Electrostatics and Magnetostatics

Maxwell's equations describe electrodynamics as follows:

$$\begin{array}{c}\nabla \cdot D=\rho \\ \nabla \cdot B=0\\ \nabla \times E=-\frac{\partial B}{\partial t}\\ \nabla \times H=\frac{\partial D}{\partial t}+J\end{array}$$

The electric flux density D is related to the electric field E, $$D=\epsilon E$$, where ε is the electrical permittivity of the material.

The magnetic flux density B is related to the magnetic field H, $$B=\mu H$$, where µ is the magnetic permeability of the material.

Also, here J is the electric current density, and ρ is the electric charge density.

For electrostatic problems, Maxwell's equations simplify to this form:

$$\begin{array}{l}\nabla \cdot \left(\epsilon E\right)=\rho \\ \nabla \times E=0\end{array}$$

Since the electric field E is the gradient of the electric potential V, $$E=-\nabla V$$, the first equation yields the following PDE:

$$-\nabla \cdot \left(\epsilon \nabla V\right)=\rho$$

For electrostatic problems, Dirichlet boundary conditions specify the electric potential V on the boundary.

For magnetostatic problems, Maxwell's equations simplify to this form:

$$\begin{array}{l}\nabla \cdot B=0\\ \nabla \times H=J\end{array}$$

Since $$\nabla \cdot B=0$$, there exists a magnetic vector potential A, such that

$$\begin{array}{l}B=\nabla \times A\\ \nabla \times \left(\frac{1}{\mu }\nabla \times A\right)=J\end{array}$$

Using the identity

$$\nabla \times \left(\nabla \times A\right)=\nabla \left(\nabla \cdot A\right)-{\nabla }^{2}A$$

and the Coulomb gauge $$\nabla ·A=0$$, simplify the equation for A in terms of J to the following PDE:

$$-{\nabla }^{2}A=-\nabla \cdot \nabla A=\mu J$$

For magnetostatic problems, Dirichlet boundary conditions specify the magnetic potential on the boundary.