JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Electrostatics

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle

θ

at P, as shown in the figure. When

θ = 3 0^{\circ}

, then the electric flux through the curved surface of the cylinder is

Φ

. If

θ = 6 0^{\circ}

, then the electric flux through the curved surface becomes

Φ/ n

, where the value of n is ____.

Answer: 3

Solution

The solid angle subtended by the two plane surfaces (top and bottom) is:

Ω_{planes} = 2 \times 2 π (1 - cos θ) = 4 π (1 - cos θ)

The solid angle subtended by the curved surface is:

Ω_{curved} = 4 π - Ω_{planes} = 4 π cos θ

The flux through the curved surface is:

ϕ = \frac{Ω _{curved}}{4 π} \cdot \frac{Q}{ε _{0}} = cos θ \cdot \frac{Q}{ε _{0}}

For

θ = 30°

ϕ_{30} = cos 30° \cdot \frac{Q}{ε _{0}} = \frac{3}{2} \cdot \frac{Q}{ε _{0}} = Φ

For

θ = 60°

ϕ_{60} = cos 60° \cdot \frac{Q}{ε _{0}} = \frac{1}{2} \cdot \frac{Q}{ε _{0}}

The ratio:

\frac{ϕ _{30}}{ϕ _{60}} = \frac{cos 30°}{cos 60°} = \frac{3 /2}{1/2} = 3

Therefore:

ϕ_{60} = \frac{Φ}{3}

, so

n = 3

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer