question_answer

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

A)            \[\text{1}00V{{m}^{\text{1}}}\] along X-axis

B)            \[\text{1}00V{{m}^{\text{1}}}\] along Y-axis

C)            \[200V{{m}^{\text{1}}}\] at an angle \[\text{12}{{0}^{\text{o}}}\] with X-axis

D)  \[50V{{m}^{\text{1}}}\]at an angle \[\text{12}{{0}^{\text{o}}}\] with X-axis

Correct Answer: C

Solution :

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