Question

The electric field for$R_1<r<R_2$ is given by $E=\frac{Q}{X\pi {\epsilon }_0{r}^2}\left[\frac{r^3-R_1^3}{R_2^3-R_1^3}\right]$. Find X?

Answer 1

Here the charge density is $\rho =\frac{Q}{\frac{4}{3}\pi (R_2^3-R_1^3)}$

By Gauss's law, the field at r where $R_1<r<R_2$ is

$E.4\pi r^2=\frac{Q_{in}}{{ϵ}_0}$

Here $Q_{in}=\frac{4}{3}\pi (r^3-R_1^3)\rho =\frac{r^3-R_1^3}{R_2^3-R_1^3}$ (substitute the value of $\rho$ )

Thus, $E=\frac{Q}{4\pi {ϵ}_0{r}^2}\left[\frac{r^3-R_1^3}{R_2^3-R_1^3}\right]$