Question 845: Electrostatics & Capacitance

Question

A charge is kept at the central point $\mathrm{P}$ of a cylindrical region. The two edges subtend a half-angle $	heta$ at $\mathrm{P}$, as shown in the figure. When $	heta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$. If $	heta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is ___.

Accepted Answer

Place a point charge $q$ at the centre $P$ of a cylindrical Gaussian surface. Because of symmetry, the electric flux across each circular end cap is the same; call this $\phi_f$. Let $\phi$ be the flux through the curved side. Applying GaussÃ¢ÂÂs law to the closed cylinder, the total flux is
$$ \phi_{	ext{total}} \;=\; 2\phi_f \;+\; \phi \;=\; \frac{q}{\varepsilon_0}. $$
The solid angle subtended at point $P$ by the flat surface (spherical cap of half-angle $	heta$ ) is $\Omega_f=2 \pi(1-\cos 	heta$ ). The total solid angle is $\Omega_t=4 \pi$. Thus, the flux through one flat surface is
$$ \phi_f=\phi_t\left(\frac{\Omega_f}{\Omega_t}ight)=\frac{\phi_t(1-\cos 	heta)}{2} $$
Solve the above equations to get
$$ \phi=\frac{q \cos 	heta}{\epsilon_0}=\frac{q \cos 30^{\circ}}{\epsilon_0}=\frac{\sqrt{3} q}{2 \epsilon_0} $$
The flux through the curved surface for $	heta=60^{\circ}$ is
$$ \phi^{\prime}=\frac{q \cos 60^{\circ}}{\epsilon_0}=\frac{q}{2 \epsilon_0}=\frac{\phi}{\sqrt{3}} $$

Electrostatics & Capacitance

Integer / Numeric Answer Type