The potential is varying with distance as V=12(y2−4x) volt. The electric field at x=1m & y=1m is

E=−(dvdr)V=12(y2−4x)E=−(dvdr)=−[2y2j^−42i]Put y=2E=−2j^+2i^ = 2i^−2j^

The potential is varying with distance as V=12(y2−4x) volt. The electric field at x=1m & y=1m is

E=−(dvdr)V=12(y2−4x)E=−(dvdr)=−[2y2j^−42i]Put y=2E=−2j^+2i^ = 2i^−2j^

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The potential is varying with distance as $V = \frac{1}{2} (y^{2} - 4 x)$ volt. The electric field at x=1m & y=1m is

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$- 2 \hat{i} + 2 \hat{j} v / m$

$2 \hat{i} + \hat{j} v / m$

$- 2 \hat{i} + \hat{j} \frac{v}{m}$

$2 \hat{i} - \hat{j} v / m$

answer is C.

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$\begin{array}{l} E = - (\frac{d v}{d r}) \\ V = \frac{1}{2} (y^{2} - 4 x) \\ E = - (\frac{d v}{d r}) = - [\frac{2 y}{2} \hat{j} - \frac{4}{2} i] \\ P u t y = 2 \\ E = - 2 \hat{j} + 2 \hat{i} \\ = 2 \hat{i} - 2 \hat{j} \end{array}$

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The potential is varying with distance as V=12(y2−4x) volt. The electric field at x=1m & y=1m is

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The potential is varying with distance as $V = \frac{1}{2} (y^{2} - 4 x)$ volt. The electric field at x=1m & y=1m is