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The net electric flux due to a uniform electric field of $3 \times 10^3 \hat{1} \mathrm{NC}^{-1}$ through a cube of side $20 \mathrm{~cm}$ oriented such that its faces are parallel to the coordinate planes is
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Verified Answer
The correct answer is:
$0$
Uniform electric field, $\mathrm{E}=3 \times 10^3 \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}$ side of cube, $a=20 \mathrm{~cm}=0.20 \mathrm{~m}$
Four face makes the angle between area vector and electric field is $90^{\circ}$.
$\phi=(\mathrm{EdA} \cos 90) \times 4=0$
One face makes the angle between area vector and electric field is $0^{\circ}$
$\begin{aligned}
& \phi_1=\mathrm{E} \cdot \mathrm{A} \cos 0 \\
& =\mathrm{EA}
\end{aligned}$
Other face makes the angle between area vector and electric field is $180^{\circ}$
$\phi_2=\mathrm{EA} \cos 180=-\mathrm{EA}$
The net electric flux is
$\begin{aligned}
& \phi_{\text {net }}=\phi_1+\phi_2 \\
& =\mathrm{E} A-\mathrm{EA}=0
\end{aligned}$
Four face makes the angle between area vector and electric field is $90^{\circ}$.
$\phi=(\mathrm{EdA} \cos 90) \times 4=0$
One face makes the angle between area vector and electric field is $0^{\circ}$
$\begin{aligned}
& \phi_1=\mathrm{E} \cdot \mathrm{A} \cos 0 \\
& =\mathrm{EA}
\end{aligned}$
Other face makes the angle between area vector and electric field is $180^{\circ}$
$\phi_2=\mathrm{EA} \cos 180=-\mathrm{EA}$
The net electric flux is
$\begin{aligned}
& \phi_{\text {net }}=\phi_1+\phi_2 \\
& =\mathrm{E} A-\mathrm{EA}=0
\end{aligned}$
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