JEE Advanced · Physics · 17. Electrostatics
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 \(\Phi / \sqrt{n}\), where the value of \(n\) is _______ .

- A 3
- B 5
- C 7
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation

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 \(=4 \pi-\Omega\)
\(\begin{aligned}
& =4 \pi-(4 \pi-4 \pi \cos \theta)=4 \pi \cos \theta \\
& \phi_{30^{\circ}}=\phi=\frac{4 \pi \cos 30^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 30^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \\
& \phi_{60}=\frac{4 \pi \cos 60^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 60^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \\
& \frac{\phi_{30}}{\phi_{60}}=\frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\sqrt{3} \\
& \frac{\phi}{\phi_{60}}=\sqrt{3} \\
& \phi_{60}=\frac{\phi}{\sqrt{3}} \Rightarrow \mathrm{n}=3
\end{aligned}\)
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