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(C) The electric field $E$ is related to the potential difference $dV$ and displacement $dr$ by the relation $dV = -\vec{E} \cdot d\vec{r}$.For a unif

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$200 \, Vm^{-1}$ at an angle $120^\circ$ with $X$-axis

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(C) The electric field $E$ is related to the potential difference $dV$ and displacement $dr$ by the relation $dV = -\vec{E} \cdot d\vec{r}$.For a uniform electric field,this can be written as $\Delta V = -E \Delta r \cos 	heta$,where $	heta$ is the angle between the electric field vector and the displacement vector.The magnitude of the electric field is $E = \frac{|\Delta V|}{\Delta r \cos \alpha}$,where $\alpha$ is the angle between the normal to the equipotential surface and the direction of displacement.From the figure,the distance between the $10 \, V$ and $20 \, V$ equipotential surfaces along the $X$-axis is $\Delta x = 10 \, cm = 0.1 \, m$.The perpendicular distance between these surfaces is $\Delta r = \Delta x \sin 30^\circ = 0.1 	imes 0.5 = 0.05 \, m$.The magnitude of the electric field is $E = \frac{\Delta V}{\Delta r} = \frac{20 - 10}{0.05} = \frac{10}{0.05} = 200 \, V/m$.The electric field is always perpendicular to the equipotential surfaces. Since the surfaces make an angle of $30^\circ$ with the $X$-axis,the normal to these surfaces makes an angle of $90^\circ + 30^\circ = 120^\circ$ with the positive $X$-axis.

Equipotential surfaces are shown in the figure. Then the electric field strength will be

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