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) $- 24\pi a\varepsilon_{0}r$
B) $- 6a\varepsilon_{0}r$
C) $- 24\pi a\varepsilon_{0}$
D) $- 6a\varepsilon_{0}$
Explanation
(d) $\phi = ar^{2} + b$
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$E = - \frac{d\phi}{dt} = - 2ar$
$\oint_{}^{}{\overrightarrow{E}.d\overrightarrow{s}} = \frac{q}{\varepsilon_{0}}$
$- 2ar.4\pi r^{2} = \frac{q}{\varepsilon_{0}}$
$q = - 8\varepsilon_{0}a\pi r^{3}$
$\rho = \frac{q}{\frac{4}{3}\pi r^{3}} \Rightarrow \rho = - 6a\varepsilon_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.