Electrostatics and Magnetostatics

Maxwell's equations describe electrodynamics as follows:

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

The electric flux densityDis related to the electric fieldE, $D = ε E$ , whereεis the electrical permittivity of the material.

The magnetic flux densityBis related to the magnetic fieldH, $B = μ H$ , whereµis the magnetic permeability of the material.

Also, hereJis the electric current density, andρis the electric charge density.

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

$\begin{array}{l} \nabla \cdot (ε E) = ρ \\ \nabla \times E = 0 \end{array}$

Since the electric fieldEis the gradient of the electric potentialV, $E = - \nabla V$ , the first equation yields the following PDE:

$- \nabla \cdot (ε \nabla V) = ρ$

For electrostatic problems, Dirichlet boundary conditions specify the electric potentialVon 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 potentialA, such that

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

Using the identity

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

and the Coulomb gauge $\nabla \cdot A = 0$ , simplify the equation forAin terms ofJto the following PDE:

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

静磁问题,狄利克雷边界ditions specify the magnetic potential on the boundary.

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