JEE Advanced 2022 Paper 2 · Q06 · Gauss's Law
A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\dfrac{nq}{6\epsilon_0}$ (in SI units). The value of $n$ is _____ .
[Figure: An inverted cone with apex pointing down, base radius R at top, height h. A hemisphere of radius R sits on top of the cone, covering it. Charge q is shown inside, near the base/equator level between the cone and hemisphere.]
Reveal answer + step-by-step solution
Correct answer:3
Solution
The closed surface encloses charge $q$, so by Gauss's law total flux through (cone + hemisphere) is $q/\epsilon_0$. The hemisphere subtends a solid angle of $2\pi$ at the centre of its great circle where the charge effectively sits (the flux through a hemispherical cap from a charge at the centre of its flat face is $q/(2\epsilon_0)$). Hence flux through cone $= q/\epsilon_0 - q/(2\epsilon_0) = q/(2\epsilon_0) = 3q/(6\epsilon_0)$. Thus $n = 3$.
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