JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Electrostatics

A charged shell of radius R carries a total charge Q. Given

Φ

as the flux of electric field through a closed cylindrical surface of height h, radius r and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? [

ϵ_{0}

is the permittivity of free space]

(A)If $h > 2 R$ and $r > R$ then $Φ = \frac{Q}{ϵ _{0}}$ ✓ Correct
(B)If $h < \frac{8 R}{5}$ and $r = \frac{3 R}{5}$ then $Φ = 0$ ✓ Correct
(C)If $h > 2 R$ and $r = \frac{4 R}{5}$ then $Φ = \frac{Q}{5 ϵ _{0}}$
(D)If $h > 2 R$ and $r = \frac{3 R}{5}$ then $Φ = \frac{Q}{5 ϵ _{0}}$ ✓ Correct

Answer: (A), (B), (D)

Solution

The flux through the closed cylinder equals the enclosed charge divided by

ϵ_{0}

, and the enclosed charge is the part of the spherical shell lying within the cylinder's radius

r

of the axis, i.e. the shell points with

R sin θ < r

. For

h > 2 R

and

r > R

the whole shell is enclosed, so

Φ = \frac{Q}{ϵ _{0}}

. For

h < \frac{8 R}{5}

and

r = \frac{3 R}{5}

the cylinder lies entirely inside the shell and encloses no charge, so

Φ = 0

. For

r = \frac{3 R}{5}

(with

h > 2 R

) the enclosed part is the two polar caps where

∣ cos θ ∣ > \frac{4}{5}

, a fraction

1 - \frac{4}{5} = \frac{1}{5}

of the shell, giving

Φ = \frac{Q}{5 ϵ _{0}}

. For

r = \frac{4 R}{5}

the caps have

∣ cos θ ∣ > \frac{3}{5}

, a fraction

1 - \frac{3}{5} = \frac{2}{5}

, so

Φ = \frac{2 Q}{5 ϵ _{0}}

and not

\frac{Q}{5 ϵ _{0}}

, so that case is not correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Electrostatics

Solution

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