Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho =k{r}^a$, where $k$ and a are constants and $r$ is the distance from its centre. If the electric field at $r=R/2$ is $1/8$ times that at $r=R$, find the value of $a$.

Answer 1

Using Gauss's law : electric field at $r=R/2$ is $E_1.4\pi (R/2{)}^2=\frac{Q_{en}}{{ϵ}_0}$

or $E_1=\frac{Q_{en}}{4\pi {ϵ}_0(R/2{)}^2}$

Here $Q_{en}={\int }_0^{R/2}\rho 4\pi r^{2}dr=4\pi {\int }_0^{R/2}k{r}^a{r}^{2}dr=4\pi k{\int }_0^{R/2}{r}^{a+2}dr=4\pi k\frac{(R/2{)}^{a+3}}{a+3}$

now $E_1=\frac{4\pi k\frac{(R/2{)}^{a+3}}{a+3}}{4\pi {ϵ}_0(R/2{)}^2}=\frac{k}{(a+3){ϵ}_0}(R/2{)}^{a+1}$

Similarly electric field at $r=R$ is $E_2=\frac{Q_{en}}{4\pi {ϵ}_0{R}^2}$

where $Q_{en}={\int }_0^{R}4\pi k{r}^{a+2}dr=4\pi k\frac{R^{a+3}}{a+3}$

now $E_2=\frac{k}{(a+3){ϵ}_0}{R}^{a+1}$

As $E_1=\left(1/8\right)E_2\Rightarrow \frac{k}{(a+3){ϵ}_0}(R/2{)}^{a+1}=(1/8)\frac{k}{(a+3){ϵ}_0}{R}^{a+1}$

or $(1/2{)}^{a+1}=(1/2{)}^3$

or $a+1=3\Rightarrow a=3-1=2$