Q.
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
Explanation
(c) Using $dV = - \overrightarrow{E}.d\overrightarrow{r}$
$\Rightarrow \mathrm{\Delta}V = - E.\mathrm{\Delta}r\cos\theta$
[image:1_image377.emf]
$\Rightarrow E = \frac{- \mathrm{\Delta}V}{\mathrm{\Delta}r\cos\theta}$
$\Rightarrow E = \frac{- (20 - 10)}{10 \times 10^{- 2}\cos{120{^\circ}}}$
$= \frac{- 10}{10 \times 10^{- 2}( - \sin{30{^\circ}})}$
$= \frac{- 10^{2}}{- 1/2} = 200V/m$
Direction of $E$ be perpendicular to the equipotential surface $i.e.$ at $120{^\circ}$ with $x$-axis
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.