Skip to content
dronacharya.ai
← JEE Advanced 2022 · Paper 2
PhysicsInteger+3 / -1Question 24 of 54

JEE Advanced 2022 · Paper 2, Physics question 6

A charge qq is surrounded by a closed surface consisting of an inverted cone of height hh and base radius RR, and a hemisphere of radius RR as shown in the figure. The electric flux through the conical surface is nq6ϵ0\dfrac{nq}{6\epsilon_0} (in SI units). The value of nn is _____.

A closed surface: a hemisphere of radius R sitting on top of an inverted cone of height h with base radius R; the flat circular face they share is not part of the surface. The charge q is at the centre of that circle.
Answer and worked solution

Official answer: 3

The charge sits at the centre of the circle shared by the cone and the hemisphere, so it lies in the plane of that circle.

Split the flux. The total flux through the closed surface is qϵ0\frac q{\epsilon_0}. The field of a point charge is symmetric about any plane through it, so half the flux leaves on each side of the plane: half through the hemisphere and half through the cone.

Answer.
Φcone=q2ϵ0=3q6ϵ0,n=3.\Phi_{\text{cone}}=\frac q{2\epsilon_0}=\frac{3q}{6\epsilon_0},\qquad n=3.

Topics:Electric flux and Gauss's law
← Previous questionNext question →