Question

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

• A. 100 Vm-1 along X-axis
• B. 100 Vm-1 along Y-axis
• C. 200 Vm-1 at an angle ${120}^0$ wirh X-axis
• D. 50 Vm-1 at an angle ${180}^0$ wirh X-axis

Answer 1

Answer: (C)

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.