Q 12-01-152JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
A solid ball of radius $R$ has a charge density $\rho$ given by $\rho = \rho_0\left(1 - \dfrac rR\right)$ for $0 \le r \le R$. The electric field outside the ball is:
Answer: (D) $\dfrac{\rho_0R^3}{12\varepsilon_0r^2}$
Total charge:
$$Q = \int_0^R\rho_0\left(1 - \frac rR\right)4\pi r^2\,dr = 4\pi\rho_0\left(\frac{R^3}{3} - \frac{R^3}{4}\right) = \frac{\pi\rho_0R^3}{3}$$
Outside the ball, by Gauss's law,
$$E = \frac{Q}{4\pi\varepsilon_0r^2} = \frac{\rho_0R^3}{12\varepsilon_0r^2}$$
Solution by Sreeraj P, M.Sc Physics