The electrostatic potential inside a charged spherical ball is given by $\phi=\alpha r^{2}+b$ where $r$ is the distance from the centre; $a, b$ are constants. Then the charge density inside ball is

(B) $-24 \pi a \varepsilon_{0} r$

(D) $-24 \pi a \varepsilon_{0} r$

JEE Main · 2011 · 10 Apr 2011 · Offline · Electrostatics

The electrostatic potential inside a charged spherical ball is given by

ϕ = α r^{2} + b

where

r

is the distance from the centre;

a, b

are constants. Then the charge density inside ball is

Answer: (C)

The field inside follows from

E_{r} = - \frac{d ϕ}{d r} = - 2 a r

, taking

ϕ = a r^{2} + b

. Applying Gauss's law to a sphere of radius

r

, the flux

E_{r} (4 π r^{2}) = (- 2 a r) (4 π r^{2}) = - 8 π a r^{3}

equals the enclosed charge over

ε_{0}

\frac{1}{ε _{0}} ρ \cdot \frac{4}{3} π r^{3}

. Equating and solving gives a constant charge density

ρ = - 6 a ε_{0}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option