Electrostatic Boundary Value Problems

From HvWiki

1 Derivation of Laplace Equation from Maxwell's Equations
2 General solution of the Poisson Equation
3 Coaxial transmission line
4 Dielectric sphere in uniform field
5 Fringing field near parallel plane capacitor

[edit]

Derivation of Laplace Equation from Maxwell's Equations

Consider the first of Maxwell's Equations

$\nabla\cdot\textbf{D} = \rho$ .

Remember that $\textbf{D} = \epsilon \textbf{E}$ . Since, for static electric fields, we also recognise that

$E = -\nabla\Phi$ .

Putting these two relationships together yields

$\nabla\cdot\nabla\Phi = \nabla^2\Phi = -\frac{\rho}{\epsilon}$

This is Poisson's Equation, which, together with suitable boundary conditions, yields the potential in terms of the distribution of charge density $ρ$ .

In regions where the charge density is zero, Poisson's Equation becomes the Laplace Equation

$\nabla^2\Phi = 0$ .

The Laplace Equation is very useful problems where potentials are defined on boundaries and one wishes to compute the field in a source-free region. The Poisson Equation is convenient for problems where charge distributions are used to compute fields.

[edit]

General solution of the Poisson Equation

See Dirac Delta Function.

Without sacrificing any generality, we can solve for the "impulse response" of the electrostatic system by deriving a solution for

$\nabla^2\Phi_\delta = - \delta(\textbf{r})$ .

By integrating this equation over a spherical volume that encloses the origin and applying the Divergence Theorem, we see that

$\oint_S \nabla\Phi_\delta \cdot \hat{\textbf{n}}\, dS = -1$ .

If the sphere is "small", the surface integral on the left-hand-side becomes

$4\pi r^2 \frac{\partial \Phi_\delta}{\partial r} = -1$ .

Rearranging yields

$\frac{\partial \Phi_\delta}{\partial r} = \frac{-1}{4\pi r^2}$ .

This is easily integrable in r and we see that the final result for the impulse response potential is

$\Phi_\delta = \frac{1}{4\pi r} + V_c\,$ .

$V_c\,$ is a constant that describes the reference potential (usually zero). Given the spherical symmetry of the delta function source, we know that this is the complete solution.

By shifting the origin to $\textbf{r}^\prime\,$ , the impulse response can be generalised to any point in space:

$\Phi_\delta = \frac{1}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert} + V_c$ .

It is left to the reader to show that this is the solution of

$\nabla^2\Phi_\delta = -\delta(\textbf{r}-\textbf{r}^\prime)$ .

(Hint: use the Divergence Theorem on the $\nabla^2$ operator when $\textbf{r}=\textbf{r}^\prime$ .)

The next stage is to show that any distribution of charge $\rho( \textbf{r}^\prime)$ uniquely defines, within a constant value, a potential everywhere in space.

By using Green's Theorem,

$\int_V \left (\Phi_\delta \nabla^2\Phi - \Phi\nabla^2\Phi_\delta\right ) \, dV = \oint_{\partial V}\left (\Phi\nabla\Phi_\delta - \Phi_\delta\nabla\Phi\right ) \cdot\hat{\textbf{n}}\, dS$ ,

we can illustrate what the expression for potential must look like. Assume $\Phi\,$ is the unknown potential and $\Phi_\delta\,$ is the "impulse response" function that we found in the first part of this chapter. All we know is that the function $\Phi\,$ satisfies Poisson's Equation as well as any necessary boundary conditions on conductors as well as at infinity (where it must go to zero). Sources (charges) must be defined over a finite volume.

Inserting what we know and changing the integration to primed (source) coordinates yields

$\int_V \left (-\frac{1}{4\pi \vert\textbf{r}- \textbf{r}^\prime\vert} \frac{\rho}{\epsilon} +\delta(\textbf{r}-\textbf{r}^\prime) \Phi\right ) \, dV^\prime = \oint_{\partial V}\left (-\Phi\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3} - \frac{1}{ 4\pi \vert\textbf{r}-\textbf{r}\vert} \nabla\Phi\right ) \cdot\hat{\textbf{n}}\, dS^\prime$ .

Carrying out the volume integration on the left hand side and the surface integration on the right gives

$-\int_V \frac{\rho}{4\pi\epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\Phi(\textbf{r}) = -\oint_{\partial V}\Phi\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3}\cdot\hat{\textbf{n}}\, dS +\oint_{\partial V} \frac{\rho_s}{ 4\pi\epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime$ .

The surface charge density $\rho_s\,$ in the final surface integral comes about if we recognise that

$\textbf{D}\cdot\hat{\textbf{n}} = -\epsilon\nabla\Phi\cdot \hat{\textbf{n}} = \rho_s$

The first integral on the left did not change. However, the integration of the product of the delta function and the unknown potential over the problem volume just yields back the potential at the obervation point in the second term. The surface integral is a bit trickier to understand. At infinity, contributions to the surface integral vanish. Only surfaces (like conductors, dielectrics) at finite distances in the problem domain contribute to the potential (since we assume a finite charge distribution). The first surface integral term (over surfaces where potential is fixed) actually collapses to a constant value on metallic boundaries, because the conducting boundary must be an equipotential. As a result, the surface integral

$I=\oint_{\partial V}\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3}\cdot\hat{\textbf{n}}\, dS$

is $-1\,$ if $\textbf{r}$ (the observation point) lies on a surface source and 0 if the observation point is not on the surface.

Hence, we find that

$-\int_V \frac{\rho}{4\pi\epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\Phi(\textbf{r}) = \oint_{\partial V} \frac{\rho_s}{ 4\pi\epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime$ ,

where the first surface integral term vanishes for observation points away from surface sources.

If we rearrange, the potential anywhere in space can be written as

$\Phi(\textbf{r})= \int_V \frac{\rho}{4\pi \epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\oint_{\partial V} \frac{\rho_s}{4\pi \epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime$ .

In source-free regions (where Laplace's equation is valid) only the surface integral term is needed. Many useful numerical methods have been developed using

$\Phi = \oint_{\partial V} \frac{\rho_s}{4\pi \epsilon\vert\textbf{r}-\textbf{r}^\prime\vert} \, dS^\prime$

as the basis for solving many complicated problems. Furthermore, in source-free regions, we can make the philosophically important observation that knowing the field along a surface is enough to know the field everywhere in space!

By finding the "impulse response" (known as a "Green's function") of the potential function, we can use what signal-processing people will recognise as convolution to generate solutions for general potentials based on a known distribution of charge. Special techniques can also be used to reconstruct the charge distribution from a known potential distribution. These form the basis of a wide class of computer simulation techniques known as boundary element models.

[edit]

Coaxial transmission line

One of the most useful and common transmission lines is the coaxial line. Figure 1 shows the geometry, which consists of a cylindrical central conductor surrounded by a conductive cylindrical shield. Here, we shall describe how to find the capacitance per unit length of this line by using the the translational and rotational symmetries to simplify the problem in cylindrical coordinates.

Illustration of a circular coaxial transmission line. Potential is fixed on the center conductor and the conducting shield.

We know that the region of space between the conductors satisfies the Laplace equation, because there are no charges present. In cylindrical coordinates, this means

$\nabla^2\Phi = \frac{1}{r}\frac{\partial}{\partial r}\left ( r\frac{\partial \Phi}{\partial r}\right ) + \frac{1}{r^2} \frac{\partial^2\Phi}{\partial \phi^2} + \frac{\partial^2\Phi}{\partial z^2} = 0$

Rotational symmetry in the $φ$ coordinate and translation symmetry in z mean that there is no variation in potential with respect to the azimuthal or longitudinal variables $z$ and $φ$ . we are then left with

$\frac{1}{r}\frac{\partial}{\partial r}\left ( r\frac{\partial \Phi}{\partial r}\right ) = 0$

This differential equation has the solution

$\Phi = C_1 \ln r + C_2 \,$ ,

where $C_1, C_2\,$ are constants to be found using the boundary conditions present on the center conductor and the shield.

On the center conductor (r=a), $Φ = 1$ . This means that

$1 = C_1 \ln a + C_2\,$ .

On the outer shield, we have $\Phi = 0\,$ , which gives

$0 = C_1 \ln b + C_2\,$ .

We can now solve for $C 1$ and $C 2$ :

$C_1 = \frac{1}{\ln a - \ln b}\,$

$C_2 = \frac{-\ln b}{\ln a - \ln b}\,$ .

Rearranging and using the properties of logarithms, we find the final expression for the potential within the coaxial conductor as

$\Phi(r) = \frac{\ln\frac{r}{b}}{\ln\frac{a}{b}}$

The electric field is purely radial and equals

$E_r(r) = -\frac{1}{\ln\frac{a}{b}}\frac{1}{r}$

The charge per unit length present on the center conductor is proportional to the perpendicular electric field present at the surface of the conductor (found using Gauss's Law), i.e.:

$Q/l = \frac{2\pi\epsilon}{\ln\frac{b}{a}}$ ,

where $\epsilon\,$ is the dielectric permittivity of the insulating material filling the region between the conductors.

The capacitance per unit length is then found to be

$C/l = \frac{Q/l}{1} = \frac{2\pi\epsilon}{\ln\frac{b}{a}}$ ,

since we chose the potential difference between conductors to be 1 at the start. (Although this in not important in capacitance calculations. The capacitance does not depend at all on the choice of voltage.)

[edit]

Dielectric sphere in uniform field

A classic problem that illustrates the mechanics of matching boundary conditions across an interface is the dielectric sphere of relative permittivity $ε r = ε / ε 0 > 1.0$ immersed in an otherwise constant electric field (Seen in Figure 2).

Illustration of a dielectric sphere immersed in an otherwise constant electric field. The electric field is directed along the z-axis.

Given the symmetry of the sphere about the z-axis, variations in the azimuthal angle $φ$ can be neglected. The Laplace Equation in the remaining spherical coordinates is

$\frac{1}{r^2}\frac{\partial^2}{\partial r^2} (r^2\Phi)+\frac{1}{r\sin \theta} \frac{\partial}{\partial \theta}\left (\sin \theta\frac{\partial \Phi}{\partial \theta}\right ) = 0$

The potential can be expressed as a product

$\Phi = R\Theta \,$

where $R\,$ is a function of radial distance r alone and $\Theta\,$ is a function of elevation angle $\theta\,$ alone. Taking the derivatives and rearranging, we get

$\frac{r^2 R^{\prime\prime}}{R} + \frac{2rR^\prime}{R}= -\frac{\Theta^{\prime\prime}}{\Theta}-\frac{\Theta^\prime}{\Theta\tan\theta}$ .

This expression can hold only if both sides are equal to the same constant, which for convenience, we can call $m(m+1)\,$ . Hence,

$r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}-m(m+1)R = 0$

$\frac{d^2\Theta}{d\theta^2} + \frac{1}{\tan \theta}\frac{d\Theta}{d\theta} +m(m+1)\Theta = 0$

The equation for $R\,$ has the solution

$R = C_1 r^m + C_2 r^{-m-1} \,$ .

The equation for $\Theta\,$ is not so obvious. However, by using a series solution for $\Theta\,$ , one can verify that the solutions are given by the Legendre polynomials

$P_m(\psi) = \frac{1}{2^m}\left ( \frac{d}{d\psi}\right )^m(\psi^2-1)^m,$

where $\psi=\cos\theta\,$ . The first few are tabulated below:

$P_0 = 1\,$

$P_1 = \psi\,$

$P_2 = \frac{1}{2}\left (3\psi^2 - 1\right )\,$

$P_3 = \frac{1}{2}\left (5\psi^3 - 3\psi\right )\,$

$P_4 = \frac{1}{8}\left (35\psi^4 - 30\psi^2 + 3\right )\,$

Going back to the problem at hand, we recgnise that the constant E-field directed along the z-axis has a potential (that must represent the behaviour at infinity)

$\Phi_c = -E_0 z = -E_0 r\cos\theta\,$ .

The contribution to the field due to the presence of the dielectric sphere must be computed in two parts: one for the region outside the sphere and another for the potential inside the sphere. Outside the sphere, we select the solution such that the perturbation in field vanishes as distance tends to infinity:

$\Phi = \sum_{m=0}^\infty b_m \left (\frac{a}{r}\right )^{m+1} P_m(\cos\theta) -E_0r P_1(\cos\theta)$ .

Note that this solution has the correct behaviour at infinity.

Within the sphere, if the solution is to "behave" as $r\rightarrow 0$ , we have

$\Phi_s = \sum_{m=0}^\infty c_m \left (\frac{r}{a}\right )^m P_m(\cos\theta)$

The key to the solution is to "match" the potential and the normal electric flux density across the air-dielectric boundary, such that

$\Phi_{\rm inside} = \Phi_{\rm outside}\,$ at $r=a\,$

and, since there are no surface charges on the sphere,

$\epsilon_r \left . \frac{\partial \Phi}{\partial r}\right \vert_{\rm inside} = \left . \frac{\partial\Phi}{\partial r}\right \vert_{\rm outside}$ at $r = a\,$ .

Equating terms with the same $\theta\,$ dependence gives

$\begin{matrix} b_0 = c_0 \\ b_1 - E_0a = c_1 \\ b_2 = c_2 \\ \vdots \end{matrix}$

and

$\begin{matrix} b_0 = 0 \\ -E_0 - 2\frac{b_1}{a} = \epsilon_r \frac{c_1}{a} \\ 3b_2 = 2c_2 \\ \vdots \end{matrix}$

The zeroth coefficient is clearly zero. Coefficients with indices greater than 1 are also zero. We are left with a contribution only from the first coefficients

$b_1 = E_0 a\frac{\epsilon_r - 1}{\epsilon_r + 2}$

$c_1 = -E_0 a\frac{3}{\epsilon_r + 2}$

The total potential outside the sphere is

$\Phi = E_0 r\left (\frac{\epsilon_r - 1}{\epsilon_r + 2} \frac{a^3}{r^3} - 1\right )\cos\theta$ .

Notice how the the presence of the dielectric sphere introduces a "dipole field" contribution (the term that varies as $1 / r 2$ ) to the overall field solution outside the sphere. Indeed, the dielectric sphere will look like an electric dipole as it is becomes polarised as a result of the external field.

The potential inside the sphere is found to be

$\Phi = -E_0 r\frac{3}{\epsilon_r + 2}\cos\theta$ .

The figures show the equipotentials and the field lines for spheres of dielectric constant 2.2 and 10.

Equipotential contours and flux lines for a sphere of relative dielectric constant 10.

Equipotential contours and flux lines for a sphere of relative dielectric constant of 2.2. In comparison to the sphere of dielectric constant of 10, the confinement of the flux lines is significantly less.

Observe how the higher dielectric constant causes a strong concentration of flux lines within the sphere. In fact, it is easy to see the the flux achieves a maximum of

$\textbf{D} = 3\epsilon_0E_0 \hat{\textbf{z}}\,$

inside the sphere as the relative dielectric constant $\epsilon_r \rightarrow\infty\,$ .

[edit]

Fringing field near parallel plane capacitor

Here, we consider two parallel planes separated by an air dielectric ( $\epsilon_r=1\,$ . This problem lacks the elegant (and convenient) symmetry of the coaxial line or dielectric sphere, so a closed form solution is difficult to find. (Although this problem can be solved exactly using conformal transformation methods []). Our approach will be to apply a computer method to generate the potential distribution, the fields and finally, the capacitance per unit length.

Figure 2 shows the geometry of the problem. The top plate is at a fixed positive potential (say, 1V) and the lower at -1V.

Geometry of the parallel plane capacitor.

How do we find the resulting distribution of charge on the plates? The answer lies in using the generalised integral equation solution for the Laplace Equation (except, here, in two dimensions) that was introduced earlier

$V(\textbf{r}) = \frac{1}{2\pi\epsilon_0} \oint_C \rho_s\ln\left (\vert\textbf{r}- \textbf{r}^\prime\vert\right ) dl^\prime$ .

Notice, however, because we are solving a two-dimensional problem, instead of the $1/\vert\textbf{r}-\textbf{r}^\prime\vert$ Green's function previously encountered, we use a logarithmic relationship (which, in fact, represents the spatial impulse response for a line source).