Electrostatic Boundary Value Problem

Electrostatic Boundary Value Problem

Electrostatic boundary value problem

In this section we consider the solution for field and potential in a region where the electrostatic conditions are known only at the boundaries. Finding solution to such problems requires solving Laplace’s or Poisson’s equation satisfying the specified boundary condition. These types of problems are usually referred to as boundary value problem. Boundary value problems for potential functions can be classified as:

• Dirichlet problem where the potential is specified everywhere in the boundary

• Neumann problem in which the normal derivatives of the potential function are specified everywhere in the boundary

• Mixed boundary value problem where the potential is specified over some boundaries and the normal derivative of the potential is specified over the remaining ones.

Usually the boundary value problems are solved using the method of separation of variables, which is illustrated, for different types of coordinate systems.

Boundary Value Problems In Cartesian Coordinates:

Laplace's equation in cartesian coordinate can be written as,

...... (2.116)

Assuming that V(x,y,z) can be expressed as V(x,y,z) = X(x)Y(y)Z(z) where X(x), Y(y) and Z(z) are functions of x, y and z respectively, we can write:

...... (2.117)

Dividing throughout by X(x)Y(y)Z(z) we get


In the above equation, all terms are function of one coordinate variably only, hence we can write:

...... (2.118)

kx, ky and kz are seperation constant satisfying

All the second order differential equations above has the same form, therefore we discuss the possible form of solution for the first one only.

Depending upon the value of kx, we can have different possible solutions.

...... (2.119)

A,B,C and D are constants.

Similarly we have the possible solutions for other two differential equations. The particular solutions for X(x), Y(y) and Z(z) to be used will be chosen based on the nature of the problems and the constants A,B etc are evaluated from the boundary conditions. We illustrate the method discussed above with some examples.

Example 1: As shown in the figure 2.31, let us consider two electrodes, one grounded and the other maintained at a potential V. The region between the plates is filled up with two dielectric layers. Further, we assume that there is no variation of the field along x or y and there is no free charge at the interface.

Fig 2.31

The Laplace equation can be written as:

for which we can write the solution in general form as V= Az+B.

Considering the two regions,

...... (2.120)

Applying the boundary conditions V=0 for z=0 and V=V0 for z = d, we can write

...... (2.121)

Both the solution should give the same potential at z =a . Therefore,

...... (2.122a)

or, ...... (2.122b)

Another equation relating to A1 and A2 is required to solve for the two constants A1 and A2.

We note that at the dielectric boundary at z = a , the normal component of electric flux density vector is constant. Noting that , we can write

...... (2.123a)

or, ...... (2.123b)

Solving for A1 and A2, we find that

...... (2.124a)

...... (2.124b)

Using the above expressions for A1 and A2, we can find and .

Example-2: In the earlier example, we considered a situation where the potential function was a function of only one coordinate. In the figure 2.32 we consider a problem where the potential is a function of two coordinate variables. We consider two number of grounded semi-infinite parallel plate electrodes separated by a distance d. A third electrode maintained at potential V0 and insulated from the grounded conductors is placed as shown in the figure 2.32. The potential function is to be determined in the region enclosed by the electrodes.

Fig 2.32: Potential Function Enclosed by the Electrodes

As discussed above, we can write V(x,y,z) = X(x) Y(y) Z(z). We now construct the solution for V such that it satisfies the stated conditions at the boundary.

We find that V is independent of x. Therefore, X(x) = C1 , where C1is a constant.

Therefore, ...... (2.125)

We observe that the potential has to be equal to V0 at y = 0 and as the potential function becomes zero. Therefore, the y dependence can be expressed as an exponential function. Similarly, since V = 0 for z = 0 and z = d, the z dependence will be a sinusoidal function with kz = k, k being a real number.

Therefore, ...... (2.126)

...... (2.127)

...... (2.128)

Combining all these terms we can write:

, where C is a constant.

Using the condition that V(y,d) =0, we can write , n=1,2,3,....

For a particular n, . Therefore we can write the general solution as

, where Cns are yet to be determined.

Given that . Therefore . Multiplying both sides by on both sides and integrating from z =0 to z = d we can write,

...... (2.129)

Using orthogonality conditions and evaluating the integrals we can show that

...... (2.130)

The general solution for the potential function can therefore be written as
...... (2.131)
n the region y > 0 and 0< zd .

Boundary Value Problem in Cylindrical and Spherical Polar Coordinates

As in the case of Cartesian coordinates, method of separation of variables can be used to obtain the general solution for boundary value problems in the cylindrical and spherical polar coordinates also. Here we illustrate the case of solution in cylindrical coordinates with a simple example where the potential is a function of one coordinate variable only.

Example 3: Potential distribution in a coaxial conductor

Let us consider a very long coaxial conductor as shown in the figure 2.33, the inner conductor of which is maintained at potential V0 and outer conductor is grounded.


Fig 2.33: Potential Distribution in a Coaxial Conductor

From the symmetry of the problem, we observe that the potential is independent of variation. Similarly, we assume the potential is not a function of z. Thus the Laplace’s equation can be written as

...... (2.132)

Integrating the above equation two times we can write

...... (2.133)

whereC1 and C2 are constants, which can be evaluated from the boundary conditions

...... (2.134)

Evaluating these constants we can write the expression for the potential as

...... (2.135)

Furthure, ...... (2.136)

Therefore, ...... (2.137)

If we consider a length L of the coaxial conductor, the capacitance

...... (2.138)

Example-4: Concentric conducting spherical shells

As shown in the figure 2.34, let us consider a pair of concentric spherical conduction shells, outer one being grounded and the inner one maintained at potential V0.