Question

The electric field at $r>c$ is $E=\frac{X}{4\pi {\epsilon }_0}\frac{2Q}{r^2}$. Find $X$?

Answer 1

As inner solid sphere has charge $+3Q$ so the inner surface of shell will get charge $-3Q$ and the outer surface will get charge $+3Q$ such that total induced charge is zero. So, net charge on outer surface is $+3Q-Q=2Q$

When consider Gaussian surface of radius $r>c$ , the all spheres are inside the Gaussian surface. Thus total charge inside it is $Q_{in}=3Q-3Q+2Q=2Q$

By Gauss's law, the field at distance $r>c$ from center is

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

or $E=\frac{2Q}{4\pi {ϵ}_0{r}^2}$