Q. The electrostatic potential inside a charged spherical ball is given by $\phi = ar^{2} + b$ where $r$ is the distance from the centre a, b are constants. Then the charge density inside the ball is
A) $- 6a\varepsilon_{0}r$
B) $- 24\pi a\varepsilon_{0}$
C) $- 6a\varepsilon_{0}$
D) $- 24\pi a\varepsilon_{0}r$
Explanation
(c) Electric field
$E = - \frac{d\phi}{dt} = - 2ar$
By Gauss’s theorem
$E\left( 4\pi r^{2} \right) = \frac{q}{\varepsilon_{0}}$
$q = - 8\pi\varepsilon_{0}{ar}^{2}$
$\rho = \frac{dq}{dV} = \frac{dq}{dr} \times \frac{dr}{dV}$
$= \left( - 24\pi\varepsilon_{0}ar^{2} \right) \times \frac{1}{4\pi r^{2}}$
$\rho = - 6\varepsilon_{0}a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.