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 \fracn q6 \epsilon_0 (in SI units). The value of n is \underline\hspace2cm .

Electrostatics PYQ — JEE Advanced 2022 · Shift 2 · Q42 with Solution

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Correct answer: 3

The given closed surface is made of:

A point charge $q$ is enclosed by this closed surface.

By Gauss's law,

\Phi_{\text{total}}=\frac{q}{\epsilon_0}

So,

\Phi_{\text{cone}}+\Phi_{\text{hemisphere}}=\frac{q}{\epsilon_0}

For a point charge at the center of the hemisphere, the flux through any open surface subtending solid angle $\Omega$ at the charge is

\Phi = \frac{q}{4\pi\epsilon_0}\,\Omega

A hemisphere subtends solid angle

\Omega = 2\pi

Therefore,

\Phi_{\text{hemisphere}}=\frac{q}{4\pi\epsilon_0}(2\pi)=\frac{q}{2\epsilon_0}

Now using Gauss's law,

\Phi_{\text{cone}}=\frac{q}{\epsilon_0}-\frac{q}{2\epsilon_0}

\Phi_{\text{cone}}=\frac{q}{2\epsilon_0}

Given,

\Phi_{\text{cone}}=\frac{nq}{6\epsilon_0}

So,

\frac{nq}{6\epsilon_0}=\frac{q}{2\epsilon_0}

Cancelling $\dfrac{q}{\epsilon_0}$ ,

\frac{n}{6}=\frac{1}{2}

n=3

\boxed{3}

The derived answer matches the stored correct answer.