[YMP/EM-03006] Potential of a solid sphere

Let a point P be at a distance r from the center of the sphere (radius=R) with charge density $\rho$
Consider a small thin shell of radius \(x\) and thickness \(dx\)

Volume of the shell = $4x^2\pi ax$
Charge contained in the shell =$4\pi x^2 \rho$

Case $r>R$Potential at a point P outside the sphere at a distance r due to the shell is $\phi_{shell}$ where\begin{eqnarray}\phi_\text{shell}&=&\frac{4\pi x^2 dx \rho}{4\pi\epsilon_0r} \label{eq29}\end{eqnarray}Therefore the total potential $\phi$ is \begin{eqnarray}\phi=\int_0^R\frac{x^2dx\rho}{\epsilon_0r}\\=\frac{\rho R^3}{3\epsilon_0r} \label{eq30}\end{eqnarray}If total charge on the sphere is Q , then \begin{eqnarray}Q=\frac{4}{3}\pi R^3  \label{eq31}\end{eqnarray} Thus the expression for the  potential $\phi$ becomes \begin{eqnarray}\phi(\vec{r})=\frac{Q}{4\pi\epsilon_0r}, \qquad r>R. \label{eq32}
\end{eqnarray}
Case $r<R$When the point where potential ($\phi$) is to be computed lies inside the sphere, we divide the sphere into two parts We take the standard result about the potential of a  sphere of radius \(R\) \begin{eqnarray} \phi = \frac{\rho r^3}{3\epsilon_0 r}=\frac{Q}{4\pi\epsilon_0}\frac{r^2}{R^2}, \qquad r < R. \label{eq33}\end{eqnarray}