Q 12-02-048JEE MainJEE Main 2026 (24 Jan, Shift 1)Medium
The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^3 + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha\times\pi a\in_0$. The value of $\alpha$ is ______ .
(permittivity of vacuum is $\epsilon_0$ )
Answer: (A) -12
Radial field: $E = -\dfrac{dV}{dr} = -3ar^2$.
By Gauss's law, for a sphere of radius $r = 1$:
$$q = \varepsilon_0\oint\vec E\cdot d\vec A = \varepsilon_0(-3a)(1)^2\cdot4\pi(1)^2 = -12\pi a\varepsilon_0$$
So $\alpha = -12$.
Solution by Sreeraj P, M.Sc Physics