3.1.POISSON’S EQUATION
The Laplacian is a scalar differential operator is a differential equation, known as poisson’s equation
The operator involves differentiation with respect to more than one variable. Hence, Poisson’s equation is a partial differential equation that may be solved once we know the functional defendence of (x,y,z) and the appropriate boundary conditions.
The operator , just like the , . , and x, makes no reference to any particular coordinate system. In order to solve a specific problem, we must write in terms of x,y,z or r, ɸ, or some other set of coordinates. The choice of the particular set of coordinates is orbitary, but substantial simplification of the problem is usually achieved by choosing a set compatible with the simmetry of the electrostatic problem. The form taken by in various coordinate systems is easily found by first taking the gradient of and the operating with . .
Rectangular coordinates:
Spherical coordinates:
Cylindrical coordinates:
3.2.LAPLACE’S EQUATION
Where the charge density vanishes, the poisson equation reduces to the simpler form
Which is Laplace’s equation
Theorem I. If are all solutions of Laplace’s equation, then
Where the C’s are arbitrary constants, is also a solution
Theorem II: Uniqueness theorem. Two solutions of Lapalace’s equation that satisfy the same boundary conditions differ at most by an additive constant.
We define a new function . Obviously, in V0. Furthermore, either or vanishes on the boundaries. Let us apply the divergence theorem to the vector :
3.3.LAPLACE’S EQUATION IN ONE INDEPENDENT VARIABLE
If is a function of one variable only, Laplace’s equation reduces to an ordinary differential equation. Consider the case where is (r), a function of the single rectangular coordinate x.then,
and (x) = ax + b
Is the general solution, where a and b are constants chosen to fit the boundary conditions. the situation is no more complicated in other coordinate systems where is a function of a single variable. In spherical coordinates where equals (r), Laplace’s equation and it’s general solution become
3.4.SOLUTIONS TO LAPLACE’S EQUATION IN SPHERICAL COORDINATES : ZONAL HARMONICS
For the spherical case, is (r, ), where r is the radius vector from a fixed origin O and is the polar angle
This partial differential equation will be solved by a technique known as separation of variables.
3.5.CONDUCTING SPHERE IN A UNIFORM ELECTRIC FIELD
The spherical conductor of radius a is an equipotential surface. Let us denote it’s potential by 0. Our problem is to find a solution to Laplace’s equation in the region outside the sphere, which reduces to 0 on the sphere it self and which has the correct limiting form at large distances away. The solution may be formally written as
3.6.SOLUTION TO LAPLACE’S EQUATION IN CYLINDRICAL COORDINATES: CYLINDRICAL HARMONICS
If the potential is independent of z, Laplace’s equation in cylindrical coordinates becomes
Subtitution of = Y(r)S( ) reduces the equation to
Where k again plays the role of a separation constant. The -equation is particularly simple, it has the solutions cos k1/2 and sin k1/2 . But if these solutions are to make sense physically, each must be a single-valued function of . Thus,
cos k1/2 + 2 ) = cos k1/2
sin k1/2 + 2 ) = sin k1/2
3.7.LAPLACE’S EQUATION IN RECTANGULAR COORDINATES
In rectangular coordinates, the variable may be separated by making the subtitution
(x,y,z) = f1(x)f2(y)f3(z)
whereby Laplace’s equation reduces to
The left side of this equation is a function of x and y, and the right side is a function of z only, hence both sides must be equal to the same constant, k. This constant is the first separation constant.
3.8.LAPLACE’S EQUATION IN TWO DIMENSIONS: GENERAL SOLUTION
If the potential is a funtion of only two rectangular coordinates, Laplace’s equation is written
It is possible to obtain the general solution to this equation by means of a transformation to a new set of independent variables. Nevertheless, it should be emphasized that such a transformation leads to a simplification of the original equation only in the two-dimensional case. Let
where i = is the unit imaginary number. In terms of these relationship,
and
3.9.ELECTROSTATIC IMAGES
For given set of boundary conditions the solutions to Laplace’s equation is unique, so that if one obtains a solution (x,y,z) by any means whatever, and if this satisfies all boundary conditions, then a complete solution to the problem has been effected. The methode of images is a procedure for accomplishing this result without specifically solving a differential eqution. It is universally applicable to all types of electrostatic problems, but enought interesting problems fall into this category to make it worthwhile discussing the methode here
Suppose the potential may be written in the following way:
(r) = 1(r) +
3.10.LINE CHARGES AND LINE IMAGES
Thus far, our image technique has been limited to problems involving point charges and hence point images. In this section we shall take up several problems that may be solved by means of line image charges. Consider two infinitely long, parallel, line charges, with charges λ and –λ per unit lenght,
where r1 and r2 are the perpendicular distances from the point to the two line charges.
= M
3.11.SYSTEM OF CONDUCTORS AND COEFFICIENTS OF POTENTIAL
A solution of Laplace’s equation that fits a particular set of boundary conditions is unique, therefore we have found the correct solution, (x,y,z), to our modified problem. The interesting conclusion we draw from this discussion is that the potential of each conductor is proportional to the charge Qj of conductor j. That is
i(j) = pijQj, (i = 1,2,3,....,N)
where pij is a constant that depends only on the geometry.
3.12.NUMERICAL SOLUTION OF ELECTROSTATIC PROBLEMS
The solutions are required to statisfy the boundary conditions and to be continuous across boundaries between elements. Some adjustable parameters are left in the approximation, and these parameters are determined by minimizing the energy of the system. It is a little difficult to follows this abstract description, so we shall discuss each method in terms of the solution of Laplace’s equation in one dimension. The equation and boundary conditions are
d2 /dx2 = 0, (0) = 0, (1) = 1
its exact solution is easily found to be (x) = x. We shall first use the finite difference method to find an approximate solution.
The first step is to divide the interval from x = 0 to x = 1 into N equal segments of lenght 1/N. At the ith dividing point, Laplace’s equation is approximated by
[( i+1 – i) – ( i – i-1)]/(1/N)2 = 0
3.13.SOLUTIONS OF POISSON’S EQUATION
We shall assume that the total charge is a bounded-that is, either the charge does not extend to infinity or the charge density drops off sufficiently rapidly a large radii.
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