where T<sub>amb</sub> is the ambient temperature, and h is the coefficient of convective heat transfer having units of W/(m<sup>2</sup>.K).
== Electrostatics Analysis==
At very low frequencies, as ω→0 and k→0, the [[Basic_Electromagnetic_Theory#Electric_and_Magnetic_Potentials | Helmholtz equation for the electric scalar potential]] reduces to the Poisson equation subject to specified boundary conditions:
<math>\Delta\Phi(\mathbf{r}) = \nabla^2 \Phi(\mathbf{r}) = -\frac{\rho(\mathbf{r})}{\epsilon}</math>
where Φ(<b>r</b>) is the electric scalar potential expressed in Volts [v], ρ(<b>r</b>) is the volume charge density expressed in C/m<sup>3</sup>, and ε = ε<sub>r</sub> ε<sub>0</sub> is the permittivity of the medium having the units of F/m.
The electric field boundary conditions at the interface between two material media are:
<math> \hat{\mathbf{n}} . [ \mathbf{D_2(r)} - \mathbf{D_1(r)} ] = \rho_s (\mathbf{r}) </math>
<math> \hat{\mathbf{n}} \times [ \mathbf{E_2(r)} - \mathbf{E_1(r)} ] = 0 </math>
where <math> \hat{\mathbf{n}} </math> is the unit normal vector at the interface pointing from medium 1 towards medium 2,
<b>D(r)</b> = ε<b>E(r)</b> is the electric flux density expressed in C/m<sup>2</sup>, <b>E(r)</b> is the electric field vector expressed in V/m, and ρ<sub>s</sub> is the surface charge density at the interface having the units of expressed in C/m<sup>2</sup>.
In a source-free region, ρ(<b>r</b>) = 0, and Poisson's equation reduces to the familiar Laplace equation:
<math>\Delta\Phi(\mathbf{r}) = \nabla^2 \Phi(\mathbf{r}) = 0</math>
Keep in mind that in the absence of an electric charge source, you need to specify a non-zero potential somewhere in your structure, for example, on a perfect electric conductor (PEC). Otherwise, you will get a trivial zero solution of the Laplace equation.
Once the electric scalar potential is computed, the electric field can easily be computed via the equation below:
<math> \mathbf{E(r)} = - \nabla \Phi(\mathbf{r})</math>
== The Solution of Poisson's Equation ==
The general form of Poisson's equation for a scalar quantity Ψ(<b>r</b>) can be expressed as:
<math>\Delta\Psi(\mathbf{r}) = \nabla^2 \Psi(\mathbf{r}) = -f(\mathbf{r}) </math>
where f is a general arbitrary function of the position vector <b>r</b>. In the case of a unit point source, the above equation reduces to:
<math>\Delta\Psi(\mathbf{r}) = \nabla^2 \Psi(\mathbf{r}) = -\delta(\mathbf{r}) </math>
where δ(<b>r</b>) is the three-dimensional Dirac delta function. The solution to the above equation is called the Green's function of the Poisson equation for a certain boundary value problem and is denoted by G(<b>r|r<sup>′</sup></b>). Then, the solution to the general Poisson equation can be expressed in the terms of the Green's function as follows:
<math> \Psi\mathbf{(r)} = \int\int\int_V G(\mathbf{r|r^{\prime}}) f(\mathbf{r^{\prime}}) dv^{\prime} </math>
== Free-Space Electric Field and Scalar Potential ==
There are very few boundary value problems for which you can find closed-form analytical Green's functions. One of them is the electric scalar potential due to a charge distribution in the free space. The potential Green's function for a free-space medium is given by:
<math> G_{\Phi}(\mathbf{r|r^{\prime}}) = -\frac{1}{4\pi | \mathbf{r - r^{\prime}} | } </math>
If the volume charge density is denoted by ρ(<b>r</b>), then the electric scalar potential is given by:
<math> \Phi\mathbf{(r)} = \frac{1}{4\pi\epsilon} \int\int\int_V \frac{\rho(\mathbf{r^{\prime}})}{ | \mathbf{r - r^{\prime}} | } dv^{\prime} </math>
where V is the volume containing the charge distribution.
The electric field due to the charge distribution is then given by:
<math> \mathbf{E(r)} = - \nabla \Phi(\mathbf{r}) = \frac{1}{4\pi\epsilon} \int\int\int_V \frac{\mathbf{r - r^{\prime}} }{ | \mathbf{r - r^{\prime}} |^3 } \rho(\mathbf{r^{\prime}}) dv^{\prime} </math>
== Static Fields Arising from Steady-State Conduction Currents ==
In an Ohmic conductor, the current density is related to the electric field as follows:
<math> \mathbf{J(r)} = \sigma \mathbf{E(r)} = -\sigma \nabla \Phi(\mathbf{r}) </math>
where σ is the electric conductivity. On the other hand, the continuity equation for a stationary current requires no charge buildup or decay inside a closed region. This means that
<math> \nabla . \mathbf{J(r)} = 0 </math>
These above two equations lead to the Laplace equation inside an Ohmic conductor medium:
<math>\nabla^2 \Phi(\mathbf{r}) = 0</math>
In addition, the boundary condition at a conductor-dielectric interface requires a vanishing normal derivative of the electric potential:
<math> \frac{\partial \Phi}{\partial n} = 0 </math>
At the interface between two contiguous conductors, the normal component of the current density must be continuous.
<math> \hat{\mathbf{n}} . [ \mathbf{J_2(r)} - \mathbf{J_1(r)} ] = 0 </math>
which can be written as:
<math> \sigma_1 \hat{\mathbf{n}} . \mathbf{E_1(r)} = \sigma_2 \hat{\mathbf{n}} . \mathbf{E_2(r)} </math>
== Magnetostatics Analysis==
At very low frequencies, as ω→0 and k→0, the [[Basic_Electromagnetic_Theory#Electric_and_Magnetic_Potentials | Helmholtz equation for the magnetic vector potential]] reduces to the vector Poisson equation subject to specified boundary conditions:
<math>\Delta \mathbf{A} (\mathbf{r}) = \nabla^2 \mathbf{A}(\mathbf{r}) = - \mu \mathbf{J}(\mathbf{r}) </math>
where <b>A(r)</b> is the magnetic vector potential, <b>J(r)</b> is the volume current density, and μ = μ<sub>r</sub> μ<sub>0</sub> is the permeability of the medium. The magnetic Poisson equation is vectorial in nature and involves a system of three scalar differential equations corresponding to the three components of <b>A(r)</b>.
The magnetic field boundary conditions at the interface between two material media are:
<math> \hat{\mathbf{n}} . [ \mathbf{B_2(r)} - \mathbf{B_1(r)} ] = 0 </math>
<math> \hat{\mathbf{n}} \times [ \mathbf{H_2(r)} - \mathbf{H_1(r)} ] = \mathbf{J_s(r)} </math>
where <math> \hat{\mathbf{n}} </math> is the unit normal vector at the interface pointing from medium 1 towards medium 2,
<b>B(r)</b> = μ<b>H(r)</b> is the magnetic flux density, <b>H(r)</b> is the magnetic field vector, and <b>J<sub>s</sub></b> is the surface current density at the interface.
Once the magnetic vector potential is computed, the magnetic field can easily be computed via the equation below:
<math> \mathbf{H(r)} = \frac{1}{\mu} \nabla \times \mathbf{A} (\mathbf{r}) </math>
The relationship between the magnetic flux density and magnetic field vectors is rather different inside permeable materials that have a permanent intrinsic magnetization. Examples of such materials are ferromagnetic material that are used as permanent magnets. When a permeable material has a permanent magnetization, the following relationship holds:
<math> \mathbf{B(r)} = {\mu} \big[ \mathbf{H(r)} + \mathbf{M(r)} \big] </math>
where <b>M(r)</b> is the magnetization vector. In the SI units system, the magnetic field <b>H</b> and magnetization <b>M</b> both have the same units of A/m. It can be shown that for magnetostatic analysis, the effect of the permanent magnetization can be modeled as an equivalent volume current source:
<math> \mathbf{J_{eq}(r)} = \nabla \times \mathbf{M(r)} </math>
If the magnetization vector is uniform and constant inside the volume, then its curl is zero everywhere inside the volume except on its boundary surface. In that case, the permanent magnetization can be effectively modeled by an equivalent surface current density on the surface of the permanent magnetic object:
<math> \mathbf{J_{s,eq}(r)} = \mathbf{M(r)} \times \hat{\mathbf{n}} </math>
where <math> \hat{\mathbf{n}} </math> is the unit outward normal vector at the surface of the permanent magnet object. Note that the volume of the permanent magnet still acts as a permeable material with a relative permeability of μ in the magnetostatic analysis.
== Free-Space Magnetic Field and Vector Potential ==
The magnetic vector potential stratifies the vectorial form of Poisson's equation. Therefore, the free-space Green's function for the magnetic vector potential has the same functional dependence as that of the scalar electric potential but in a dyadic form, which is given by:
<math> \mathbf{\overline{\overline{G}}_A}(\mathbf{r|r^{\prime}}) = -\frac{1}{4\pi | \mathbf{r - r^{\prime}} | } \mathbf{\overline{\overline{I}}} </math>
where is <math> \mathbf{\overline{\overline{I}}} </math> is the unit dyad.
If the volume current density is denoted by <b>J(r)</b>, then the magnetic vector potential is given by:
<math> \mathbf{A(r)} = \frac{\mu}{4\pi} \int\int\int_V \frac{ \mathbf{J(r^{\prime})}}{ | \mathbf{r - r^{\prime}} | } dv^{\prime} </math>
where V is the volume containing the current distribution.
The magnetic field due to the current distribution is then given by:
<math> \mathbf{H(r)} = \frac{1}{\mu} \nabla \times \mathbf{A} (\mathbf{r}) = \frac{1}{4\pi} \int\int\int_V \mathbf{J(r^{\prime})} \times \frac{ \mathbf{r - r^{\prime}} }{ | \mathbf{r - r^{\prime}} |^3 } dv^{\prime} </math>
== The Finite Difference Technique ==