A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle

Question

A charge is kept at the central point \(\mathrm{P}\) of a cylindrical region. The two edges subtend a half-angle \(\theta\) at \(\mathrm{P}\), 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 \(\Phi / \sqrt{\mathrm{n}}\), where the value of \(n\) is ______.

Answer 1

Correct answer: 3

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

\(=4 \pi-\Omega\)

\(=4 \pi-(4 \pi-4 \pi \cos \theta)\)

\(=4 \pi \cos \theta\)

\( \phi_{30^{\circ}}=\phi=\frac{4 \pi \cos 30^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 30^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \)

\( \phi_{60}=\frac{4 \pi \cos 60^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 60^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \)

\(\frac{\phi_{30}}{\phi_{60}}=\frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\sqrt{3} \)

\( \frac{\phi}{\phi_{60}}=\sqrt{3}\)

\( \phi_{60}=\frac{\phi}{\sqrt{3}} \)

\(\Rightarrow \mathrm{n}=3\)