Electrostatics

Gauss's law ∇·E(r) = ρ(r)/ε₀ relates the electric field to charges. Learn the basic techniques to determine the electric field E(r) and/or the electrostatic potential 𝜙(r) for given charge distributions ρ(r)!

The fundamental solution to the Poisson equation with respect to a point-like excitation is known. Such a solution is called Green's function - in three dimensions, we have the famous \(1/r\)-dependency. However, if one knows the Green's function for a differential operator (here Laplace \(\Delta\)) any problem can be solved with respect to the given boundary conditions. In this section we will learn different techniques to solve such boundary value problems, for example the method of mirror charges or direct integrations.

In the following problems we will be concerned with the movement of charged particles under the influence of an electrostatic field. Depending on the question we will have to solve Euler-Lagrange equations of motion or use similar approaches. A lot of the problems are historically motivated to which we shall refer in the respective background sections.

The step from the microscopical to the macroscopical Maxwell equations is as fundamental as it is often underestimated. In these problems we will make ourself familiar with the electrostatic version of these equations. We will assume that the electric field induces polarization charges \(\rho_\mathrm{pol}\) in a material as the sources of a polarization density \(\mathbf{P}\),\[\begin{eqnarray*}  \varepsilon_{0}\nabla\cdot\mathbf{E}\left(\mathbf{r}\right)=\rho\left(\mathbf{r}\right)&=&\rho_{\mathrm{ext}}\left(\mathbf{r}\right)+\rho_{\mathrm{pol}}\left(\mathbf{r}\right)\\&=&\rho_{\mathrm{ext}}\left(\mathbf{r}\right)-\mathrm{div}\mathbf{P}\left(\mathbf{r}\right)\ ,  \end{eqnarray*}\]such that we are able to introduce a new field, the electric displacement field \(\mathbf{D}\):\[\begin{eqnarray*}  \nabla\cdot\left[\varepsilon_{0}\mathbf{E}\left(\mathbf{r}\right)+\mathbf{P}\left(\mathbf{r}\right)\right]&\equiv&\nabla\cdot\mathbf{D}\left(\mathbf{r}\right)\\&=&\rho_{\mathrm{ext}}\left(\mathbf{r}\right) \ . \end{eqnarray*}\]Here, \(\rho_\mathrm{ext}\) corresponds to all charges that do not correspond to polarization charges. Using the boundary conditions we can derive from these equations will help us to determine the electrostatic properties of certain dielectric bodies. A special attention will be made to capacitors as these are devices we will be able to describe thoroughly.

Important quantities to characterize dielectric materials are the (relative) permittivity \(\varepsilon\) and the suszeptibility \(\chi\). Both are introduced for linear materials as follows:\[\begin{eqnarray*}  \mathbf{D}\left(\mathbf{r}\right)&=&\varepsilon_{0}\mathbf{E}\left(\mathbf{r}\right)+\mathbf{P}\left(\mathbf{r}\right)\\&=&\varepsilon_{0}\mathbf{E}\left(\mathbf{r}\right)+\varepsilon_{0}\chi\left(\mathbf{r}\right)\mathbf{E}\left(\mathbf{r}\right)\\&=&\varepsilon_{0}\varepsilon_{r}\left(\mathbf{r}\right)\mathbf{E}\left(\mathbf{r}\right)\equiv\varepsilon\left(\mathbf{r}\right)\mathbf{E}\left(\mathbf{r}\right)\ .  \end{eqnarray*}\]

The multipole moments of a charge distribution \(\rho\left(\mathbf{r}\right)\) arise naturally if we try to solve Poisson's equation for the electrostatic potential, \(\Delta\phi\left(\mathbf{r}\right)= -\rho\left(\mathbf{r}\right)/\varepsilon_{0}\):
The Green's function of the Laplace operator obeying \(\Delta_{\mathbf{r}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)\), is given by \(G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = -1/4\pi\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\), which provides us the well-known

\[\phi\left(\mathbf{r}\right)=\frac{1}{4\pi\varepsilon_{0}}\int\frac{\rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV\ .\]

The Green's function can now be expanded in a Taylor series around \(\mathbf{r}^{\prime}=0\) yielding (see also in the solution of "The Electric Field of a Dipole")

\[\begin{eqnarray*}\phi\left(\mathbf{r}\right) & = & \frac{1}{4\pi\varepsilon_{0}}\left\{\int\frac{\rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}\right|}dV^{\prime}+\int\frac{\mathbf{r}\cdot\mathbf{r}^{ \prime}\rho\left(\mathbf{r}^{\prime}\right)} {\left|\mathbf{r}\right|^{3}}dV^{\prime}+\dots\right\}\\&=& \frac{1}{4\pi\varepsilon_{0}}\left\{\frac{Q}{r}+\frac{\mathbf{p}\cdot\mathbf{r}}{r^{3}}+\dots\right\}\end{eqnarray*}\]

with \(Q=\int\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}\) as the total charge, the dipole moment \(\mathbf{p}=\int\mathbf{r}^{\prime}\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}\) and \(r=\left|\mathbf{r}\right|\). You can see that the moments we already know are just the beginning of a fundamental representation of the electrostatic potential. In the following problems we shall make ourselfs familier with the multipole moments. Enjoy!