JEE Advanced 2024 Paper 2 · Q10 · Gauss's Law

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 $\Phi/\sqrt{n}$, where the value of $n$ is _____.

[Figure: A cylinder with point charge at its centre $P$. The half-angle subtended by each circular edge of the cylinder at $P$ is $\theta$.]

[Figure: A cylinder with point charge at its centre $P$. The half-angle subtended by each circular edge of the cylinder at $P$ is $	heta$.]

SolveKar AI · Accepted Answer

By Gauss's law, total flux through the closed cylindrical surface is $q/\varepsilon_0$. Flux through each flat circular cap is the solid-angle fraction subtended at $P$: each cap subtends a cone of half-angle $	heta$ at $P$, with solid angle $\Omega = 2\pi(1-\cos	heta)$. So flux through each cap $= \dfrac{q}{4\pi\varepsilon_0}\cdot 2\pi(1-\cos	heta) = \dfrac{q(1-\cos	heta)}{2\varepsilon_0}$.

Two caps total: $\Phi_{	ext{caps}} = \dfrac{q(1-\cos	heta)}{\varepsilon_0}$.

Flux through curved surface: $\Phi_{	ext{curved}} = \dfrac{q}{\varepsilon_0} - \dfrac{q(1-\cos	heta)}{\varepsilon_0} = \dfrac{q\cos	heta}{\varepsilon_0}$.

At $	heta = 30°$: $\Phi = \dfrac{q\cos 30°}{\varepsilon_0} = \dfrac{q\sqrt{3}/2}{\varepsilon_0} = \dfrac{\sqrt{3}\,q}{2\varepsilon_0}$.

At $	heta = 60°$: $\Phi' = \dfrac{q\cos 60°}{\varepsilon_0} = \dfrac{q/2}{\varepsilon_0} = \dfrac{q}{2\varepsilon_0}$.

Ratio $\dfrac{\Phi'}{\Phi} = \dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}$, so $\Phi' = \Phi/\sqrt{3}$. Hence $n = 3$.

Units: flux in V$\cdot$m or N$\cdot$m$^2$/C, ratios dimensionless. Answer: $3$.

JEE Advanced 2024 Paper 2 · Q10 · Gauss's Law

Solution