A uniform electric field of 400 V/m is directed at $4 5^{\circ}$ above the x-axis as shown in the figure. The potential difference $V_{A} - V_{B}$ is given by :

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Answer 1

Here, the electric field, $E = E cos 45 \hat{i} + E sin 45 \hat{j} = \frac{E}{2} (\hat{i} + \hat{j})$
now, $V_{A} - V_{B} = \int_{A}^{B} E . d r = \int_{A}^{B} \frac{E}{2} (\hat{i} + \hat{j}) . (d x \hat{i} + d y \hat{j}) = \frac{E}{2} [\int_{0}^{0.03} d x + \int_{0.02}^{0} d y]$
or $V_{A} - V_{B} = \frac{4 0 0}{2} [0.03 - 0.02] = 2.8 V$

Answer 2

Here, the electric field,

E = E cos 45 \hat{i} + E sin 45 \hat{j} = \frac{E}{2} (\hat{i} + \hat{j})

now,

V_{A} - V_{B} = \int_{A}^{B} E . d r = \int_{A}^{B} \frac{E}{2} (\hat{i} + \hat{j}) . (d x \hat{i} + d y \hat{j}) = \frac{E}{2} [\int_{0}^{0.03} d x + \int_{0.02}^{0} d y]

or

V_{A} - V_{B} = \frac{4 0 0}{2} [0.03 - 0.02] = 2.8 V