JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Electrostatics

Let a total charge 2 Q be distributed in a sphere of radius R , with the charge density given by

r (r) = k r

, where

r

is the distance from the centre. Two charges

A

and

B

, of

- Q

each, are placed on diametrically opposite points, at equal distance, a , from the centre. If A and B do not experience any force, then :

Answer: (C)

Solution

Build up the enclosed charge first. With

ρ = k r

, the charge inside radius

r

\int_{0}^{r} k r^{'} 4 π r^{'2} d r^{'} = π k r^{4}

, and setting the whole sphere (

r = R

) equal to

2 Q

gives

π k R^{4} = 2 Q

. So the charge enclosed within radius

a

q (a) = 2 Q (\frac{a}{R})^{4}

. Now balance the two pulls on the charge at

A

. The enclosed charge attracts it inward with

\frac{1}{4 π ε _{0}} \frac{q ( a ) Q}{a ^{2}}

, while the equal charge

B

a distance

2 a

away repels it outward with

\frac{1}{4 π ε _{0}} \frac{Q Q}{( 2 a ) ^{2}}

. Equating these,

q (a) = \frac{Q}{4}

, so

2 Q (\frac{a}{R})^{4} = \frac{Q}{4}

, giving

(\frac{a}{R})^{4} = \frac{1}{8}

and hence

a = 8^{- 1/4} R

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option