If V₀ be the potential at origin in an electric field $E = E_{x} i + E_{y} j$ the potential at (x,y) is

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$\sqrt{x^{2} + y^{2}} \sqrt{E_{x}^{2} + E_{y}^{2} - V_{0}}$

V₀ – xE_x – yE_y

xE_x + yE_y – V₀

V₀ + xE_x + yE_y

answer is A.

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Detailed Solution

Potential difference will be same
$\begin{array}{l} dV = - \vec{E} \cdot \vec{dr}; V - V_{o} = - (E_{x} i + E_{y} j) \cdot (x \hat{i} + y \hat{j}) \\ V = V_{o} - {xE}_{x} - {yE}_{y} \end{array}$

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