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

(a) -6 a ε_{0} r

(b) -24 π a ε_{0} r

(c) -6 a ε_{0}

(d) -24 π a ε_{0}

Ans:(c)

Sol: Potential inside a charged spherical ball

φ = a r^{2} + b

According to Gauss’s theorem ,

E (4πr^{2}) = q/ε_{0}

Differentiating ,

; Where ρ = volume charge density

ρ = -6 a ε_{0}