A uniform electric field of 400 ,v/m is directed 45^o above x

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2. Electric Potential and Capacitance

hard

A uniform electric field of $400 \,v/m$ is directed $45^o$ above the $x$ - axis. The potential difference $V_A - V_B$ is -.....$V$

$0$

$4$

$6.4$

$2.8$

Solution

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 $V_{A}-V_{B}=\frac{400}{\sqrt{2}}[0.03-0.02]=2.8 V$

Standard 12

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