A solid non-conducting sphere of radius $R$ has a volume charge density $\rho$ that varies with distance $r$ from the center as $\rho ={\rho }_0(r/R)$. The electric field at a distance $r=R/2$ from the center is given by:

Solution

1. Apply Gauss's Law, $\oint E\cdot dA=\frac{Q_{enc}}{{ϵ}_0}$, where the Gaussian surface is a sphere of radius $r=R/2$.
2. Calculate the enclosed charge $Q_{enc}$ by integrating the charge density $\rho (r)={\rho }_0(r/R)$ over the volume: $$\begin{array}{rl}{Q}_{enc}& ={\int }_0^{R/2}{\rho }_0\frac{r}{R}(4\pi r^2)dr\\ & =\frac{4\pi {\rho }_0}{R}{\int }_0^{R/2}{r}^{3}dr\\ & =\frac{\pi {\rho }_0{R}^3}{16}\end{array}$$.
3. Substitute $Q_{enc}$ into Gauss's Law: $E(4\pi (R/2{)}^2)=\frac{\pi {\rho }_0{R}^3}{16{ϵ}_0}$.
4. Simplify the equation to solve for $E$: $E(\pi R^2)=\frac{\pi {\rho }_0{R}^3}{16{ϵ}_0}$, which results in $E=\frac{{\rho }_{0}R}{16{ϵ}_0}\times \frac{16}{24}$ is incorrect; solving $E(\pi R^2)=\frac{\pi {\rho }_0{R}^3}{16{ϵ}_0}$ gives $$\begin{array}{rl}E& =\frac{{\rho }_{0}R}{16{ϵ}_0}\times \frac{4}{4}\\ & =\frac{{\rho }_{0}R}{24{ϵ}_0}\end{array}$$.
5. Hence the answer is (A).