JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Electrostatics

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 ____.

Answer: 3

Solution

By Gauss's law, the total flux through the closed surface is

\frac{q}{ϵ _{0}}

.

For a point charge, the flux through a surface subtending solid angle

Ω

ϕ = \frac{q Ω}{4 π ϵ _{0}}

.

The hemisphere subtends a solid angle of

2 π

steradians (half the total

4 π

), so:

ϕ_{hemisphere} = \frac{q \cdot 2 π}{4 π ϵ _{0}} = \frac{q}{2 ϵ _{0}}

By symmetry and flux conservation:

ϕ_{cone} = \frac{q}{ϵ _{0}} - \frac{q}{2 ϵ _{0}} = \frac{q}{2 ϵ _{0}} = \frac{3 q}{6 ϵ _{0}}

Therefore,

n = 3

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer