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A circular plate sheet of radius $10 \mathrm{~cm}$ is placed in a uniform electric field of $2 \sqrt{3} \times 10^5 \mathrm{NC}^{-1}$, making an angle of $60^{\circ}$ with the field. Then, find the electric flux through the sheet.
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Verified Answer
The correct answer is:
$9.42 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Given that, radius of plate,
$$
\begin{aligned}
R & =10 \mathrm{~cm} \\
& =10 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Uniform electric field, $E=2 \sqrt{3} \times 10^5 \mathrm{NC}^{-1}$
Since, the angle between plate and electric field is $60^{\circ}$.
Then, angle between normal to the plate and electric field, $\theta=90^{\circ}-60^{\circ}=30^{\circ}$.
By using expression of electric flux,
$$
\begin{aligned}
\phi & =E A \cos \theta=2 \sqrt{3} \times 10^5 \times \pi\left(10 \times 10^{-2}\right)^2 \times \cos 30^{\circ} \\
& =9.42 \times 10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
R & =10 \mathrm{~cm} \\
& =10 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Uniform electric field, $E=2 \sqrt{3} \times 10^5 \mathrm{NC}^{-1}$
Since, the angle between plate and electric field is $60^{\circ}$.
Then, angle between normal to the plate and electric field, $\theta=90^{\circ}-60^{\circ}=30^{\circ}$.
By using expression of electric flux,
$$
\begin{aligned}
\phi & =E A \cos \theta=2 \sqrt{3} \times 10^5 \times \pi\left(10 \times 10^{-2}\right)^2 \times \cos 30^{\circ} \\
& =9.42 \times 10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\end{aligned}
$$
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