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Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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We have seen that in the region of space where there are no sources of charge, Laplace’s equation gives the potential. In this lecture, we will discuss solutions of Laplace’s equation subject to some boundary conditions. 

Formal Solution in One Dimension

The solution of Laplace’s equation in one dimension gives a linear potential,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

has the solution  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)where m and c are constants. The solution is featureless because it is a monotonically increasing or a decreasing function of x. Further, since the potential at x is the average of the potentials at x+a and x-a, it has no minimum or maximum.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Formal Solution in Two Dimension

In two dimensions , the Laplace’s equation is  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Since the derivatives are partial, we must have

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where a is a constant. Thus the solution to the equation is

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Before looking at this solution let us look at a closely resembling Poisson’s equation in 2 dimensions,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which has the solution

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solution looks like cup having a minimum at x = 0, y = 0.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

On the other hand, if you look at the solution (1) of the Laplace’s equation, the graph of the potential as a function of x and y looks like the following :

 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Note that the function does not have a minimum or a maximum. The point x=0, y=0 is a “saddle point” where the function rises in one direction and decreases in another direction, much like a saddle on a horse where the seat slopes downwards along the back but rises along its length both forward and backward.

[Mathematically, in two dimensions, if a function f partial derivatives  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) such that  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) the function has a saddle point. If the expression has a positive value, the function has a maximum or a minimum depending on whether  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) are both negative or both positive respectively. For eqn. (1) at the point (0,0)the expression is Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This is true in any dimension because for a minimum the second partial derivative with respect to each variable must be negative so that their sum cannot add to zero as is required by Laplace’s equation.

Earnshaw’s Theorem

A consequence of solutions Laplace’s equation not having a minimum is that a charge cannot be held in equilibrium by electrostatic forces alone (i.e., if we are to keep a charge in equilibrium, we need forces other than just electrostatic forces; or for that matter, magnetostatic or static gravitational forces). This is known as the Earnshaw’s theorem.

The proof is elementary. In a region of space containing no charge, Laplace’s equation is valid for the potential φ If a charge is to be kept in this potential, its potential energy qφ also satisfies the Laplace’s equation. Since the solutions of Laplace’s equation do not have minima, the charge cannot be in static equilibrium.

In the remaining part of this lecture we will discuss the solutions of Laplace’s equation in three dimensions in different coordinate systems.

Solution in Cartesian Coordinates :

The Laplace’s equation is Cartesian coordinates is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We will look into the solution of this equation in a region defined by a parallelepiped of dimensions Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) In principle, one could choose boundary conditions such as prescribing the potential on each of the six surfaces of the parallelepiped. However, we consider a situation in which five walls of the parallelepiped are conductors which are joined together so that each is maintained at the same potential, which we take to be zero. The sixth plate of the parallelepiped at z = L3 is made with a different sheet of metal and is kept at a potential which is a known function  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) of position on the plate.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Summarizing the boundary conditions, we have,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We will use a technique known as separation of variables to solve the equation. We assume that the solution Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) can be written as a product of three independent functions, X, Y and Z, the first one depending on the variable x alone, the second on variable y alone and the third on the variable z alone. It is not that such a trick will always work, but in cases where it does, the uniqueness theorem would guarantee us that that would be the only solution. Keeping this in view, let us attempt such a solution:

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substituting this in eqn. (3) and dividing both sides by the product Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

In this equation there are three terms, the first depending on x alone, the second on y alone and the third on z alone. Since x,y,z are arbitrary, the equation can be satisfied only if each of the terms is a constant and the three constants are such that they add up to zero.

Let us write

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Where Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) are constants, which satisfy  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The boundary conditions on five faces of the parallelepiped where Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) can be written as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

However, on the sixth face we must have  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) because the potential on this face depends only on x,y whereas Z can only depend on z.

Consider the first equation, viz., Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The solution of this equation is well known to be linear combination of sine and cosine functions. Keeping the boundary conditions Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in mind, we can write the solution to be given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) where m is an integer not equal to zero (as it would make the function identically vanish) and A is a constant .

In a very similar way the solution for Y is written as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

With B being a constant and Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Once again n is a non-zero integer.

The solution for Z is a little more complicated because of the constraint Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)are positive integers Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)   is also positive. 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Then has solutions which are hyperbolic functions,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

with  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)  Thus, we can write the expression for the potential as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solution above satisfy the first five boundary conditions. Let us see the effect of the last boundary condition.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The constants Cmn can be derived from above equation because f(x,y) is a known function. For this we use the orthogonal property of the trigonometric functions,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This gives,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Take for instance, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The integral above is then product of two integrals,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus Cmn is non-zero only when both m and n are odd and has the value

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

In the next lecture we will obtain solutions of Laplace’s equation in spherical coordinates.

 

Tutorial Assignment

1. Find an expression for the potential in the region between two infinite parallel planes z = 0 and z = 1 the potential on the planes being given by the following :

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

2. Obtain a solution to Laplace’s equation in two dimensions in Cartesian coordinates assuming that the principle of variable separation holds.

3. Find the solution of two dimensional Laplace’s equation inside a rectangular region bounded by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)The potential has value zero on the first three boundaries and takes value  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) on the fourth side.

Solution to Tutorial Assignment

1. We know that the general solution is a product of linear combination of either sine, cosine or hyperbolic sine and hyperbolic cosine functions with a relationship between the arguments. . In this case, since on both plates the dependence on x and y is given as  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) the general solution is

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Where the argument of the hyperbolic function is obtained from the relation  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Substituting the boundary condition Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

2. We start with  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get, on substituting and dividing throughout by XY,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solution is then a product of  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) cosh ky. We can reduce the number of constants from 5 to 4 by redefining the constants. For instance, if  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we define the constants as  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and write,
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The constants A, B, C and k are to be determined from boundary conditions.

3. We have seen that the general solution is a product of linear combination of sine and cosine in one of the variables and hyperbolic sine and cosine in the other variable. Since the potential is to be zero for x = a for all values of y, this is possible only if we choose the solution for the variable x to be the trigonometric function and for y to be hyperbolic function. This is because a sine function can become zero at values of its argument other than zero, a hyperbolic function cannot. Further, only sine and sinh functions need to be considered so that at x= 0 and at y=0, the function vanishes. Thus we write 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

In this form, each term of the series automatically satisfies the boundary conditions at x= 0 and at y=0. Since the potential is to become zero for all values of y for x= 1, this is possible if Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) if  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus the potential function becomes

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We have to satisfy the only remaining boundary condition, viz., for Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we have,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

To determine the coefficients An, we multiply both sides of the expression by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and integrate over x from 0 to a
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The integral on the left gives  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) while that on the right gives Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This completely determines the potential.

Self Assessment Quiz

1. Does the function   Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) satisfy Laplace’s equation? 2. Find an expression for the potential in the region between two infinite

2. Find an expression for the potential in the region between two infinite parallel planes z = 0 and z π/2 on which the potentials are respectively given by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and 0.

3. Find an expression for the potential in the region between two infinite parallel planes, the potential on which are given by the following :

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solution to self assessment Quiz

1. Yes, a direct calculation shows

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

2. In this case also we have solutions which are written as products of separate variables. The solution can be written as (since X and Y are hyperbolic functions, the function Z is a linear combination of sine and cosine),

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The constants A and B can be determined by the boundary conditions. For  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) gives Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) gives A=0. Thus

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

3. The z dependence of the potential is a linear combination of hyperbolic sine and cosine function with argument Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus the general expression is

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

For z=0, the potential being zero, the cosh term vanishes, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) gives  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) so that  

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Laplace’s Equation in Spherical Coordinates :

In spherical coordinates the equation can be written as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Note that we are using the notation Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Inserting this into the Laplace’s equation and dividing throughout by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The left hand side depends only on (r, θ) while the right hand side on  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Note that since the only dependence is on Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This requires m to be an integer. This allows us to restrict the domain of Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We now rewrite (2) as,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) which is quite general as we have not said what l is. Thus we have,

 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We can simplify this equation by making a variable transformation, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) using which we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we look at it. This implies that our solutions must be Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)independent, i.e. m=0.

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Take = 0 : The equation becomes,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The equation is

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

It is straightforward to check that the solution is Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

It can be verified that the solution is Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We are now left with only the radial equation,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get, on substituting into the radial equation,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Equating the coefficient of rn to zero, we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which gives the value of  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus, the function R(n) has the form Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Example 1: A sphere of radius R has a potential Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) cannot exist as it would make the potential diverge at infinity. Setting Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get the solution to be

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Comparing this with the expression

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus the potential outside the sphere is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where the constants have been appropriately redefined. The functions Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) introduced above are known as “Spherical Harmonics”.These are essentially products of associated Legendre polynomials introduced earlier and functions Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The functions are normalized as follows : 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

For a given l , the spherical harmonics are polynomials of degree Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Some of the lower order Spherical harmonics are listed below.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Example 2 : A sphere of radius R has a surface charge density given by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Consider the general expression for the potential given earlier,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and outside the surface, the potential should vanish at infinity, requiring Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus, we have

Inside : Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Outside : Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE).  Since the angle part is identical in the expressions for the potential inside and outside the sphere, we have,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

These two equations allow us to solve for  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus, the potential in this case is given by, 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Find the potential outside the sphere

2. A spherical shell of radius R has a charge density  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) On the surface, the potential can be written as Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Comparing this with the general expression for the potential, we have, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus the potential function is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The charge density expression contains only  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) term on the right. Clearly, only the Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

From the continuity of the potential, we had  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus we have, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus the potential is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Self Assessment Quiz

1. The surface of a sphere of radius R has a potential  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in terms of Legendre polynomials. It can be checked that  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The general expression for the potential outside the sphereSolutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Comparing this with the boundary condition at r=R, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) all other coefficients are zero. Thus Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus only Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) terms are there in the expression for the potential. We have

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

At large distances from the conductor, the above expression must match with  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) are the only non-zero terms,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Comparing both sides, we have, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
The potential outside the sphere is thus given by,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The charge on the lower hemisphere is Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) You may recall that the potential due to a dipole placed at the origin is Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus the sphere behaves like a dipole of dipole moment Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) while the lower is maintained at Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) .We are required to find the potential within the sphere.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Using this we have,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Since the potential is constant in two hemispheres, we split this integral from  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) corresponding to Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) corresponding to Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We know that  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) the integrals within the square bracket adds up to φ0, so that we get  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Laplace’s Equation in Cylindrical Coordinates :

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substituting this into Laplace’s equation and dividing both sides of the equation by Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which has the solution,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

(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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Multiplying throughout by p2, we rewrite this equation as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which has the solution

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

The radial equation now reads

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Some of the limiting values of the function are as follows:
For  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) the function oscillates and has the form

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)123
k0m02.4065.5208.654
k1m13.8327.01610.173
k2m25.1368.41711.620


Higher roots are approximately located at  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Bessel functions satisfy orthogonality condition,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Using orthogonality property of Bessel functions, we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) variation of the Neumann function are as shown below.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solution of the radial equation can be written as

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

(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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

A separation of variables of the form  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) gives us, on expanding the first term,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solution for Q(θ), is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

has a solution

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

And for n=0

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Clearly  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Substituting these, we get,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The charge density on the surface of the conductor is the normal component of the electric field multiplied with  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) has no sources of charge on it. The potential on the rim Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)  is within the cylinder, Neumann functions are excluded from the solution and we have,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The potential at  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) also vanishes, giving  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

using which we have, substituting Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Multiply both sides of eqn. (I) with  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and integrate from 0 t0 R,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

2. Laplace’s equation in polar coordinates is  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) separation of variables Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which have the solution  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Clearly, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) all coefficients other than B4  are also zero and B4 = 2.

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Using the orthogonality property of the Legendre polynomial,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where, Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The integral can be looked up from standard tables,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Using standard table for Legendre polynomials we have, 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) etc. while all odd orders are zero. Thus,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Self Assessment Quiz

1. A conducting sphere of radius R is kept in an electrostatic field given by  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) has no sources of charge on it. The electric field on the rim  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is given by  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and at  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 

Solutions to Self Assessment Quiz

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The boundary conditions to be satisfied are

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The surface charge density is given by

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

 

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The radial equation has the solution Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Integrate the first expression from  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) and the second expression from Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) 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 - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Thus the potential expression at p = R becomes

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

We are left with

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

These integrals can be done by elementary methods,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

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

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The integral can be easily done If m ≠ n,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We have the final expression for the potential,

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

To verify that this is consistent with the given boundary conditions, let us evaluate the potential at  Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The infinite series has a value Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Similarly, one can check that Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 

The document Solutions to Laplace’s Equations - Notes | Study Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Electromagnetic Fields Theory (EMFT).
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