A superconducting wire expells a magnetic field.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 model for a homogeneously charged sphere.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 potential 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.


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?

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.

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