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Physics Question on Electrostatic potential

The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b\phi = ar^2 + b where r is the distance from the centre a, b are constants. Then the charge density inside the ball is

A

−6aε0r - 6 a \varepsilon_0 r

B

−24πaε0- 24 \pi a \varepsilon_0

C

−6aε0- 6 a \varepsilon_0

D

−24πaε0r - 24 \pi a \varepsilon_0 r

Answer

−6aε0- 6 a \varepsilon_0

Explanation

Solution

Electric field, E=−dϕdt=−2arE = - \frac{d \phi}{dt} = - 2ar ...(i) By Gauss's theorem E(4πr2)=qε0E\left(4\pi r^{2}\right) = \frac{q}{\varepsilon_{0}} ...(ii) From (i) and (ii), ⇒q=−8πε0ar3\Rightarrow q = - 8 \pi\varepsilon_{0 }ar^{3}