Question

Equipotential surfaces are shown in figure. 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 $120^{\circ}$ with $X$ -axis
• D. $50\, Vm ^{-1}$ at an angle $120^{\circ}$ with $X$ -axis

Answer 1

Answer: (C)

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