Question

A uniform electric field of 400 V/m is directed at ${45}^{\circ }$ above the x-axis as shown in the figure. The potential difference $V_A-V_B$ is given by :

• A. 0
• B. 4 V
• C. 6.4 V
• D. 2.8 V

Answer 1

Answer: (D)

Here, the electric field, $\overrightarrow{E}=E\cos 45\hat{i}+E\sin 45\hat{j}=\frac{E}{\sqrt{2}}(\hat{i}+\hat{j})$
now, $V_A-V_B={\int }_A^B\overrightarrow{E}.\overrightarrow{dr}={\int }_A^B\frac{E}{\sqrt{2}}(\hat{i}+\hat{j}).(dx\hat{i}+dy\hat{j})=\frac{E}{\sqrt{2}}[{\int }_0^{0.03}dx+{\int }_{0.02}^{0}dy]$
or $V_A-V_B=\frac{400}{\sqrt{2}}[0.03-0.02]=2.8V$