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(C) The electric field $\vec{E}$ is related to the potential gradient by the relation $\vec{E} = -
abla V$. The magnitude of the potential gradient i

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$\frac{\pi}{2}$

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(C) The electric field $\vec{E}$ is related to the potential gradient by the relation $\vec{E} = -
abla V$. The magnitude of the potential gradient is maximum in the direction of the electric field. By definition,an equipotential surface is a surface where the electric potential $V$ is constant at every point. For any displacement $d\vec{r}$ along the equipotential surface,the change in potential $dV = -\vec{E} \cdot d\vec{r} = 0$. This implies that the electric field vector $\vec{E}$ must be perpendicular to the surface at every point. Since the maximum potential gradient points in the direction of the electric field,the angle between the maximum potential gradient and the equipotential surface is $90^{\circ}$ or $\frac{\pi}{2}$ radians.

What is the angle between the maximum value of the potential gradient and the equipotential surface?

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