A uniform electric field of $400 \,v/m$ is directed $45^o$ above the $x$ - axis. The potential difference $V_A - V_B$ is -.....$V$
Diffcult
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Here, the electric field, $\vec{E}=E \cos 45 \hat{i}+E \sin 45 \hat{j}=\frac{E}{\sqrt{2}}(\hat{i}+\hat{j})$
now, $V_{A}-V_{B}=\int_{A}^{B} \vec{E} \cdot \overrightarrow{d r}=\int_{A}^{B} \frac{E}{\sqrt{2}}(\hat{i}+\hat{j}) \cdot(d x \hat{i}+d y \hat{j})=\frac{E}{\sqrt{2}}\left[\int_{0}^{0.03} d x+\int_{0.02}^{0} d y\right]$
or $V_{A}-V_{B}=\frac{400}{\sqrt{2}}[0.03-0.02]=2.8 V$
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