Physics - Electrostatics Question with Solution | TestHub

To find the electric field at $r=R/2$, we use Gauss's Law: $E(4\pi r^2)=Q_{enc}/{ϵ}_0$. The enclosed charge $Q_{enc}$ for $r<R$ is given by $Q_{enc}={\int }_0^r\rho (r^{\prime })(4\pi r^{\prime 2})d{r}^{\prime }=4\pi {\rho }_0{\int }_0^r(r^{\prime 2}-r^{\prime 3}/R)d{r}^{\prime }=4\pi {\rho }_0[r^{\prime 3}/3-r^{\prime 4}/(4R){]}_0^r=4\pi {\rho }_0(r^3/3-r^4/(4R))$. Substituting this into Gauss's Law, $E(4\pi r^2)=\frac{4\pi {\rho }_0}{{ϵ}_0}(r^3/3-r^4/(4R))$, which simplifies to $E=\frac{{\rho }_0}{{ϵ}_0}(r/3-r^2/(4R))$. For $r=R/2$, the electric field is $E=\frac{{\rho }_0}{{ϵ}_0}((R/2)/3-(R/2{)}^2/(4R))=\frac{{\rho }_0}{{ϵ}_0}(R/6-R^2/(16R))=\frac{{\rho }_0}{{ϵ}_0}(R/6-R/16)=\frac{{\rho }_{0}R}{{ϵ}_0}(\frac{8-3}{48})=\frac{5{\rho }_{0}R}{48{ϵ}_0}$.