \frac{1}{4 {πε}_{0}} \frac{3 \sqrt{3} q}{d} \end{array}$

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