The electrical potential V at any point O ( $x, y, z$ all in meters) in space is given by $V = 4 x^{2} volt$ . The electric field at the point $(1 m, 0, 2 m)$ in volt metre is

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16 along positive $Z$ – axis

8 along positive $X$ – axis

16 along negative $X$ – axis

8 along negative $X$ – axis

answer is A.

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Detailed Solution

The electric potential $V (x, y, z) = 4 x^{2} volt$

$\vec{E} = - (\hat{i} \frac{\partial v}{\partial x} + \hat{j} \frac{\partial v}{\partial y} + \hat{k} \frac{\partial v}{\partial z})$

Hence $\vec{E} = - 8 x \hat{i},$ so at point $(1 m, 0, 2 m)$

$\vec{E} = - 8 i volt / metre$

or 8 along negative X – axis

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