The electric potential in a certain region of space is given by $V = - 3 x^{2} + 4 x,$ where x is in metres and V is in volts. In this region.

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Electric field in the region is given by $- (6 x - 4) \hat{i}$

Electric field in the region is given by $- (6 x + 4) \hat{i}$

Equipotential surfaces are planes parallel to XY plane

Equipotential surfaces are planes parallel to YZ plane

answer is B.

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Detailed Solution

$\vec{E} = - \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j} - \frac{\partial V}{\partial z} \hat{k} = - (- 6 x + 4) \hat{i}$

So, the equipotential surfaces would be planes parallel to YZ plane, as $\vec{E}$ is perpendicular to the equipotential surface.

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