question_answer

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

A) \[100\text{ }V{{m}^{-1}}\] along X-axis

B) \[100\text{ }V{{m}^{-1}}\] along Y-axis

C) \[200\text{ }V{{m}^{-1}}\]at an angle \[120{}^\circ \]with X-axis

D) \[50\text{ }V{{m}^{-1}}\] at an angle \[120{}^\circ \] with X-axis

Correct Answer: C

Solution :

[c] Using \[dV=-\vec{E}.d\vec{r}\]. \[\Rightarrow \Delta C=-E\Delta r\cos \theta \] \[\Rightarrow E=\frac{-\Delta V}{\Delta r\cos \theta }\] \[\Rightarrow E=\frac{-\left( 20-10 \right)}{10\times {{10}^{-2}}\cos 120{}^\circ }\] \[=\frac{-10}{10\times {{10}^{-2}}\left( -\sin 30{}^\circ  \right)}=200V/m\]