Q 12-01-069JEE MainJEE Main 2023 (29 Jan, Shift 2)Medium
For a charged spherical ball, electrostatic potential inside the ball varies with $r$ as $V=2ar^2+b$. Here, $a$ and $b$ are constant and $r$ is the distance from the center. The volume charge density inside the ball is $-\lambda a\varepsilon$. The value of $\lambda$ is ______. ($\varepsilon=$ permittivity of medium)
Numerical value type. Enter your answer.
Answer: 12
$E=-\dfrac{dV}{dr}=-4ar$.
By Gauss's law for a uniformly charged ball, $E=\dfrac{\rho r}{3\varepsilon}$. Equating:
$$\frac{\rho r}{3\varepsilon}=-4ar\ \Rightarrow\ \rho=-12a\varepsilon$$
So $\lambda=12$.
Solution by Sreeraj P, M.Sc Physics