Chapter 23, Problem 20P

(III) Repeat Problem 19 assuming the charge density ρ_E increases as the square of the distance from the center of the sphere, and ρ_E = 0 at the center.

We assume the total charge is still Q, and let ρ_E = kr². We evaluate the constant k by calculating the total charge, in the manner of Example 22-5.

$Q={\displaystyle \int {\rho }_{\text{E}}dV}={\displaystyle \underset{0}{\overset{r_0}{\int }}k{r}^2(4\pi r^{2}dr)}=\frac{4}{5}k\pi r_0^5\to k=\frac{5Q}{4\pi r_0^5}$

(a) The electric field outside a charged, spherically symmetric volume is the same as that for a point charge of the same magnitude of charge. Integrating the electric field from infinity to the radius of interest gives the potential at that radius.

$E(r\ge r_0)=\frac{Q}{4\pi {\epsilon }_0{r}^2};V(r\ge r_0)=-{\displaystyle \underset{\infty }{\overset{r}{\int }}\frac{Q}{4\pi {\epsilon }_0{r}^2}}dr={\frac{Q}{4\pi {\epsilon }_{0}r}|}_{\infty }^r=\boxed{\frac{Q}{4\pi {\epsilon }_{0}r}}$

(b) Inside the sphere the electric field is obtained from Gauss’s Law using the charge enclosed by a sphere of radius r.

$4\pi r^{2}E=\frac{Q_{\text{encl}}}{{\epsilon }_0};Q_{\text{encl}}={\displaystyle \int {\rho }_{\text{E}}dV}=\frac{5Q}{4\pi r_0^5}{\displaystyle \underset{0}{\overset{r}{\int }}{r}^2(4\pi r^{2}dr)}=\frac{5Q}{4\pi r_0^5}\frac{4}{5}\pi r^5=\frac{Q{r}^5}{r_0^5}\to$

$E(r<r_0)=\frac{Q_{\text{encl}}}{4\pi {\epsilon }_0{r}^2}=\frac{Q{r}^3}{4\pi {\epsilon }_0{r}_0^5}$

Integrating the electric field from the surface to r < r₀ gives the electric potential inside the sphere.

$V(r<r_0)=V(r_0)-{\displaystyle \underset{r_0}{\overset{r}{\int }}\frac{Q{r}^3}{4\pi {\epsilon }_0{r}_0^5}}dr={\frac{Q}{4\pi {\epsilon }_0{r}_0}-\frac{Q{r}^4}{16\pi {\epsilon }_0{r}_0^5}|}_{r_0}^r=\frac{Q}{16\pi {\epsilon }_0{r}_0}\left(5-\frac{r^4}{r_0^4}\right)$

(c) To plot, we first calculate $V_0=V(r=r_0)=\frac{Q}{4\pi {\epsilon }_0{r}_0}$ and $E_0=E(r=r_0)=\frac{Q}{4\pi {\epsilon }_0{r}_0^2}$. Then we plot V/V₀ and E/E₀ as functions of r/r₀.

$\begin{array}{c}\text{For}r<r_0:V/V_0=\frac{\frac{Q}{16\pi {\epsilon }_0{r}_0}\left(5-\frac{r^4}{r_0^4}\right)}{\frac{Q}{4\pi {\epsilon }_0{r}_0}}=\frac{1}{4}\left(5-\frac{r^4}{r_0^4}\right);E/E_0=\frac{\frac{Q{r}^3}{4\pi {\epsilon }_0{r}_0^5}}{\frac{Q}{4\pi {\epsilon }_0{r}_0^2}}=\frac{r^3}{r_0^3}\\ \text{For}r>r_0:V/V_0=\frac{\frac{Q}{4\pi {\epsilon }_{0}r}}{\frac{Q}{4\pi {\epsilon }_0{r}_0}}=\frac{r_0}{r}{\left(r/r_0\right)}^{-1};E/E_0=\frac{\frac{Q}{4\pi {\epsilon }_0{r}^2}}{\frac{Q}{4\pi {\epsilon }_0{r}_0^2}}=\frac{r_0^2}{r^2}{\left(r/r_0\right)}^{-2}\end{array}$