Question

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

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

Answer 1

Answer: (C)

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