To find the electrostatic potential of a general charge distribution can be quite complicated. However, if **symmetries** are present, the problem can be boiled down to the solution of much simple(r) equation(s). Let us employ the **spherical symmetry** of the homogeneously charged sphere to determine its potential and electric field!

## Problem Statement

A homogeneously charged sphere of radius \(R\) is described by the **charge distribution**

\[\rho\left(\mathbf{r}\right) = \begin{cases}\rho_{0} & r<R\\0 & r\geq R\end{cases}\ .\]

The charge density \(\rho_0\) can be understood as the number of homogeneously distributed charges vs. the volume of the sphere, see right figure.

Find the **electrostatic potentia**l of the homogeneously charged sphere and its **electric field**. You may want to follow these steps:

- Rewrite the
**Laplace equation**for the electrostatic potential in a suitable coordinate system. **Simplify**the equation using the**spherical symmetry**of the charge distribution.- Try to
**integrate**the equation. - Determine the
**constants**of integration to find the**electrostatic potential**. - Finally, calculate the
**electric field**.

## Hints

If a charge distribution is **spherically symmetric**, the same holds for the electrostatic potential, i.e. \(\phi\left( \mathbf{r} \right) = \phi\left( r \right)\).

What is the **Laplace operator** in spherical coordinates?

## 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. We found the electrostatic potential outside of the sphere. Note how only one parameter - the** total charge Q of the sphere** - was important!

## 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

- checking if they are physically meaningful (finiteness) and
- using the boundary conditions.

We finally found the electric field by just one differentiation.

Interestingly the field outside of the homogeneously charged sphere was the same as a point charge. You probably have guessed that in the first place!