Physics - Electrostatics Question with Solution | TestHub

To find the electric field $E$ inside the sphere ($r<R$), we use Gauss's Law:

$$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{ϵ}_0}$$

Due to spherical symmetry, the electric field is radial and constant in magnitude over a spherical Gaussian surface of radius $r$.

So, $E(4\pi r^2)=\frac{Q_{enc}}{{ϵ}_0}$.

The enclosed charge $Q_{enc}$ is found by integrating the charge density over the volume of the Gaussian sphere:

$$Q_{enc}={\int }_0^r\rho dV={\int }_0^r(k{r}^{1/2})(4\pi r^{2}dr)$$

$$Q_{enc}=4\pi k{\int }_0^r{r}^{5/2}dr=4\pi k{\left[\frac{r^{7/2}}{7/2}\right]}_0^r$$

$$Q_{enc}=4\pi k\frac{2}{7}{r}^{7/2}=\frac{8\pi k}{7}{r}^{7/2}$$

Substitute $Q_{enc}$ back into Gauss's Law:

$$E(4\pi r^2)=\frac{1}{{ϵ}_0}\left(\frac{8\pi k}{7}{r}^{7/2}\right)$$

$$E=\frac{1}{4\pi r^2{ϵ}_0}\left(\frac{8\pi k}{7}{r}^{7/2}\right)$$

$$E=\frac{2k}{7{ϵ}_0}{r}^{(7/2-2)}=\frac{2k}{7{ϵ}_0}{r}^{3/2}$$

The final answer is $\boxed{\frac{2k}{7{ϵ}_0}(r{)}^{3/2}}$.