Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

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Laplace’s Equation in Spherical Coordinates :

We continue with our discussion of solutions of Laplace’s equation in spherical coordinates by giving some more examples.

Example 3 : A conducting sphere in a uniform electric field This is a classic problem. Let the sphere be placed in a uniform electric field in z direction Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Though we expect the field near the conductor to be modified, at large distances from the conductor, the field should be uniform and we have,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Clearly, we have azimuthal symmetry because of the sphere but the direction of the electric field will bring in polar angle dependence. As a result, the potential at large distances has the following form :

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where C is a constant. The conductor being a equipotential, the field lines strike its surface normally.

The lines of force near the sphere look as follows :

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

As shown here, there will be charges induced on the surface. On the surface, the potential is constant (it is a conductor) φ0. Since there are no sources, outside the sphere, the potential satisfies the Laplace’s equation. As there is azimuthal symmetry in the problem, the potential is given by,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Note that the potential expression cannot contain l = 0 term in the second term because a potential Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev implies a delta function source.

At large distances from the conductor, the above expression must match with  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev which we had obtained earlier. As all terms containing Bl vanish at large distances, we need to compare with the terms containing  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev are the only non-zero terms,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

We will now use the boundary condition on the surface of the conductor, which is a constant, say  φ0  This implies

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Comparing both sides, we have, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev
The potential outside the sphere is thus given by,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The electric field is given by the negative gradient of the potential,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The charge density on the sphere is given by the normal component (i.e. radial component) of the electric field on the surface (r=R) and is given by

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The charge induced on the upper hemisphere is (recall that θ is measured from the north pole of the sphere as z direction points that way,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The charge on the lower hemisphere is Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev so that the total charge is zero, as expected.

Note that an uncharged sphere in an electric field modifies the potential by  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev You may recall that the potential due to a dipole placed at the origin is Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the sphere behaves like a dipole of dipole moment Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

If the sphere had a charge Q, it would modify the potential by an additional Coulomb term.

Example 4 : Conducting hemispherical shells joined at the equator :

Consider two conducting hemispherical shells which are joined at the equator with negligible separation between them. The upper hemisphere is mained at a potential of  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev while the lower is maintained at Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev .We are required to find the potential within the sphere.

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the system gas azimuthal symmetry, we can expand the potential in terms of Legendre polynomials,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using the argument we have given earlier, the relevant equation for the potential inside and outside the sphere are as follows :

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

We will determine a few of the coefficients in these expansions. To find the coefficients, we use the orthogonality property of the Legendre polynomials,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using this we have,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the potential is constant in two hemispheres, we split this integral from  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev corresponding to Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev corresponding to Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

We know that  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev is a polynomial in μ, containing only odd powers of μ if m is off and only even powers of μ if m is even, the degree of polynomial being m. It can be easily seen that for even values of m, the above integral vanishes and only m that give non zero value are those which are odd. We will calculate a few such coefficients.

Take m=1 for which  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev the integrals within the square bracket adds up to φ0, so that we get  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Take m=3 for which Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The integrals add up to  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev so that we get  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Laplace’s Equation in Cylindrical Coordinates :

In cylindrical coordinates , Laplace’s equation has the following form :

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

As before, we will attempt a separation of variables, by writing,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Substituting this into Laplace’s equation and dividing both sides of the equation by Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev we get,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where, as before, we have used the fact that the first two terms depend on p and θ while the third term depends on z alone. Here -k2 is a constant and we have not yet specified its nature. We can easily solve the z equation,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

which has the solution,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

(Remember that we have not specified k, if it turns out to be imaginary, our solution would be sine and cosine functions. In the following discussion we choose the hyperbolic form.) 

Let us now look at the equation for R and Q, Expanding the first term,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Multiplying throughout by p2, we rewrite this equation as

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Once again, we notice that the left hand side depends only on ρ while the right hand side depends on θ only. Thus we must have,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where v2 is an yet unspecified constant. The angle equation can be easily solved,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

which has the solution

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The domain of θ is [0:2π]. Since the potential must be single valued, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev which requires ν must be an integer.

The radial equation now reads

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The above can be written in a compact form by defining  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev in terms of which the equation reads,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

This equation is known as the Bessel equation and its solutions are known as Bessel Functions. We will not solve this equation but will point out the nature of its solutions. Being a second order equation, there are two independent solutions, known as Bessel functions of the first and the second kind. The first kind is usually referred to as Bessel functions whereas the second kind is also known as Neumann functions.

The Bessel functions of order ν is given by the power series,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Some of the limiting values of the function are as follows:
For  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev the function oscillates and has the form

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

When ν is an integer,  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev are not independent and are related by Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The variation of the Bessel function of some of the integral orders are given in the following figure.

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

It can be seen that other that the zeroth order Bessel function, all Bessel functions vanish at the origin. In addition the Bessel functions have zeros at different values of their argument. The following table lists the zeros of Bessel functions. In the following knm denotes the m-th zero of Bessel function of order n.

 

 Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev123
k0m02.4065.5208.654
k1m13.8327.01610.173
k2m25.1368.41711.620


Higher roots are approximately located at  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev
Bessel functions satisfy orthogonality condition,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Another usefulness of Bessel function lies in the fact that a piecewise continuous function f defined in [0,a] with f(a)=0 can be expanded as follows :

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using orthogonality property of Bessel functions, we get,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The second kind of Bessel function, viz., the Neumann function, diverges at the origin and oscillates for larger values of its argument ,Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Like the Bessel function of the first kind, the Neumann functions are also not independent for integral values of n  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev variation of the Neumann function are as shown below.

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The solution of the radial equation can be written as

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

(If in the solution of z equation we had chosen k to be imaginary, the argument of the Bessel functions would be imaginary)
The complete solution is obtained by summing over all values of k and ν.

Special Case : A system with potential independent of z: In problems such as a long conductor, the symmetry of the problem makes the potential independent of z coordinates. In such cases, the problem is essentially two dimensional, the Laplace’s equation being,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

A separation of variables of the form  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev gives us, on expanding the first term,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The solution for Q(θ), is given by

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where, as before, singlevaluedness of the potential requires that n is an integer. The radial equation,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

has a solution

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

And for n=0

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Example : An uncharged cylinder in a uniform electric field.

Like the problem of sphere in an electric field, we will discuss how potential function for a uniform electric field is modified in the presence of a cylinder. Let us take the electric field directed along the x-axis and we further assume that the potential does not have any z dependence.

As before, at large distances, the potential is that corresponding to a constant electric field in the x direction, i.e  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The charges induced on the conductor surface produce their own potential which superpose with this potential to satisfy the boundary condition on the surface of the conductor. Since there is no z dependence, the general solution of the Laplace’s equation, as explained above is,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Here we have not taken n=0 term because its behavior is logarithmic which diverges at infinite distances. At large distances, the asymptotic behavior of this should match with the potential corresponding to the uniform field. Thus we choose the term proportional to  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Clearly  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev We need to determine the constant D1.  If the potential on the surface of the cylinder is taken to be zero, we must have for all angles θ, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Substituting these, we get,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The electric field is given by the gradient of the potential, and is

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The charge density on the surface of the conductor is the normal component of the electric field multiplied with  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Like in the case of the sphere, you can check that the total charge density is zero.

 

Tutorial Assignment

1. A cylindrical shell of radius R and length L has its top cap maintained at a constant potential φ0, its bottom cap and the curved surface are grounded. Obtain an expression for the potential within the cylinder.

2. A unit disk Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev has no sources of charge on it. The potential on the rim Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev is given by  2 sin 4θ, where θ is the polar angle. Obtain an expression for potential inside the disk.

3. Consider two hemispherical shells of radius R where the bottom half is kept at zero potential and the top half is maintained at a constant potential φ0, Obtain an expression for the potential inside the shell.

 

Solutions to Tutorial Assignment

1. Take the bottom cap in the x-y plane with its centre at the origin and the z axis along the axis of the cylinder. Since the potential at z=0 is zero, we take Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev  is within the cylinder, Neumann functions are excluded from the solution and we have,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the potential on the curved surface is zero irrespective of the azimuthal angle, we must v = 0. This gives,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The potential at  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev also vanishes, giving  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev This determines the values of  kn being given by the zeros of Bessel function of order zero. Finally, the boundary condition on the top cap,Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The coefficients An are determined by the orthogonal property of Bessel functions,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

using which we have, substituting Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev
Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Multiply both sides of eqn. (I) with  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev and integrate from 0 t0 R,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

2. Laplace’s equation in polar coordinates is  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev separation of variables Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

which have the solution  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Here we have used the singlevaluedness of the potential to conclude that λ is an integer. The solution of the radial equation is  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The constant D must be zero because the function must be well behaved at the origin. Likewise the solution of the radial equation for n=0 is a logarithmic function the coefficient of which also vanishes for the same reason. Thus the solution is of the form,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

We can determine the coefficients by the boundary condition given at r=1,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Clearly, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev all coefficients other than B4  are also zero and B4 = 2.

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

3. See example 4 of lecture notes. Inside the sphere the potential is given by

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using the orthogonality property of the Legendre polynomial,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Using the fact that only one half of the sphere is maintained at constant potential while the other half is at zero potential, we have,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The integral can be looked up from standard tables,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Using standard table for Legendre polynomials we have, 

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev etc. while all odd orders are zero. Thus,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Self Assessment Quiz

1. A conducting sphere of radius R is kept in an electrostatic field given by  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Obtain an expression for the potential in the region outside the sphere and obtain the charge density on the sphere

2. A unit disk  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev has no sources of charge on it. The electric field on the rim  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev is given by  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev is the polar angle. Obtain an expression for potential inside the disk.

3. (Hard Problem) A cylindrical conductor of infinite length has its curved surfaces divided into two equal halves, the part with polar angle  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev has a constant potential φ1 while the remaining half is at a potential φ2.Obtain an expression for the potential inside the cylinder. Verify your result by calculating the potential on the surface at   Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev and at  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

 

Solutions to Self Assessment Quiz

1. The potential corresponding to the electric field is  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRevSolutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Using the table for Spherical harmonics, the potential can be expressed as Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The general expression for solution of Laplace’s equation is

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The boundary conditions to be satisfied are

(i) For all  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev the potential is constant, which we take to be zero. This implies  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev for all m. This makes the potential expression to be

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

(ii) For large distances, Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev in this limit,, the second term for the expression for potential goes to zero and we are left with,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Comparing, we get only non-zero value of to be 2 and the corresponding m values as +2. Thus, the potential has the form

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The surface charge density is given by

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

2. The solution of Laplace’s equation on a unit disk is given by (see tutorial problem)

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The electric field is given by the negative gradient of potential. Since the field is given to be radial,

 

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

This gives all coefficients other than B4 to be zero and  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The coefficient A0, however remains undetermined.

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

3. Because of symmetry along the z axis, the potential can only depend on the polar coordinates (p, θ) . By using a separation of variables, we can show, as shown in the lecture, the angular part Q(θ) =  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev The radial equation has the solution Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev we have ignored n=0 solution because it gives rise to logarithm in radial coordinate which diverges at origin. Thus the general solution for the potential is

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

where K is a constant. Further, the constant D must be zero otherwise the potential would diverge at p = 0. Let us now apply the boundary conditions, For p = R,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Integrate the first expression from  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev and the second expression from Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev and add. The integration over the second term on the right hand side of the two expressions is equal to an integration from 0 to 2π and the integral vanishes. We are left with

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the potential expression at p = R becomes

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

To evaluate the coefficients An, multiply both sides by cos mθ and integrate from 0 to 2π,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Similarly, the first term on the right also gives zero.

We are left with

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

These integrals can be done by elementary methods,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

To determine the coefficients Bn, multiply both sides of (I) by sin mθ and integrate from 0 to 2π,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The first term on the right gives zero.
We are left with

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The integral can be easily done If m ≠ n,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

We have the final expression for the potential,

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

To verify that this is consistent with the given boundary conditions, let us evaluate the potential at  Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

The infinite series has a value Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev Similarly, one can check that Solutions to Laplace’s Equations (Part - 3) Electrical Engineering (EE) Notes | EduRev

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