JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

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 $\frac{n q}{6 ε _{0}}$ (in SI units). The value of $n$ is _________.

Your answer

Solution

The charge $q$ sits on the flat circular disc that forms the common base of the hemisphere and the cone, at its center.

By symmetry, exactly half of the total flux $\frac{q}{ε _{0}}$ from the point charge goes into each side of the plane of the disc.

Therefore the flux through the conical (downward) side is $Φ_{cone} = \frac{q}{2 ε _{0}} = \frac{3 q}{6 ε _{0}}$ .

Matching with $\frac{n q}{6 ε _{0}}$ gives $n = 3$ .

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