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A charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $P$, as shown in the figure. When $\theta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$ If $\theta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is. . . . . . .
$4$
$2$
$3$
$5$
Solution
Solid angle made by plane surfaces $\Omega=2 \times 2 \pi(1-\cos \theta)$
$\Rightarrow \Omega=4 \pi-4 \pi \cos \theta$
So solid angle made by curved surface $=4 \pi-\Omega$
$=4 \pi-(4 \pi-4 \pi \cos \theta)=4 \pi \cos \theta$
$\phi_{30^{\circ}}=\phi=\frac{4 \pi \cos 30^{\circ}}{4 \pi} \frac{ Q }{\epsilon_0}=\cos 30^{\circ} \frac{ Q }{\epsilon_0}$
$\phi_{60}=\frac{4 \pi \cos 60^{\circ}}{4 \pi} \frac{ Q }{\epsilon_0}=\cos 60^{\circ} \frac{ Q }{\epsilon_0}$
$\frac{\phi_{30}}{\phi_{60}}=\frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\sqrt{3}$
$\frac{\phi}{\phi_{60}}=\sqrt{3}$
$\phi_{60}=\frac{\phi}{\sqrt{3}} \Rightarrow n =3$
