Electrostatics Ques 78
- Consider an electric field $\mathbf{E}=E_{0} \hat{\mathbf{x}}$, where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
(2011)
(a) $2 E_{0} a^{2}$
(b) $\sqrt{2} E_{0} a^{2}$
(c) $E_{0} a^{2}$
(d) $\frac{E_{0} a^{2}}{\sqrt{2}}$
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Answer:
Correct Answer: 78.(c)
Solution:
Formula:
- Electric flux, $\varphi=\mathbf{E} \cdot \mathbf{S}$ or $\varphi=E S \cos \theta$
Here, $\theta$ is the angle between $\mathbf{E}$ and $\mathbf{S}$.
In this question $\theta=45^{\circ}$, because $\mathbf{S}$ is perpendicular to surface.
$ \begin{aligned} E & =E_{0} \\ S & =(\sqrt{2} a)(a)=\sqrt{2} a^{2} \\ \therefore \quad \varphi & =\left(E_{0}\right)\left(\sqrt{2} a^{2}\right) \cos 45^{\circ}=E_{0} a^{2} \end{aligned} $
$\therefore$ Correct option is (c).
BETA
