Question

The electric field in the region $r<a$ is $E=\frac{YQr}{a^3}$. Find $Y$?

Answer 1

Here the net charge of solid sphere of radius a is 3Q. So the charge density will be $\rho =\frac{3Q}{\left(4/3\right)\pi a^3}$

Thus, net charge enclosed by surface of radius r where $r<a$ is

$Q_{en}=\rho \frac{4}{3}\pi r^3=\frac{3Q}{\frac{4}{3}\pi a^3}\times \frac{4}{3}\pi r^3=\frac{3Q{r}^3}{a^3}$

By Gauss's law, the electric field in the region $r<a$ is

$E.4\pi r^2=\frac{Q_{en}}{{ϵ}_0}=\frac{3Q{r}^3}{a^3{ϵ}_0}$

or $E=\frac{3Qr}{4\pi {ϵ}_0{a}^3}$