## Show Solution

Let us formally state the problem we have to solve in mathematical terms, i.e. relate the electrostatic potential and the given charge distribution via the **Laplace equation**:\[\begin{eqnarray*} \Delta\phi\left(\mathbf{r}\right)&=&-\frac{\rho\left(\mathbf{r}\right)}{\varepsilon_{0}}\\\rho\left(\mathbf{r}\right)&=&\begin{cases}\rho_{0} & r<R\\0 & r\geq R\end{cases}\ . \end{eqnarray*}\]We notice that is an entirely radially symmetric charge distribution and hence the potential can only be a function of the radius. This symmetry will hugely simplify our calculations. Let us start with the field outside the sphere.

## Electrostatic Potential outside the Sphere

Following **Gauss's law**, we know already that the field outside of the sphere has to be that of a point charge:\[\begin{eqnarray*} \oint_{\partial V}\mathbf{E}\left(\mathbf{r}\right)\cdot d\mathbf{A}&=&\frac{1}{\varepsilon_{0}}\int_{V}\rho\left(\mathbf{r}\right)dV \end{eqnarray*}\]implies that for a charge distribution with spherical symmetry it does not matter how the charge inside the volume is distributed - only the magnitude of the charge counts. The magnitude is given by the **volume integral** over the whole charge density of the sphere. So we find the potential outside as\[\begin{eqnarray*} \phi\left(r\right)&=&\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}\ \text{for}\ r\geq R\ \text{with}\\Q&=&\int_{0}^{2\pi}\int_{0}^{\pi}\int_{r=0}^{R}\rho_{0}dV^{\prime}\\&=&\rho_{0}\frac{4}{3}\pi R^{3}\ ,\ \text{so}\\\phi\left(r\right)&=&\frac{1}{3\varepsilon_{0}}\frac{\rho_{0}R^{3}}{r}\ \text{for}\ r\geq R\ . \end{eqnarray*}\]Ok, that was straight forward.

## Solution inside of the Sphere

Now to the inner part which is a little more challenging. The **Laplace operator** in spherical coordinates is given by\[\begin{eqnarray*} \Delta f\left(r,\theta.\varphi\right)&=&\frac{1}{r^{2}}\partial_{r}\left(r^{2}\partial_{r}f\left(r,\theta,\varphi\right)\right) + \frac{1}{r^{2}\sin\theta}\partial_{\theta}\left(\sin\theta\partial_{\theta}f\left(r,\theta,\varphi\right)\right)\\ & & + \frac{1}{r^{2}\sin^{2}\theta}\partial_{\varphi\varphi}f\left(r,\theta,\varphi\right)\ . \end{eqnarray*}\]If now the potential is just a function of the radius, \(\phi\left(\mathbf{r}\right)=\phi\left(r\right) \), we find that the Laplace equation inside of the sphere reduces to an **ordinary differential equation**:\[\begin{eqnarray*} \Delta\phi\left(r\right)&=&\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d}{dr}\phi\left(r\right)\right)=-\frac{\rho_{0}}{\varepsilon_{0}}\ ,\ r<R\ . \end{eqnarray*}\]We may solve this equation by** integration**. The first step yields\[\begin{eqnarray*} \int\frac{d}{dr}\left(r^{2}\frac{d}{dr}\phi\left(r\right)\right)dr&=&-\int r^{2}\frac{\rho_{0}}{\varepsilon_{0}}dr\ ,\\r^{2}\frac{d}{dr}\phi\left(r\right)&=&-\frac{1}{3}r^{3}\frac{\rho_{0}}{\varepsilon_{0}}+c_{1} \end{eqnarray*}\]where now \(c_{1}\) is a **constant** we have to determine later. We proceed with the next integration:\[\begin{eqnarray*} \int\frac{d}{dr}\phi\left(r\right)dr&=&\int-\frac{1}{3}r\frac{\rho_{0}}{\varepsilon_{0}}+\frac{c_{1}}{r^{2}}dr\ ,\\\phi\left(r\right)&=&-\frac{1}{6}r^{2}\frac{\rho_{0}}{\varepsilon_{0}}-\frac{c_{1}}{r}+c_{2}\ . \end{eqnarray*}\]Now, we have to **determine the constants**. First of all we demand the potential to be finite at the origin, such that\[c_{1} = 0\ .\]Second, we also demand the potential to be continuous at the termination of the sphere:\[\begin{eqnarray*} \frac{1}{3\varepsilon_{0}}\frac{\rho_{0}R^{3}}{R}&\overset{!}{=}&-\frac{1}{6}R^{2}\frac{\rho_{0}}{\varepsilon_{0}}+c_{2}\ ,\\c_{2}&=&\left(\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\right)\frac{\rho_{0}R^{2}}{\varepsilon_{0}}\ . \end{eqnarray*}\]In the end we put everything together and find that for the homogeneously charged sphere the electrostatic potential is given by\[\begin{eqnarray*} \phi\left(r\right)&=&\begin{cases}\frac{1}{3\varepsilon_{0}}\frac{\rho_{0}R^{3}}{r} & r\geq R\\-\frac{1}{6}\frac{\rho_{0}}{\varepsilon_{0}}\left(r^{2}-3R^{2}\right) & r<R\end{cases}\ . \end{eqnarray*}\]

## The Electric Field

The electrostatic potential varies only with respect to the radial coordinate. So, \(\mathbf{E}\left(\mathbf{r}\right)=-\nabla\phi\left(r\right)\) can only have a component in radial direction. We can find the radial component by **differentiation** of the potential in the different areas:\[\begin{eqnarray*} E_{r}\left(r\right)&=&-\frac{d}{dr}\phi\left(r\right)\\&=&\begin{cases}\frac{1}{3\varepsilon_{0}}\frac{\rho_{0}R^{3}}{r^{2}} & r\geq R\\\frac{1}{3}\frac{\rho_{0}}{\varepsilon_{0}}r & r<R\end{cases}=\begin{cases}\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}} & r\geq R\\\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{R^{3}}r & r<R\end{cases}\ . \end{eqnarray*}\]To solve this problem we have made enormous use of the rotational symmetry of the problem. We could boil down the Laplace equation to just an ordinary differential equation that could be solved by two successive integration steps. We ended up having some constants which could be found by a) checking if they are physically meaningful (finiteness) and b) using the boundary conditions. We finally found the electric field by just one differentiation.