Question

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 \(\theta = 30\) , then the electric flux through the curved surface of the cylinder is \(Φ\). If \(\theta= 60\degree\) , then the electric flux through the curved surface becomes \(Φ/√𝑛\), where the value of n is______.

 

Answer 1

Correct Answer: 3

Flux Through the Cone's Surface 

For a cone, the solid angle subtended at the center is:

\[ \Omega = 2\pi \left( 1 - \cos\theta \right) \]

The flux through each plane surface is:

\[ \varphi = \frac{\Omega}{4\pi \epsilon_0} Q = \frac{Q}{2\epsilon_0} (1 - \cos\theta) \]

Flux through both plane surfaces:

The flux through the curved surface is:

\[ \Phi_{\text{curved}} = \frac{Q}{\epsilon_0} \cos\theta \]

When \( \theta = 30^\circ \):

\[ 2\varphi = \frac{Q}{\epsilon_0} (1 - \cos\theta) \] \[ \Phi = \frac{Q}{\epsilon_0} \cdot \frac{\sqrt{3}}{2} \]

When \( \theta = 60^\circ \):

Quick Tip:

\[ \Phi' = \frac{Q}{\epsilon_0} \cdot \frac{1}{2} \] \[ \sqrt{n} = \sqrt{3} \quad \Rightarrow \quad n = 3 \]