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...

The correct numerical answer is 3. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Physics Question 10 Solution

The solid angle subtended by a disk (end cap of the cylinder) with half-angle $\theta$ at point P is $\Omega = 2\pi(1 - \cos\theta)$ .

The flux through one end cap is $\Phi_{\text{cap}} = \frac{q}{\epsilon_0} \frac{\Omega}{4\pi} = \frac{q}{2\epsilon_0}(1 - \cos\theta)$ .

Since there are two symmetric end caps, the total flux through the ends is $2\Phi_{\text{cap}} = \frac{q}{\epsilon_0}(1 - \cos\theta)$ .

The flux through the curved surface $\Phi_{\text{curved}}$ is the total flux minus the flux through the ends:

$\Phi_{\text{curved}} = \frac{q}{\epsilon_0} - \frac{q}{\epsilon_0}(1 - \cos\theta) = \frac{q}{\epsilon_0}\cos\theta$

For $\theta = 30^\circ$ : $\Phi = \frac{q}{\epsilon_0}\cos 30^\circ = \frac{q\sqrt{3}}{2\epsilon_0}$ .

For $\theta = 60^\circ$ : $\Phi' = \frac{q}{\epsilon_0}\cos 60^\circ = \frac{q}{2\epsilon_0}$ .

Comparing $\Phi'$ and $\Phi$ :

$\frac{\Phi'}{\Phi} = \frac{q/2\epsilon_0}{q\sqrt{3}/2\epsilon_0} = \frac{1}{\sqrt{3}}$

Given $\Phi' = \Phi/\sqrt{n}$ , we have $\sqrt{n} = \sqrt{3}$ , so $n = 3$ .