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

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,

has the solution  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.

Formal Solution in Two Dimension

In two dimensions , the Laplace’s equation is 

Since the derivatives are partial, we must have

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

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

which has the solution

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

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 :

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   such that   the function has a saddle point. If the expression has a positive value, the function has a maximum or a minimum depending on whether   are both negative or both positive respectively. For eqn. (1) at the point (0,0)the expression is

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

We will look into the solution of this equation in a region defined by a parallelepiped of dimensions  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   of position on the plate.

Summarizing the boundary conditions, we have,

We will use a technique known as separation of variables to solve the equation. We assume that the solution  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:

Substituting this in eqn. (3) and dividing both sides by the product  we get,

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

Where  are constants, which satisfy 

The boundary conditions on five faces of the parallelepiped where  can be written as

However, on the sixth face we must have   because the potential on this face depends only on x,y whereas Z can only depend on z.

Consider the first equation, viz.,  The solution of this equation is well known to be linear combination of sine and cosine functions. Keeping the boundary conditions  in mind, we can write the solution to be given by

where  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

With B being a constant and  Once again n is a non-zero integer.

The solution for Z is a little more complicated because of the constraint are positive integers    is also positive. 

Then has solutions which are hyperbolic functions,

with    Thus, we can write the expression for the potential as

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

The constants C_mn 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,

This gives,

Take for instance,

The integral above is then product of two integrals,

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

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 :

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 The potential has value zero on the first three boundaries and takes value   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   the general solution is

Where the argument of the hyperbolic function is obtained from the relation   Substituting the boundary condition

2. We start with   we get, on substituting and dividing throughout by XY,

The solution is then a product of   cosh ky. We can reduce the number of constants from 5 to 4 by redefining the constants. For instance, if   we define the constants as   and write,

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 

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  if   Thus the potential function becomes

We have to satisfy the only remaining boundary condition, viz., for  we have,

To determine the coefficients A_n, we multiply both sides of the expression by  and integrate over x from 0 to a

The integral on the left gives   while that on the right gives  Thus,

This completely determines the potential.

Self Assessment Quiz

1. Does the function    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  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 :

Solution to self assessment Quiz

1. Yes, a direct calculation shows

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),

The constants A and B can be determined by the boundary conditions. For   gives  gives A=0. Thus

3. The z dependence of the potential is a linear combination of hyperbolic sine and cosine function with argument  Thus the general expression is

For z=0, the potential being zero, the cosh term vanishes,  gives   so that  

Laplace’s Equation in Spherical Coordinates :

In spherical coordinates the equation can be written as

Note that we are using the notation  denote the azimuthal angle and φ to denote the potential function.
As with the rectangular coordinates, we will attempt a separation of variable, writing

Inserting this into the Laplace’s equation and dividing throughout by  we get,

The left hand side depends only on (r, θ) while the right hand side on   Thus each of the terms must be equated to a constant, which we take as m². Writing the right hand side as

Note that since the only dependence is on  we need not write the partial derivative and have replaced it by ordinary derivative. The solution of this equation is  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,

This requires m to be an integer. This allows us to restrict the domain of

We now rewrite (2) as,

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  which is quite general as we have not said what l is. Thus we have,

We can simplify this equation by making a variable transformation,  using which we get,

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   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   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

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  we look at it. This implies that our solutions must be independent, i.e. m=0.

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

The solutions of this equation are known as ordinary “Legendre Polynomials” and are denoted by   Let us look at some of the lower order Legendre polynomials.

Take = 0 : The equation becomes,

It is trivial to check that the solution is a constant. We take the constant to be 1 for normalization purpose.  

 The equation is

It is straightforward to check that the solution is

It can be verified that the solution is

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

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

We are now left with only the radial equation,

A simple inspection tells us that the solutions are power series in r. Taking the solution to be of the form  we get, on substituting into the radial equation,

Equating the coefficient of rⁿ to zero, we get,

which gives the value of  

Thus, the function R(n) has the form  Substituting the solutions obtained for Rand P, we get the complete solution for the potential for the azimuthally symmetric case (m=0) to be,

Example 1: A sphere of radius R has a potential  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   cannot exist as it would make the potential diverge at infinity. Setting  we get the solution to be

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

Comparing this with the expression

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

Thus the potential outside the sphere is given by

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

where the constants have been appropriately redefined. The functions  introduced above are known as “Spherical Harmonics”.These are essentially products of associated Legendre polynomials introduced earlier and functions  which form a complete set for expansion of an arbitrary function on the surface of a sphere. The normalized spherical harmonics are given by

The functions are normalized as follows : 

For a given l , the spherical harmonics are polynomials of degree

Some of the lower order Spherical harmonics are listed below.

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

Example 2 : A sphere of radius R has a surface charge density given by  Determine the potential both inside and outside the sphere.

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

where

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

Consider the general expression for the potential given earlier,

We need to take the derivative of this expression just inside and just outside the surface. Inside the surface, the origin being included,  and outside the surface, the potential should vanish at infinity, requiring  Thus, we have

Inside :

Outside :

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

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

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  .  Since the angle part is identical in the expressions for the potential inside and outside the sphere, we have,

These two equations allow us to solve for   and we get,

Thus, the potential in this case is given by, 

Tutorial Assignment

1. A sphere of radius R, centered at the origin, has a potential on its surface given by  
Find the potential outside the sphere

2. A spherical shell of radius R has a charge density   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  On the surface, the potential can be written as  Comparing this with the general expression for the potential, we have,  Thus the potential function is given by

2. General expressions for potential inside and outside are given by,   The surface charge density is given by the discontinuity of normal component of the potential at r=R, ie.,

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

Thus  The charge density expression contains only   term on the right. Clearly, only the  term should be considered in the expressions and all other coefficients must add up so as to give zero. We have,

From the continuity of the potential, we had   Thus we have,  Thus the potential is given by

Self Assessment Quiz

1. The surface of a sphere of radius R has a potential   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   Obtain an expression for the potential inside the sphere.

Solutions to Self Assessment Quiz 

1. One has to first express  in terms of Legendre polynomials. It can be checked that   The general expression for the potential outside the sphere Comparing this with the boundary condition at r=R,  all other coefficients are zero. Thus

2. The potential inside has the form    the potential can be expressed in terms of spherical harmonics as

Thus only  terms are there in the expression for the potential. We have

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  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,

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 :

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 :

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,

Note that the potential expression cannot contain l = 0 term in the second term because a potential  implies a delta function source.

At large distances from the conductor, the above expression must match with   which we had obtained earlier. As all terms containing B_l vanish at large distances, we need to compare with the terms containing   are the only non-zero terms,

Thus,

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

Comparing both sides, we have,
The potential outside the sphere is thus given by,

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

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

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,

The charge on the lower hemisphere is  so that the total charge is zero, as expected.

Note that an uncharged sphere in an electric field modifies the potential by   You may recall that the potential due to a dipole placed at the origin is

Thus the sphere behaves like a dipole of dipole moment

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   while the lower is maintained at  .We are required to find the potential within the sphere.

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

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

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

Using this we have,

Thus,

Since the potential is constant in two hemispheres, we split this integral from   corresponding to  corresponding to

We know that   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   the integrals within the square bracket adds up to φ₀, so that we get  

Take m=3 for which  The integrals add up to   so that we get  

Laplace’s Equation in Cylindrical Coordinates :

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

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

Substituting this into Laplace’s equation and dividing both sides of the equation by  we get,

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,

which has the solution,

(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,

Multiplying throughout by p², we rewrite this equation as

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,

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

which has the solution

The domain of θ is [0:2π]. Since the potential must be single valued,  which requires ν must be an integer.

The radial equation now reads

The above can be written in a compact form by defining   in terms of which the equation reads,

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,

Some of the limiting values of the function are as follows:
For   the function oscillates and has the form

When ν is an integer,   are not independent and are related by  The variation of the Bessel function of some of the integral orders are given in the following figure.

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 kn_m denotes the m-th zero of Bessel function of order n.

		1	2	3
k0m	0	2.406	5.520	8.654
k1m	1	3.832	7.016	10.173
k2m	2	5.136	8.417	11.620

Higher roots are approximately located at  
Bessel functions satisfy orthogonality condition,

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 :

Using orthogonality property of Bessel functions, we get,

The second kind of Bessel function, viz., the Neumann function, diverges at the origin and oscillates for larger values of its argument ,

where  Like the Bessel function of the first kind, the Neumann functions are also not independent for integral values of n   variation of the Neumann function are as shown below.

The solution of the radial equation can be written as

(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,

A separation of variables of the form   gives us, on expanding the first term,

The solution for Q(θ), is given by

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

has a solution

And for n=0

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   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,

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  

Clearly   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,  Substituting these, we get,

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

The charge density on the surface of the conductor is the normal component of the electric field multiplied with  

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  has no sources of charge on it. The potential on the rim  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   is within the cylinder, Neumann functions are excluded from the solution and we have,

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

The potential at   also vanishes, giving   This determines the values of  k_n being given by the zeros of Bessel function of order zero. Finally, the boundary condition on the top cap,

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

using which we have, substituting

Multiply both sides of eqn. (I) with   and integrate from 0 t0 R,

2. Laplace’s equation in polar coordinates is   separation of variables

which have the solution   Here we have used the singlevaluedness of the potential to conclude that λ is an integer. The solution of the radial equation is   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,

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

Clearly,  all coefficients other than B₄  are also zero and B₄ = 2.

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

Using the orthogonality property of the Legendre polynomial,

where,  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,

The integral can be looked up from standard tables,

Thus,

Using standard table for Legendre polynomials we have, 

 etc. while all odd orders are zero. Thus,

Self Assessment Quiz

1. A conducting sphere of radius R is kept in an electrostatic field given by   Obtain an expression for the potential in the region outside the sphere and obtain the charge density on the sphere

2. A unit disk   has no sources of charge on it. The electric field on the rim   is given by   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   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    and at  

Solutions to Self Assessment Quiz

1. The potential corresponding to the electric field is   Using the table for Spherical harmonics, the potential can be expressed as  The general expression for solution of Laplace’s equation is

The boundary conditions to be satisfied are

(i) For all   the potential is constant, which we take to be zero. This implies   for all m. This makes the potential expression to be

(ii) For large distances,  in this limit,, the second term for the expression for potential goes to zero and we are left with,

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

The surface charge density is given by

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

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

This gives all coefficients other than B₄ to be zero and   The coefficient A₀, however remains undetermined.

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) =   The radial equation has the solution  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

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,

Integrate the first expression from   and the second expression from  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

Thus the potential expression at p = R becomes

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

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

We are left with

These integrals can be done by elementary methods,

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

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

The integral can be easily done If m ≠ n,

We have the final expression for the potential,

To verify that this is consistent with the given boundary conditions, let us evaluate the potential at  

The infinite series has a value  Similarly, one can check that