Solution of Poisson's Equation for Specified ChargeDistributions

The superposition principle is now used to find the solution of Poisson's equation for any given charge distribution

(. The argument presented in the previous section for singular charge distributions suggests the approach.r)

Figure 4.5.1r ^{'}gives rise to a potential at the observer positionr.For the purpose of representing the arbitrary charge density distribution as a sum of "elementary" charge distributions, we subdivide the space occupied by the charge density into elementary volumes of size

dx'dy'dz'. Each of these elements is denoted by the Cartesian coordinates(x',y',z'), as shown in Fig. 4.5.1. The charge contained in one of these elementary volumes, the one with the coordinates(x',y',z'), is

We now express the total potential due to the charge density as the superposition of the potentials

ddue to the differential elements of charge, (1), positioned at the points. Note that each of these elementary charge distributions has zero charge density at all points outside of the volume elementr^{'}dv'situated at. Thus, they represent point charges of magnitudesr^{'}dqgiven by (1). Provided that|is taken as the distance between the point of observationr-r^{'}|and the position of one incremental charger, the potential associated with this incremental charge is given by (4.4.1).r^{'}

where in Cartesian coordinates

Note that (2) is a function of two sets of Cartesian coordinates: the (observer) coordinates

(x,y,z)of the pointat which the potential is evaluated and the (source) coordinatesr(x',y',z')of the pointat which the incremental charge is positioned.r^{'}

According to the superposition principle, we obtain the total potential produced by the sum of the differential charges by adding over all differential potentials, keeping the observation point

(x,y,z)fixed. The sum over the differential volume elements becomes a volume integral over the coordinates(x.^{'}, y^{'}, z^{'})

This is the

superposition integralfor the electroquasistatic potential.The evaluation of the potential requires that a triple integration be carried out. With the help of a computer, or even a programmable calculator, this is a straightforward process. There are few examples where the three successive integrations are carried out analytically without considerable difficulty.

There are special representations of (3), appropriate in cases where the charge distribution is confined to surfaces, lines, or where the distribution is two dimensional. For these, the number of integrations is reduced to two or even one, and the difficulties in obtaining analytical expressions are greatly reduced.

Three-dimensional charge distributions can be represented as the superposition of lines and sheets of charge and, by exploiting the potentials found analytically for these distributions, the numerical integration that might be required to determine the potential for a three-dimensional charge distribution can be reduced to two or even one numerical integration.

## Superposition Integral for Surface Charge Density

If the charge density is confined to regions that can be described by surfaces having a very small thickness , then one of the three integrations of (3) can be carried out in general. The situation is as pictured in Fig. 4.5.2, where the distance to the observation point is large compared to the thickness over which the charge is distributed. As the integration of (3) is carried out over this thickness , the distance between source and observer,

|, varies little. Thus, with used to denote a coordinate that is locally perpendicular to the surface, the general superposition integral, (3), reduces tor-r^{'}|

Figure 4.5.2An element of surface charge at the locationr^{'}gives rise to a potential at the observer pointr.The integral on is by definition the surface charge density. Thus, (4) becomes a form of the superposition integral applicable where the charge distribution can be modeled as being on a surface.

The following example illustrates the application of this integral.

## Example 4.5.1. Potential of a Uniformly Charged Disk

The disk shown in Fig. 4.5.3 has a radius

Rand carries a uniform surface charge density. The following steps lead to the potential and field on the axis of the disk._{o}

Figure 4.5.3A uniformly charged disk with coordinates for finding the potential along thezaxis.The distance

|between the pointr-r^{'}|at radius and angle (in cylindrical coordinates) and the pointr^{'}on the axis of the disk (therzaxis) is given by

It follows that (5) is expressible in terms of the following double integral

where we have allowed for both positive

z, the case illustrated in the figure, and negativez. Note that these are points on opposite sides of the disk.

The axial field intensity

Ecan be found by taking the gradient of (7) in the_{z}zdirection.

The upper sign applies to positive

z, the lower sign to negativez.

The potential distribution of (8) can be checked in two limiting cases for which answers are easily obtained by inspection: the potential at a distance

|z| R, and the field at|z| \ll R.

observation from the disk, the radius of the disk

Ris small compared to|z|, and the potential of the disk must approach the potential of a point charge of magnitude equal to the total charge of the disk,. The potential given by (7) can be expanded in powers of_{o}R^{2}R/z

to find that indeed approaches the potential function

of a point charge at distance

|z|from the observation point.

(a)At a very large distance |z|of the point of(b)At |z| R, on either side of the disk, the field of the disk must approach that of a charge sheet of very large (infinite) extent. But that field is. We find, indeed, that in the limit_{o}/2_{o}|z| 0, (8) yields this limiting result.

## Superposition Integral for Line Charge Density

Another special case of the general superposition integral, (3), pertains to fields from charge distributions that are confined to the neighborhoods of lines. In practice, dimensions of interest are large compared to the cross-sectional dimensions of the areaA'of the charge distribution. In that case, the situation is as depicted in Fig. 4.5.4, and in the integration over the cross-section the distance from source to observer is essentially constant. Thus, the superposition integral, (3), becomes

Figure 4.5.4An element of line charge at the positionr^{'}gives rise to a potential at the observer locationr.In view of the definition of the line charge density, (1.3.10), this expression becomes

## Example 4.5.2. Field of Collinear Line Charges of Opposite Polarity

A positive line charge density of magnitude

is uniformly distributed along the_{o}zaxis between the pointsz = dandz = 3d. Negative charge of the same magnitude is distributed betweenz = -dandz = -3d. The axial symmetry suggests the use of the cylindrical coordinates defined in Fig. 4.5.5.

Figure 4.5.5Collinear positive and negative line elements of charge symmetrically located on thezaxis.The distance from an element of charge

to an arbitrary observer point_{o}dz^{'}(r, z)is

Thus, the line charge form of the superposition integral, (12), becomes

These integrations are carried out to obtain the desired potential distribution

Here, lengths have been normalized to

d, so thatandz= z/d. Also, the potential has been normalized such thatr= r/d

Figure 4.5.6Cross-section of equipotential surfaces and lines of electric field intensity for the configuration of Fig. 4.5.5A programmable calculator can be used to evaluate (15), given values of

(. The equipotentials in Fig. 4.5.6 were, in fact, obtained in this way, making it possible to sketch the lines of field intensity shown. Remember, the configuration is axisymmetric, so the equipotentials are surfaces generated by rotating the cross-section shown about ther,z)zaxis.

## Two-Dimensional Charge and Field Distributions

In two-dimensional configurations, where the charge distribution uniformly extends from

z = -toz = +, one of the three integrations of the general superposition integral is carried out by representing the charge by a superposition of line charges, each extending fromz = -toz = +. The fundamental element of charge, shown in Fig. 4.5.7, is not the point charge of (1) but rather an infinitely long line charge. The associated potential is not that of a point charge but rather of a line charge.

Figure 4.5.7For two-dimensional charge distributions, the elementary charge takes the form of a line charge of infinite length. The observer and source position vectors,randr^{'}, are two-dimensional vectors.With the line charge distributed along the

zaxis, the electric field is given by (1.3.13) as

and integration of this expression gives the potential

where

ris a reference radius brought in as a constant of integration. Thus, with_{o}dadenoting an area element in the plane upon which the source and field depend andandrthe vector positions of the observer and source respectively in that plane, the potential for the incremental line charge of Fig. 4.5.7 is written by making the identificationsr^{'}

Integration over the given two-dimensional source distribution then gives as the two-dimensional superposition integral

In dealing with charge distributions that extend to infinity in the

zdirection, the potential at infinity can not be taken as a reference. The potential at an arbitrary finite position can be defined as zero by adding an integration constant to (20).The following example leads to a result that will be found useful in solving boundary value problems in Sec. 4.8.

## Example 4.5.3. Two-Dimensional Potential of Uniformly Charged Sheet

A uniformly charged strip lying in the

y = 0plane betweenx = xand_{2}x = xextends from_{1}z = +toz = -, as shown in Fig. 4.5.8. Because the thickness of the sheet in theydirection is very small compared to other dimensions of interest, the integrand of (20) is essentially constant as the integration is carried out in theydirection. Thus, theyintegration amounts to a multiplication by the thickness of the sheet

Figure 4.5.8Strip of uniformly charged material stretches to infinity in thezdirections, giving rise to two-dimensional potential distribution.

and (20) is written in terms of the surface charge density

as_{s}

If the distance between source and observer is written in terms of the Cartesian coordinates of Fig. 4.5.8, and it is recognized that the surface charge density is uniform so that

is a constant, (22) becomes_{s}=_{o}

Introduction of the integration variable

u = x - xconverts this integral to an expression that is readily integrated.^{'}

Two-dimensional distributions of surface charge can be piece-wise approximated by uniformly charged planar segments. The associated potentials are then represented by superpositions of the potential given by (24).

## Potential of Uniform Dipole Layer

The potential produced by a dipole of charges

qspaced a vector distanceapart has been found to be given by (4.4.11)d

where

A

dipole layer, shown in Fig. 4.5.9, consists of a pair of surface charge distributionsspaced a distance_{s}apart. An area elementddof such a layer, with the direction ofad(pointing from the negative charge density to the positive one), can be regarded as a differential dipole producing a (differential) potentialad

Denote the surface dipole density by

where_{s}

and the potential produced by a surface dipole distribution over the surface

Sis given by

This potential can be interpreted particularly simply if the dipole density is constant. Then

can be pulled out from under the integral, and there is equal to_{s}times the integral_{s}/(4_{o})

This integral is dimensionless and has a simple geometric interpretation. As shown in Fig. 4.5.9,

is the area element projected into the direction connecting the source point to the point of observation. Division byi_{r' r}da|reduces this projected area element onto the unit sphere. Thus, the integrand is the differential solid angle subtended byr-r^{'}|^{2}das seen by an observer ata. The integral, (29), is equal to the solid angle subtended by the surfacerSwhen viewed from the point of observation. In terms of this solid angle,r

Figure 4.5.9The differential solid angle subtended by dipole layer of areada.Next consider the discontinuity of potential in passing through the surface

Scontaining the dipole layer. Suppose that the surfaceSis approached from the+side; then, from Fig. 4.5.10, the surface is viewed under the solid angle\Omega. Approached from the other side, the surface subtends the solid angle_{o}-(4 - \Omega. Thus, there is a discontinuity of potential across the surface of_{o})

Because the dipole layer contains an infinite surface charge density

, the field within the layer is infinite. The "fringing" field, i.e., the external field of the dipole layer, is finite and hence negligible in the evaluation of the internal field of the dipole layer. Thus, the internal field follows directly from Gauss' law under the assumption that the field exists solely between the two layers of opposite charge density (see Prob. 4.5.12). Because contributions to (28) are dominated by_{s}in the immediate vicinity of a point_{s}as it approaches the surface, the discontinuity of potential is given by (31) even ifris a function of position. In this case, the tangential_{s}is not continuous across the interface (Prob. 4.5.12).E

Figure 4.5.10The solid angle from opposite sides of dipole layer.