Electrostatics and Magnetostatics

Maxwell's equations describe electrodynamics as:

$$\begin{array}{c}\epsilon \nabla \cdot E=\rho ,\\ \nabla \cdot H=0,\\ \nabla \times E=-\mu \frac{\partial H}{\partial t},\\ \nabla \times H=\epsilon \frac{\partial E}{\partial t}+J.\end{array}$$

Here, E and H are the electric and magnetic fields, ε and µ are the electrical permittivity and magnetic permeability of the material, and ρ and J are the electric charge and current densities.

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 this 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:

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

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

$$\begin{array}{l}H={\mu }^{-1}\nabla \times A,\\ \nabla \times \left({\mu }^{-1}\nabla \times A\right)=J.\end{array}$$

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 this PDE:

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

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