Use Eq. 2.29 to calculate the potential inside a uniformly charged solid sphere of radius R and total charge q. Compare your answer to Prob. 2.21.
V( r )=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left( r ^{\prime}\right)}{η} d \tau^{\prime} (2.29)
Orient axes so P is on z axis.
V=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho}{η } d \tau. \left\{\begin{array}{l}\text { Here } \rho \text { is constant, } d \tau=r^{2} \sin \theta d r d \theta d \phi ,\\η =\sqrt{z^{2}+r^{2}-2 r z \cos \theta}.\end{array}\right.
V=\frac{\rho}{4 \pi \epsilon_{0}} \int \frac{r^{2} \sin \theta d r d \theta d \phi}{\sqrt{z^{2}+r^{2}-2 r z \cos \theta}} ; \int_{0}^{2 \pi} d \phi=2 \pi.
\int_{0}^{\pi} \frac{\sin \theta}{\sqrt{z^{2}+r^{2}-2 r z \cos \theta}} d \theta=\left.\frac{1}{r z}\left(\sqrt{r^{2}+z^{2}-2 r z \cos \theta}\right)\right|_{0} ^{\pi}=\frac{1}{r z}\left(\sqrt{r^{2}+z^{2}+2 r z}-\sqrt{r^{2}+z^{2}-2 r z}\right)
=\frac{1}{r z}(r+z-|r-z|)=\left\{\begin{array}{l}2 / z, \text { if } r<z ,\\2 / r, \text { if } r>z.\end{array}\right\}
\therefore V=\frac{\rho}{4 \pi \epsilon_{0}} \cdot 2 \pi \cdot 2\left\{\int_{0}^{z} \frac{1}{z} r^{2} d r+\int_{z}^{R} \frac{1}{r} r^{2} d r\right\}=\frac{\rho}{\epsilon_{0}}\left\{\frac{1}{z} \frac{z^{3}}{3}+\frac{R^{2}-z^{2}}{2}\right\}=\frac{\rho}{2 \epsilon_{0}}\left(R^{2}-\frac{z^{2}}{3}\right).
But \rho=\frac{q}{\frac{4}{3} \pi R^{3}}, \text { so } V(z)=\frac{1}{2 \epsilon_{0}} \frac{3 q}{4 \pi R^{3}}\left(R^{2}-\frac{z^{2}}{3}\right)=\frac{q}{8 \pi \epsilon_{0} R}\left(3-\frac{z^{2}}{R^{2}}\right) ; V(r)=\frac{q}{8 \pi \epsilon_{0} R}\left(3-\frac{r^{2}}{R^{2}}\right).
