Electric Flux Through Curved Surface of Cylinder

To find the electric flux through the curved surface of the cylinder, we can apply Gauss's Law and the concept of solid angles.

1. Total Electric Flux

Let $q$ be the point charge located at the central point $P$ of the cylinder. According to Gauss's Law, the total electric flux emanating from the charge through any closed surface enclosing it is: $\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$

2. Flux Through the Flat Circular Faces

The solid angle $\Omega$ subtended by a circular disk at a point on its axis, which subtends a half-angle $\theta$ at that point, is given by: $\Omega = 2\pi (1 - \cos\theta)$

The electric flux $\Phi_{\text{top}}$ passing through the top flat circular face of the cylinder is proportional to this solid angle: $\Phi_{\text{top}} = \frac{q}{4\pi\varepsilon_0} \cdot \Omega = \frac{q}{4\pi\varepsilon_0} \cdot 2\pi(1 - \cos\theta) = \frac{q}{2\varepsilon_0}(1 - \cos\theta)$

By symmetry, the electric flux passing through the bottom flat circular face is identical: $\Phi_{\text{bottom}} = \frac{q}{2\varepsilon_0}(1 - \cos\theta)$

3. Flux Through the Curved Surface

The total flux through the closed cylinder is the sum of the fluxes through the top face, bottom face, and the curved surface: $\Phi_{\text{curved}} + \Phi_{\text{top}} + \Phi_{\text{bottom}} = \Phi_{\text{total}}$

Substituting the values into the equation: $\Phi_{\text{curved}} + \frac{q}{2\varepsilon_0}(1 - \cos\theta) + \frac{q}{2\varepsilon_0}(1 - \cos\theta) = \frac{q}{\varepsilon_0}$

$\Phi_{\text{curved}} + \frac{q}{\varepsilon_0}(1 - \cos\theta) = \frac{q}{\varepsilon_0}$

$\Phi_{\text{curved}} = \frac{q}{\varepsilon_0}\cos\theta$

4. Calculating the Value of $n$

• For $\theta = 30^\circ$, the flux through the curved surface is: $\Phi = \frac{q}{\varepsilon_0} \cos 30^\circ = \frac{q}{\varepsilon_0} \frac{\sqrt{3}}{2}$

• For $\theta = 60^\circ$, the flux through the curved surface becomes: $\Phi' = \frac{q}{\varepsilon_0} \cos 60^\circ = \frac{q}{\varepsilon_0} \frac{1}{2}$

We are given that $\Phi' = \frac{\Phi}{\sqrt{n}}$. Thus: $\frac{\Phi}{\Phi'} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

Comparing this with $\frac{\Phi}{\Phi'} = \sqrt{n}$, we get: $\sqrt{n} = \sqrt{3} \implies n = 3$

Final Answer: The value of $n$ is 3.