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Consider an electric field $\mathbf{E}=E_0 \hat{\mathbf{x}}$ where $\bar{E}_0$ is a constant. The flux through the shaded area ( as shown in the figure) due to this field is
Options:
Solution:
1719 Upvotes
Verified Answer
The correct answer is:
$E_0 a^2$
$E_0 a^2$
Electric flux, $\mathbf{E} \cdot \mathbf{S}$, or $\phi=E S \cos \theta$ Here, $\theta$ is the angle between $\mathbf{E}$ and $\mathbf{S}$. In this question $\theta=45^{\circ}$, because $\mathbf{S}$ is perpendicular to the surface.
$$
\begin{gathered}
E=E_0 \\
\Rightarrow \quad S=(\sqrt{2 a})(a)=\sqrt{2} a^2 \\
\therefore \phi=\left(E_0\right)\left(\sqrt{2} a^2\right) \cos 45^{\circ}=E_0 a^2
\end{gathered}
$$
$\therefore$ Correct option is (c).
Analysis of Question
(i) Question is moderately tough.
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is $a$ and other side is $\sqrt{2} a$.
(iii) Electric field is uniform, whose magnitude is $E_0$ and direction is positive $x$. In uniform electric field we can use, $\phi=E S \cos \theta$
$$
\begin{gathered}
E=E_0 \\
\Rightarrow \quad S=(\sqrt{2 a})(a)=\sqrt{2} a^2 \\
\therefore \phi=\left(E_0\right)\left(\sqrt{2} a^2\right) \cos 45^{\circ}=E_0 a^2
\end{gathered}
$$
$\therefore$ Correct option is (c).
Analysis of Question
(i) Question is moderately tough.
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is $a$ and other side is $\sqrt{2} a$.
(iii) Electric field is uniform, whose magnitude is $E_0$ and direction is positive $x$. In uniform electric field we can use, $\phi=E S \cos \theta$
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