Question

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

• A. $100{\text{Vm}}^{-1}$ along X-axis
• B. $100{\text{Vm}}^{-1}$ along Y-axis
• C. $200{\text{Vm}}^{-1}$ at an angle ${120}^{\circ }$ with X-axis
• D. $50{\text{Vm}}^{-1}$ at an angle ${120}^{\circ }$ with X-axis

Answer 1

The correct option is C $200{\text{Vm}}^{-1}$ at an angle ${120}^{\circ }$ with X-axis


Using $dV=-\overrightarrow{E}.\overrightarrow{dr}$
$\Rightarrow \Delta V=-E.\Delta r\cos \theta$
$\Rightarrow E=\frac{-\Delta V}{\Delta r\cos \theta }$
$\Rightarrow E=\frac{-(20-10)}{10\times {10}^{-2}\cos {120}^{\circ }}$
$=\frac{-10}{10\times {10}^{-2}(-\sin {30}^{\circ })}=\frac{-{10}^2}{-1/2}=200\text{V/m}$
Direction of $E$ is perpendicular to the equipotential surface.
i.e., at ${120}^{\circ }$ with x-axis.