An electric charge 10⁻³ µC is placed at the origin (0, 0 ) of X-Y coordinate system. Two points A and B are situated at $(\sqrt{2}, \sqrt{2})$ and (2, 0) respectively. The potential difference between the points A and B will be

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9 V

zero

2 V

4.5 V

answer is B.

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Detailed Solution

$\begin{array}{l} V_{A} = \frac{1}{4 {πε}_{0}} \frac{10^{- 3} μC}{(2)} \\ V_{B} = \frac{1}{4 {πε}_{0}} \frac{10^{- 3} μC}{(2)} \\ V_{A} - V_{B} = 0 \end{array}$

As the two points are at equal distance from the point charge, they are at equipotential points

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