Question

Figure shows two equipotential surfaces $V_1$ and $V_2$ The component of electric field in x and y direction will be -

• A. $E_x\hat{i}\text{and}{E}_x\hat{j}$
• B. $-E_x\hat{i}\text{and}{E}_x\hat{j}$
• C. $E_x\hat{i}\text{and}-E_x\hat{j}$
• D. $-E_x\hat{i}\text{and}-E_x\hat{j}$

Answer 1

The correct option is B $-E_x\hat{i}\text{and}{E}_x\hat{j}$

$\begin{array}{c}\left|\overrightarrow{E}\right|=\frac{2v}{2\cos {45}^o}\\ =\frac{2\times \sqrt{2}}{2\times 1}=\sqrt{2}V{m}^{-1}\\ \overrightarrow{E}=-Ex\hat{i}+Ey\hat{j}\\ \Rightarrow \overrightarrow{E}=-Ex\hat{i}+Ex\hat{j}\\ bothcomponentwillbeequal\\ Hence,\\ optionBiscorrectanswer.\end{array}$