Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

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Laplace’s Equation in Spherical Coordinates :

In spherical coordinates the equation can be written as

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Note that we are using the notation Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev denote the azimuthal angle and φ to denote the potential function. 
As with the rectangular coordinates, we will attempt a separation of variable, writing

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Inserting this into the Laplace’s equation and dividing throughout by Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev we get,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The left hand side depends only on (r, θ) while the right hand side on  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus each of the terms must be equated to a constant, which we take as m2. Writing the right hand side as

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Note that since the only dependence is on Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev we need not write the partial derivative and have replaced it by ordinary derivative. The solution of this equation is Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev where A is a constant. Note that the potential function, and hence, F is single valued. Thus if we increase the azimuthal angle by 2π, we must have the same value for F, so that,

  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

This requires m to be an integer. This allows us to restrict the domain of Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We now rewrite (2) as,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Using identical argument as above, because the left hand side is a function of r alone while the right hand side is a function of θ alone, we must equate each side of (3) to a constant. For reasons that will become clear later, we write this constant as Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev which is quite general as we have not said what l is. Thus we have,

 

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We can simplify this equation by making a variable transformation, Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev using which we get,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The domain θ of being [0:π], the range of μ is [-1:+1]. We will not attempt to solve this equation as it turns out that the equation is a rather well known equation in the theory of differential equations and the solutions are known to be polynomials in μ.

They are known as “Associated Legendre Polynomials” and are denoted by  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev We will point out the nature of the solutions.

It turns out that unless l happens to be an integer, the solutions of the above equation will diverge for  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus, physically meaningful solutions exist for integral values of l only. Let us look at some special cases of the solutions. For a given l, m takes integral values from Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

A particularly simple class of solution occur when the system has azimuthal symmetry, i.e., the system looks the same in the xy plane no matter from which angle Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev we look at it. This implies that our solutions must be Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevindependent, i.e. m=0.

In such a case, the equation for the associated Legendre polynomial takes the form,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The solutions of this equation are known as ordinary “Legendre Polynomials” and are denoted by  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Let us look at some of the lower order Legendre polynomials.

Take = 0 : The equation becomes,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

It is trivial to check that the solution is a constant. We take the constant to be 1 for normalization purpose.  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev The equation is

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

It is straightforward to check that the solution is Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

It can be verified that the solution is Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The solutions of a few lower order polynomials are shown below.

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

It can be verified that the Legendre polynomials of different orders are orthogonal,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We are now left with only the radial equation,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

A simple inspection tells us that the solutions are power series in r. Taking the solution to be of the form Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev we get, on substituting into the radial equation,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Equating the coefficient of rn to zero, we get,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

which gives the value of  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus, the function R(n) has the form Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Substituting the solutions obtained for Rand P, we get the complete solution for the potential for the azimuthally symmetric case (m=0) to be,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Example 1: A sphere of radius R has a potential Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev on its surface. Determine the potential outside the sphere.

Since we are only interested in solutions outside the sphere, in the solution (4), the term  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev cannot exist as it would make the potential diverge at infinity. Setting Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev we get the solution to be

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We can determine the constants Bl by looking at the surface potential For this purpose, we have to reexpress the given potential in terms of Legendre polynomials.

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Comparing this with the expression

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We conclude that l = 0,2 and the corresponding coefficients are given by

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus the potential outside the sphere is given by

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Complete Solution in Spherical Polar (without azimuthal symmetry)

If we do not have azimuthal symmetry, we get the complete solution by taking the product of R, P and F. We can write the general solution as

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev
Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

where the constants have been appropriately redefined. The functions Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev introduced above are known as “Spherical Harmonics”.These are essentially products of associated Legendre polynomials introduced earlier and functions Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev which form a complete set for expansion of an arbitrary function on the surface of a sphere. The normalized spherical harmonics are given by

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The functions are normalized as follows : 

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

For a given l , the spherical harmonics are polynomials of degree Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Some of the lower order Spherical harmonics are listed below.

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We have not listed the negative m values as they are related to the corresponding positive m values by the property

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Example 2 : A sphere of radius R has a surface charge density given by Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Determine the potential both inside and outside the sphere.

Solution : Surface charge density implies a discontinuity in the normal component of the electric field

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

where Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We need to express the right hand side in terms of spherical harmonics. Using the table of spherical harmonics given above, we can see that

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Consider the general expression for the potential given earlier,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We need to take the derivative of this expression just inside and just outside the surface. Inside the surface, the origin being included, Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev and outside the surface, the potential should vanish at infinity, requiring Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus, we have

Inside : Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Outside : Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Taking derivatives with respect to r and substituting r=R,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Comparing, we notice that only l = 2 terms are required in the sum. Comparing, we get,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

We get another connection between the coefficients by using the continuity of the tangential component of the electric fields inside and outside, given by derivatives with respect to and  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev.  Since the angle part is identical in the expressions for the potential inside and outside the sphere, we have,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

These two equations allow us to solve for  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev and we get,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus, the potential in this case is given by, 

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev
Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Tutorial Assignment

1. A sphere of radius R, centered at the origin, has a potential on its surface given by  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev
Find the potential outside the sphere

2. A spherical shell of radius R has a charge density  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev glued on its surface. There are no charges either inside or outside. Find the potential both inside and outside the sphere.

Solutions to Tutorial Assignment

1. The problem has azimuthal symmetry. Since we are interested in potential outside the sphere, we put all Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev On the surface, the potential can be written as Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Comparing this with the general expression for the potential, we have, Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus the potential function is given by

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

2. General expressions for potential inside and outside are given by,  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev The surface charge density is given by the discontinuity of normal component of the potential at r=R, ie.,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

The potential must be continuous across the surface. Since the Legendre polynomials are orthogonal, we have,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev The charge density expression contains only  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev term on the right. Clearly, only the Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev term should be considered in the expressions and all other coefficients must add up so as to give zero. We have,

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

From the continuity of the potential, we had  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus we have, Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Thus the potential is given by

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Self Assessment Quiz

1. The surface of a sphere of radius R has a potential  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev If there are no charges outside the sphere, obtain an expression for the potential outside.

2. A sphere of radius R has a surface potential given by  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Obtain an expression for the potential inside the sphere.

Solutions to Self Assessment Quiz 

1. One has to first express Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev in terms of Legendre polynomials. It can be checked that  Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev The general expression for the potential outside the sphereSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev Comparing this with the boundary condition at r=R, Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev all other coefficients are zero. Thus Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

2. The potential inside has the form   Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev the potential can be expressed in terms of spherical harmonics as

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

Thus only Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev terms are there in the expression for the potential. We have

Solutions to Laplace’s Equations (Part - 2) Electrical Engineering (EE) Notes | EduRev

 

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