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 \(\theta = 30^\circ\), 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 \(\frac{\Phi}{\sqrt{n}}\), where the value of \(n\) is \_\_\_\_. \includegraphics[width=0.3\linewidth]{ph8.png}

Hint:

Use solid angles to calculate electric flux for symmetric charge distributions.

Answer 1

For a cone, the solid angle subtended at the center is: \[ \Omega = 2\pi(1 - \cos\theta). \] The flux through each plane surface: \[ \phi = \frac{\Omega}{4\pi \epsilon_0} Q = \frac{Q}{2\epsilon_0}(1 - \cos\theta). \] Flux through both plane surfaces: \[ 2\phi = \frac{Q}{\epsilon_0}(1 - \cos\theta). \] Flux through the curved surface: \[ \Phi_{\text{curved}} = \frac{Q}{\epsilon_0}\cos\theta. \] When \(\theta = 30^\circ\): \[ \Phi = \frac{Q}{\epsilon_0} \cdot \frac{\sqrt{3}}{2}. \] When \(\theta = 60^\circ\): \[ \Phi' = \frac{Q}{\epsilon_0} \cdot \frac{1}{2}. \] \[ \sqrt{n}= \sqrt{3} \quad \Rightarrow \quad n = 3. \]