Question

Equipotential surfaces are shown in fig, then the electric field strength will be

Answer 1

Using$,$ $dV=-\overrightarrow{E}\cdot \overrightarrow{dr}$

$\Delta V=-E\cdot \Delta rcs\theta$

$E=\frac{-\Delta V}{\Delta r\cos \theta }$

$E=\frac{-\left(-20-10\right)}{10\times {10}^{-2}\cos {120}^0}=\frac{-10}{10\times {10}^{-2}\left(-\sin {30}^0\right)}=\frac{-{10}^2}{-1/2}=200V/m$

direction of $E$ be perpendicular to the equipotential surface $i.e.,$ at ${120}^0$ with $x-$ axis$.$

Hence,

option $\left(C\right)$ is correct answer.