Chapter 3. Special Techniques for Calculating Potentials

Given a stationary charge distribution we can, in principle, calculate the electric field:

where . This integral involves a vector as an integrand and is, in general, difficult to calculate. In most cases it is easier to evaluate first the electrostatic potential V which is defined as

since the integrand of the integral is a scalar. The corresponding electric field can then be obtained from the gradient of V since

The electrostatic potential V can only be evaluated analytically for the simplest charge configurations. In addition, in many electrostatic problems, conductors are involved and the charge distribution ρ is not known in advance (only the total charge on each conductor is known).
A better approach to determine the electrostatic potential is to start with Poisson's equation

Very often we only want to determine the potential in a region where ρ = 0. In this region Poisson's equation reduces to Laplace's equation

There are an infinite number of functions that satisfy Laplace's equation and the appropriate solution is selected by specifying the appropriate boundary conditions. This Chapter will concentrate on the various techniques that can be used to calculate the solutions of Laplace's equation and on the boundary conditions required to uniquely determine a solution.

In one dimension the electrostatic potential V depends on only one variable x. The electrostatic potential V(x) is a solution of the one-dimensional Laplace equation

The general solution of this equation is

where s and b are arbitrary constants. These constants are fixed when the value of the potential is specified at two different positions.

Example
Consider a one-dimensional world with two point conductors located at x = 0 m and at x = 10 m. The conductor at x = 0 m is grounded (V = 0 V) and the conductor at x = 10 m is kept at a constant potential of 200 V. Determine V.

The boundary conditions for V are

and

The first boundary condition shows that b = 0 V. The second boundary condition shows that s = 20 V/m. The electrostatic potential for this system of conductors is thus

The corresponding electric field can be obtained from the gradient of V

The boundary conditions used here, can be used to specify the electrostatic potential between x = 0 m and x = 10 m but not in the region x < 0 m and x > 10 m. If the solution obtained here was the general solution for all x, then V would approach infinity when x approaches infinity and V would approach minus infinity when x approaches minus infinity. The boundary conditions therefore provide the information necessary to uniquely define a solution to Laplace's equation, but they also define the boundary of the region where this solution is valid (in this example 0 m < x < 10 m).

The following properties are true for any solution of the one-dimensional Laplace equation:

Property 1:
V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in the region between the boundary points. This property is easy to proof:

This property immediately suggests a powerful analytical method to determine the solution of Laplace's equation. If the boundary values of V are

and

then property 1 can be used to determine the value of the potential at (a + b)/2:

Next we can determine the value of the potential at x = (3 a + b)/4 and at x = (a + 3 b)/4 :

This process can be repeated and V can be calculated in this manner at any point between x = a and x = b (but not in the region x > b and x < a).

Property 2:
The solution of Laplace's equation can not have local maxima or minima. Extreme values must occur at the end points (the boundaries). This is a direct consequence of property 1.
Property 2 has an important consequence: a charged particle can not be held in stable equilibrium by electrostatic forces alone (Earnshaw's Theorem). A particle is in a stable equilibrium if it is located at a position where the potential has a minimum value. A small displacement away from the equilibrium position will increase the electrostatic potential of the particle, and a restoring force will try to move the particle back to its equilibrium position. However, since there can be no local maxima or minima in the electrostatic potential, the particle can not be held in stable equilibrium by just electrostatic forces.

In two dimensions the electrostatic potential depends on two variables x and y. Laplace's equation now becomes

This equation does not have a simple analytical solution as the one-dimensional Laplace equation does. However, the properties of solutions of the one-dimensional Laplace equation are also valid for solutions of the two-dimensional Laplace equation:

Property 1:
The value of V at a point (x, y) is equal to the average value of V around this point

where the path integral is along a circle of arbitrary radius, centered at (x, y) and with radius R.

Property 2:
V has no local maxima or minima; all extremes occur at the boundaries.

In three dimensions the electrostatic potential depends on three variables x, y, and z. Laplace's equation now becomes

This equation does not have a simple analytical solution as the one-dimensional Laplace equation does. However, the properties of solutions of the one-dimensional Laplace equation are also valid for solutions of the three-dimensional Laplace equation:

Property 1:
The value of V at a point (x, y, z) is equal to the average value of V around this point

where the surface integral is across the surface of a sphere of arbitrary radius, centered at (x,y,z) and with radius R.

Figure 3.1. Proof of property 1.

To proof this property of V consider the electrostatic potential generated by a point charge q located on the z axis, a distance r away from the center of a sphere of radius R (see Figure 3.1). The potential at P, generated by charge q, is equal to

where d is the distance between P and q. Using the cosine rule we can express d in terms of r, R and θ

The potential at P due to charge q is therefore equal to

The average potential on the surface of the sphere can be obtained by integrating across the surface of the sphere. The average potential is equal to

which is equal to the potential due to q at the center of the sphere. Applying the principle of superposition it is easy to show that the average potential generated by a collection of point charges is equal to the net potential they produce at the center of the sphere.

Property 2:
The electrostatic potential V has no local maxima or minima; all extremes occur at the boundaries.

Example: Problem 3.3
Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends only on r. Then do the same for cylindrical coordinates.

Laplace's equation in spherical coordinates is given by

If V is only a function of r then

and

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-order differential equation:

This differential equation can be rewritten as

The general solution of this first-order differential equation is

where b is a constant. If V = 0 at infinity then b must be equal to zero, and consequently

Laplace's equation in cylindrical coordinates is

If V is only a function of r then

and

Therefore, Laplace's equation can be rewritten as

The solution V of this second-order differential equation must satisfy the following first-order differential equation:

This differential equation can be rewritten as

The general solution of this first-order differential equation is

where b is a constant. The constants a and b are determined by the boundary conditions.

Consider a volume (see Figure 3.2) within which the charge density is equal to zero. Suppose that the value of the electrostatic potential is specified at every point on the surface of this volume. The first uniqueness theorem states that in this case the solution of Laplace's equation is uniquely defined.

Figure 3.2. First Uniqueness Theorem

To proof the first uniqueness theorem we will consider what happens when there are two solutions V₁ and V₂ of Laplace's equation in the volume shown in Figure 3.2. Since V₁ and V₂ are solutions of Laplace's equation we know that

and

Since both V₁ and V₂ are solutions, they must have the same value on the boundary. Thus V₁ = V₂ on the boundary of the volume. Now consider a third function V₃, which is the difference between V₁ and V₂

The function V₃ is also a solution of Laplace's equation. This can be demonstrated easily:

The value of the function V₃ is equal to zero on the boundary of the volume since V₁ = V₂ there. However, property 2 of any solution of Laplace's equation states that it can have no local maxima or minima and that the extreme values of the solution must occur at the boundaries. Since V₃ is a solution of Laplace's equation and its value is zero everywhere on the boundary of the volume, the maximum and minimum value of V₃ must be equal to zero. Therefore, V₃ must be equal to zero everywhere. This immediately implies that

everywhere. This proves that there can be no two different functions V₁ and V₂ that are solutions of Laplace's equation and satisfy the same boundary conditions. Therefore, the solution of Laplace's equation is uniquely determined if its value is a specified function on all boundaries of the region. This also indicates that it does not matter how you come by your solution: if (a) it is a solution of Laplace's equation, and (b) it has the correct value on the boundaries, then it is the right and only solution.

Figure 3.3. System with conductors.

The first uniqueness theorem can only be applied in those regions that are free of charge and surrounded by a boundary with a known potential (not necessarily constant). In the laboratory the boundaries are usually conductors connected to batteries to keep them at a fixed potential. In many other electrostatic problems we do not know the potential at the boundaries of the system. Instead we might know the total charge on the various conductors that make up the system (note: knowing the total charge on a conductor does not imply a knowledge of the charge distribution ρ since it is influenced by the presence of the other conductors). In addition to the conductors that make up the system, there might be a charge distribution ρ filling the regions between the conductors (see Figure 3.3). For this type of system the first uniqueness theorem does not apply. The second uniqueness theorem states that the electric field is uniquely determined if the total charge on each conductor is given and the charge distribution in the regions between the conductors is known.
The proof of the second uniqueness theorem is similar to the proof of the first uniqueness theorem. Suppose that there are two fields and that are solutions of Poisson's equation in the region between the conductors. Thus

and

where ρ is the charge density at the point where the electric field is evaluated. The surface integrals of and , evaluated using a surface that is just outside one of the conductors with charge Q_i, are equal to . Thus

The difference between and , , satisfies the following equations:

Consider the surface integral of , integrated over all surfaces (the surface of all conductors and the outer surface). Since the potential on the surface of any conductor is constant, the electrostatic potential associated with and must also be constant on the surface of each conductor. Therefore, will also be constant on the surface of each conductor. The surface integral of over the surface of conductor i can be written as

Since the surface integral of over the surface of conductor i is equal to zero, the surface integral of over all conductor surfaces will also be equal to zero. The surface integral of over the outer surface will also be equal to zero since on this surface. Thus

The surface integral of can be rewritten using Green's identity as

where the volume integration is over all space between the conductors and the outer surface. Since is always positive, the volume integral of can only be equal to zero if everywhere. This implies immediately that everywhere, and proves the second uniqueness theorem.

Consider a point charge q held as a distance d above an infinite grounded conducting plane (see Figure 3.4). The electrostatic potential of this system must satisfy the following two boundary conditions:

A direct calculation of the electrostatic potential can not be carried out since the charge distribution on the grounded conductor is unknown. Note: the charge distribution on the surface of a grounded conductor does not need to be zero.

Figure 3.4. Method of images.

Consider a second system, consisting of two point charges with charges +q and -q, located at z = d and z = -d, respectively (see Figure 3.5). The electrostatic potential generated by these two charges can be calculated directly at any point in space. At a point P = (x, y, 0) on the xy plane the electrostatic potential is equal to

The potential of this system at infinity will approach zero since the potential generated by each charge will decrease as 1/r with increasing distance r. Therefore, the electrostatic potential generated by the two charges shown in Figure 3.5 satisfies the same boundary conditions as the system shown in Figure 3.4. Since the charge distribution in the region z > 0 (bounded by the xy plane boundary and the boundary at infinity) for the two systems is identical, the corollary of the first uniqueness theorem states that the electrostatic potential in this region is uniquely defined. Therefore, if we find any function that satisfies the boundary conditions and Poisson's equation, it will be the right answer. Consider a point (x, y, z) with z > 0. The electrostatic potential at this point can be calculated easily for the charge distribution shown in Figure 3.5. It is equal to

Since this solution satisfies the boundary conditions, it must be the correct solution in the region z > 0 for the system shown in Figure 3.4. This technique of using image charges to obtain the electrostatic potential in some region of space is called the method of images.
The electrostatic potential can be used to calculate the charge distribution on the grounded conductor. Since the electric field inside the conductor is equal to zero, the boundary condition for (see Chapter 2) shows that the electric field right outside the conductor is equal to

where σ is the surface charge density and is the unit vector normal to the surface of the conductor. Expressing the electric field in terms of the electrostatic potential V we can rewrite this equation as

Substituting the solution for V in this equation we find

Only in the last step of this calculation have we substituted z = 0. The induced charge distribution is negative and the charge density is greatest at (x = 0, y = 0, z = 0). The total charge on the conductor can be calculated by surface integrating of σ:

where . Substituting the expression for σ in the integral we obtain

As a result of the induced surface charge on the conductor, the point charge q will be attracted towards the conductor. Since the electrostatic potential generated by the charge image-charge system is the same as the charge-conductor system in the region where z > 0, the associated electric field (and consequently the force on point charge q) will also be the same. The force exerted on point charge q can be obtained immediately by calculating the force exerted on the point charge by the image charge. This force is equal to

There is however one important difference between the image-charge system and the real system. This difference is the total electrostatic energy of the system. The electric field in the image-charge system is present everywhere, and the magnitude of the electric field at (x, y, z) will be the same as the magnitude of the electric field at (x, y, -z). On the other hand, in the real system the electric field will only be non zero in the region with z > 0. Since the electrostatic energy of a system is proportional to the volume integral of the electrostatic energy of the real system will be 1/2 of the electrostatic energy of the image-charge system (only 1/2 of the total volume has a non-zero electric field in the real system). The electrostatic energy of the image-charge system is equal to

The electrostatic energy of the real system is therefore equal to

The electrostatic energy of the real system can also be obtained by calculating the work required to be done to assemble the system. In order to move the charge q to its final position we will have to exert a force opposite to the force exerted on it by the grounded conductor. The work done to move the charge from infinity along the z axis to z = d is equal to

which is identical to the result obtained using the electrostatic potential energy of the image-charge system.

Example: Example 3.2 + Problem 3.7
A point charge q is situated a distance s from the center of a grounded conducting sphere of radius R (see Figure 3.6).
a) Find the potential everywhere.
b) Find the induced surface charge on the sphere, as function of q. Integrate this to get the total induced charge.
c) Calculate the electrostatic energy of this configuration.

a) Consider a system consisting of two charges q and q', located on the z axis at z = s and z = z', respectively. If the potential produced by this system is identical everywhere to the potential produced by the system shown in Figure 3.6 then the position of point charge q' must be chosen such that the potential on the surface of a sphere of radius R, centered at the origin, is equal to zero (in this case the boundary conditions for the potential generated by both systems are identical).
We will start with determining the correct position of point charge q'. The electrostatic potential at P (see Figure 3.7) is equal to

This equation can be rewritten as

The electrostatic potential at Q is equal to

This equation can be rewritten as

Combining the two expression for q' we obtain

or

This equation can be rewritten as

The position of the image charge is equal to

The value of the image charge is equal to

Now consider an arbitrary point P' on the circle. The distance between P' and charge q is d and the distance between P' and charge q' is equal to d'. Using the cosine rule (see Figure 3.7) we can express d and d' in terms of R, s, and θ:

The electrostatic potential at P' is equal to

Thus we conclude that the configuration of charge and image charge produces an electrostatic potential that is zero at any point on a sphere with radius R and centered at the origin. Therefore, this charge configuration produces an electrostatic potential that satisfies exactly the same boundary conditions as the potential produced by the charge-sphere system. In the region outside the sphere, the electrostatic potential is therefore equal to the electrostatic potential produced by the charge and image charge. Consider an arbitrary point . The distance between this point and charge q is d and the distance between this point and charge q' is equal to d'. These distances can be expressed in terms of r, s, and θ using the cosine rule:

The electrostatic potential at will therefore be equal to

b) The surface charge density σ on the sphere can be obtained from the boundary conditions of

where we have used the fact that the electric field inside the sphere is zero. This equation can be rewritten as

Substituting the general expression for V into this equation we obtain

The total charge on the sphere can be obtained by integrating σ over the surface of the sphere. The result is

c) To obtain the electrostatic energy of the system we can determine the work it takes to assemble the system by calculating the path integral of the force that we need to exert in charge q in order to move it from infinity to its final position (z = s). Charge q will feel an attractive force exerted by the induced charge on the sphere. The strength of this force is equal to the force on charge q exerted by the image charge q'. This force is equal to

The force that we must exert on q to move it from infinity to its current position is opposite to . The total work required to move the charge is therefore equal to

Example: Problem 3.10
Two semi-infinite grounded conducting planes meet at right angles. In the region between them, there is a point charge q, situated as shown in Figure 3.8. Set up the image configuration, and calculate the potential in this region. What charges do you need, and where should they be located? What is the force on q? How much work did it take to bring q in from infinity?

Consider the system of four charges shown in Figure 3.9. The electrostatic potential generated by this charge distribution is zero at every point on the yz plane and at every point on the xz plane. Therefore, the electrostatic potential generated by this image charge distribution satisfies the same boundary conditions as the electrostatic potential of the original system. The potential generated by the image charge distribution in the region where x > 0 and y > 0 will be identical to the potential of the original system. The potential at a point P = (x, y, z) is equal to

The force exerted on q can be obtained by calculating the force exerted on q by the image charges. The total force is equal to the vector sum of the forces exerted by each of the three image charges. The force exerted by the image charge located at (-a, b, 0) is directed along the negative x axis and is equal to

The force exerted by the image charge located at (a, -b, 0) is directed along the negative y axis and is equal to

The force exerted by the image charge located at (-a, -b, 0) is directed along the vector connecting (-a, -b, 0) and (a, b, 0) and is equal to

The total force on charge q is the vector sum of , and :

The electrostatic potential energy of the system can, in principle, be obtained by calculating the path integral of between infinity and (a, b, 0). However, this is not trivial since the force is a rather complex function of a and b. An easier technique is to calculate the electrostatic potential energy of the system with charge and image charges. The potential energy of this system is equal to

However, in the real system the electric field is only non-zero in the region where x > 0 and y > 0. Therefore, the total electrostatic potential energy of the real system is only 1/4 of the total electrostatic potential energy of the image charge system. Thus

A powerful technique very frequently used to solve partial differential equations is separation of variables. In this section we will demonstrate the power of this technique by discussing several examples.

Example: Example 3.3 (Griffiths)
Two infinite, grounded, metal plates lie parallel to the xz plane, one at y = 0, the other at y = π (see Figure 3.10). The left end, at x = 0, is closed off with an infinite strip insulated from the two plates and maintained at a specified potential . Find the potential inside this "slot".

Figure 3.10. Example 3.3 (Griffiths).

The electrostatic potential in the slot must satisfy the three-dimensional Laplace equation. However, since V does not have a z dependence, the three-dimensional Laplace equation reduces to the two-dimensional Laplace equation:

The boundary conditions for the solution of Laplace's equation are:

1. V(x, y = 0) = 0 (grounded bottom plate).

2. V(x, y = π) = 0 (grounded top plate).

3. V(x = 0, y) = V₀(y) (plate at x = 0).

4. V → 0 when x → ∞.

These four boundary conditions specify the value of the potential on all boundaries surrounding the slot and are therefore sufficient to uniquely determine the solution of Laplace's equation inside the slot. Therefore, if we find one solution of Laplace's equation satisfying these boundary conditions than it must be the correct one. Consider solutions of the following form:

If this is a solution of the two-dimensional Laplace equation than we must require that

This equation can be rewritten as

The first term of the left-hand side of this equation depends only on x while the second term depends only on y. Therefore, if this equation must hold for all x and y in the slot we must require that

and

The differential equation for X can be rewritten as

If C₁ is a negative number than this equation can be rewritten as

where k² = -C₁ . The most general solution of this equation is

However, this function is an oscillatory function and does not satisfy boundary condition # 4, which requires that V approaches zero when x approaches infinity. We therefore conclude that C₁ can not be a negative number. If C₁ is a positive number then the differential equation for X can be written as

The most general solution of this equation is

This solution will approach zero when x approaches infinity if A = 0. Thus

The solution for Y can be obtained by solving the following differential equation:

The most general solution of this equation is

Therefore, the general solution for the electrostatic potential V(x,y) is equal to

where we have absorbed the constant B into the constants C and D. The constants C and D must be chosen such that the remaining three boundary conditions (1, 2, and 3) are satisfied. The first boundary condition requires that V(x, y = 0) = 0:

which requires that C = 0. The second boundary condition requires that V(x, y = π) = 0:

which requires that . This condition limits the possible values of k to positive integers:

Note: negative values of k are not allowed since exp(-kx) approaches zero at infinity only if k > 0. To satisfy boundary condition # 3 we must require that

This last expression suggests that the only time at which we can find a solution of Laplace's equation that satisfies all four boundary conditions has the form is when happens to have the form . However, since k can take on an infinite number of values, there will be an infinite number of solutions of Laplace's equation satisfying boundary conditions # 1, # 2 and # 4. The most general form of the solution of Laplace's equation will be a linear superposition of all possible solutions. Thus

Boundary condition # 3 can now be written as

Multiplying both sides by sin(ny) and integrating each side between y = 0 and y = π we obtain

The integral on the left-hand side of this equation is equal to zero for all values of k except k = n. Thus

The coefficients D_k can thus be calculated easily:

The coefficients D_k are called the Fourier coefficients of . The solution of Laplace's equation in the slot is therefore equal to

where

Now consider the special case in which . In this case the coefficients D_k are equal to

The solution of Laplace's equation is thus equal to

Example: Problem 3.12
Find the potential in the infinite slot of Example 3.3 (Griffiths) if the boundary at x = 0 consists to two metal stripes: one, from y = 0 to y = π/2, is held at constant potential , and the other, from y = π/2 to y = π is at potential .

The boundary condition at x = 0 is

The Fourier coefficients of the function are equal to

The values for the first four C coefficients are

It is easy to see that C_{k + 4} = C_k and therefore we conclude that

The Fourier coefficients C_k are thus equal to

The electrostatic potential is thus equal to

Example: Problem 3.13
For the infinite slot (Example 3.3 Griffiths) determine the charge density on the strip at x=0, assuming it is a conductor at constant potential .

The electrostatic potential in the slot is equal to

The charge density at the plate at x = 0 can be obtained using the boundary condition for the electric field at a boundary:

where is directed along the positive x axis. Since this boundary condition can be rewritten as

Differentiating V(x,y) with respect to x we obtain

At the x = 0 boundary we obtain

The charge density σ on the x = 0 strip is therefore equal to

Example: Double infinite slots
The slot of example 3.3 in Griffiths and its mirror image at negative x are separated by an insulating strip at x = 0. If the charge density σ(y) on the dividing strip is given, determine the potential in the slot.

The boundary condition at x = 0 requires that

where is directed along the positive x axis. Here we have used the symmetry of the configuration which requires that the electric field in the region x < 0 is the mirror image of the field in the region x > 0. Since this boundary condition can be rewritten as

We will first determine the potential in the x > 0 region. Following the same procedure as in Example 3 we obtain for the electrostatic potential

where the constants D_k must be chosen such that the boundary condition at x = 0 is satisfied. This requires that

Thus

The constants D_k can be determined by multiplying both sides of this equation with and integrating both sides with respect to y between y = 0 and y = π. The result is

The constants C_k are thus equal to

The electrostatic potential is thus equal to

Consider a spherical symmetric system. If we want to solve Laplace's equation it is natural to use spherical coordinates. Assuming that the system has azimuthal symmetry () Laplace's equation reads

Multiplying both sides by r² we obtain

Consider the possibility that the general solution of this equation is the product of a function , which depends only on the distance r, and a function , which depends only on the angle θ:

Substituting this "solution" into Laplace's equation we obtain

Dividing each term of this equation by we obtain

The first term in this expression depends only on the distance r while the second term depends only on the angle θ. This equation can only be true for all r and θ if

and

Consider a solution for R of the following form:

where A and k are arbitrary constants. Substituting this expression in the differential equation for R(r) we obtain

Therefore, the constant k must satisfy the following relation:

This equation gives us the following expression for k

The general solution for is thus given by

where A and B are arbitrary constants.
The angle dependent part of the solution of Laplace's equation must satisfy the following equation

The solutions of this equation are known as the Legendre polynomial . The Legendre polynomials have the following properties:

1. if m is even:

2. if m is odd:

3. for all m

4. or

Combining the solutions for and we obtain the most general solution of Laplace's equation in a spherical symmetric system with azimuthal symmetry:

Example: Problem 3.18
The potential at the surface of a sphere is given by

where k is some constant. Find the potential inside and outside the sphere, as well as the surface charge density on the sphere. (Assume that there is no charge inside or outside of the sphere.)

The most general solution of Laplace's equation in spherical coordinates is

First consider the region inside the sphere (r < R). In this region since otherwise would blow up at r = 0. Thus

The potential at r = R is therefore equal to

Using trigonometric relations we can rewrite as

Substituting this expression in the equation for we obtain

This equation immediately shows that unless . If then

The electrostatic potential inside the sphere is therefore equal to

Now consider the region outsider the sphere (r > R). In this region since otherwise would blow up at infinity. The solution of Laplace's equation in this region is therefore equal to

The potential at r = R is therefore equal to

The equation immediately shows that except when . If then

The electrostatic potential outside the sphere is thus equal to

The charge density on the sphere can be obtained using the boundary conditions for the electric field at a boundary:

Since this boundary condition can be rewritten as

The first term on the left-hand side of this equation can be calculated using the electrostatic potential just obtained:

In the same manner we obtain

Therefore,

The charge density on the sphere is thus equal to

Example: Problem 3.19
Suppose the potential at the surface of a sphere is specified, and there is no charge inside or outside the sphere. Show that the charge density on the sphere is given by

where

Most of the solution of this problem is very similar to the solution of Problem 3.18. First consider the electrostatic potential inside the sphere. The electrostatic potential in this region is given by

and the boundary condition is

The coefficients can be determined by multiplying both sides of this equation by and integrating with respect to θ between θ = 0 and θ = π:

Thus

In the region outside the sphere the electrostatic potential is given by

and the boundary condition is

The coefficients are given by

The charge density on the surface of the sphere is equal to

Differentiating with respect to r in the region r > R we obtain

Differentiating with respect to r in the region r < R we obtain

The charge density is therefore equal to

Substituting the expressions for and into this equation we obtain

where

Example: Problem 3.23
Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming there is no dependence on z (cylindrical symmetry). Make sure that you find all solutions to the radial equation. Does your result accommodate the case of an infinite line charge?

For a system with cylindrical symmetry the electrostatic potential does not depend on z. This immediately implies that . Under this assumption Laplace's equation reads

Consider as a possible solution of V:

Substituting this solution into Laplace's equation we obtain

Multiplying each term in this equation by r² and dividing by we obtain

The first term in this equation depends only on r while the second term in this equation depends only on φ. This equation can therefore be only valid for every r and every φ if each term is equal to a constant. Thus we require that

and

First consider the case in which . The differential equation for can be rewritten as

The most general solution of this differential solution is

However, in cylindrical coordinates we require that any solution for a given φ is equal to the solution for φ + 2π. Obviously this condition is not satisfied for this solution, and we conclude that . The differential equation for can be rewritten as

The most general solution of this differential solution is

The condition that requires that m is an integer. Now consider the radial function . We will first consider the case in which . Consider the following solution for :

Substituting this solution into the previous differential equation we obtain

Therefore, the constant k can take on the following two values:

The most general solution for under the assumption that is therefore

Now consider the solutions for when . In this case we require that

or

This equation can be rewritten as

If then the solution of this differential equation is

If then the solution of this differential equation is

Combining the solutions obtained for with the solutions obtained for we conclude that the most general solution for is given by

Therefore, the most general solution of Laplace's equation for a system with cylindrical symmetry is

Example: Problem 3.25
A charge density

is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside the cylinder.

The electrostatic potential can be obtained using the general solution of Laplace's equation for a system with cylindrical symmetry obtained in Problem 3.24. In the region inside the cylinder the coefficient must be equal to zero since otherwise would blow up at . For the same reason . Thus

In the region outside the cylinder the coefficients must be equal to zero since otherwise would blow up at infinity. For the same reason . Thus

Since must approach 0 when r approaches infinity, we must also require that is equal to 0. The charge density on the surface of the cylinder is equal to

Differentiating in the region r > R and setting r = R we obtain

Differentiating in the region r < R and setting r = R we obtain

The charge density on the surface of the cylinder is therefore equal to

Since the charge density is proportional to we can conclude immediately that for all and that for all except . Therefore

This requires that

A second relation between and can be obtained using the condition that the electrostatic potential is continuous at any boundary. This requires that

Thus

and

We now have two equations with two unknown, and , which can be solved with the following result:

and

The electrostatic potential inside the cylinder is thus equal to

The electrostatic potential outside the cylinder is thus equal to

Example: Problem 3.37
A conducting sphere of radius a, at potential , is surrounded by a thin concentric spherical shell of radius b, over which someone has glued a surface charge

where is a constant.
a) Find the electrostatic potential in each region:
i) r > b
ii) a < r < b
b) Find the induced surface charge on the conductor.
c) What is the total charge of the system? Check that your answer is consistent with the behavior of V at large r.

a) The system has spherical symmetry and we can therefore use the most general solution of Laplace's equation in spherical coordinates:

In the region inside the sphere since otherwise would blow up at r = 0. Therefore

The boundary condition for is that it is equal to at r = a:

This immediately shows that for all except :

The electrostatic potential inside the sphere is thus given by

which should not come as a surprise.
In the region outside the shell since otherwise would blow up at infinity. Thus

In the region between the sphere and the shell the most general solution for is given by

The boundary condition for at r = a is

This equation can only be satisfied if

The requirement that the electrostatic potential is continuous at r = b requires that

or

This condition can be rewritten as

The other boundary condition for the electrostatic potential at r = b is that it must produce the charge distribution given in the problem. This requires that

This condition is satisfied if

Substituting the relation between the various coefficients obtained by applying the continuity condition we obtain

These equations show that

Using these values for we can show that

The boundary condition for V at r = a shows that

These values for immediately fix the values for :

The potential in the region outside the shell is therefore equal to

The potential in the region between the sphere and the shell is equal to

b) The charge density on the surface of the sphere can be found by calculating the slope of the electrostatic potential at this surface:

c) The total charge on the sphere is equal to

The total charge on the shell is equal to zero. Therefore the total charge of the system is equal to

The electrostatic potential at large distances will therefore be approximately equal to

This is equal to limit of the exact electrostatic potential when .

Consider a given charge distribution ρ. The potential at a point P (see Figure 3.11) is equal to

where d is the distance between P and a infinitesimal segment of the charge distribution. Figure 3.11 shows that d can be written as a function of r, r' and θ:

This equation can be rewritten as

At large distances from the charge distribution and consequently . Using the following expansion for :

we can rewrite 1/d as

Using this expansion of 1/d we can rewrite the electrostatic potential at P as

This expression is valid for all r (not only ). However, if then the potential at P will be dominated by the first non-zero term in this expansion. This expansion is known as the multipole expansion. In the limit of only the first terms in the expansion need to be considered:

The first term in this expression, proportional to 1/r, is called the monopole term. The second term in this expression, proportional to 1/r², is called the dipole term. The third term in this expression, proportional to 1/r³, is called the quadrupole term.

If the total charge of the system is non zero then the electrostatic potential at large distances is dominated by the monopole term:

where Q is the total charge of the charge distribution.
The electric field associated with the monopole term can be obtained by calculating the gradient of :

If the total charge of the charge distribution is equal to zero (Q = 0) then the monopole term in the multipole expansion will be equal to zero. In this case the dipole term will dominate the electrostatic potential at large distances

Since θ is the angle between and we can rewrite as

The electrostatic potential at P can therefore be rewritten as

In this expression is the dipole moment of the charge distribution which is defined as

The electric field associated with the dipole term can be obtained by calculating the gradient of :

Example
Consider a system of two point charges shown in Figure 3.12. The total charge of this system is zero, and therefore the monopole term is equal to zero. The dipole moment of this system is equal to

where is the vector pointing from -q to +q.

The dipole moment of a charge distribution depends on the origin of the coordinate system chosen. Consider a coordinate system S and a charge distribution ρ. The dipole moment of this charge distribution is equal to

A second coordinate system S' is displaced by with respect to S:

The dipole moment of the charge distribution in S' is equal to

This equation shows that if the total charge of the system is zero (Q = 0) then the dipole moment of the charge distribution is independent of the choice of the origin of the coordinate system.

Example: Problem 3.40
A thin insulating rod, running from z = -a to z = +a, carries the following line charges:
a)
b)
c)
In each case, find the leading term in the multipole expansion of the potential.

a) The total charge on the rod is equal to

Since , the monopole term will dominate the electrostatic potential at large distances. Thus

b) The total charge on the rod is equal to zero. Therefore, the electrostatic potential at large distances will be dominated by the dipole term (if non-zero). The dipole moment of the rod is equal to

Since the dipole moment of the rod is not equal to zero, the dipole term will dominate the electrostatic potential at large distances. Therefore

c) For this charge distribution the total charge is equal to zero and the dipole moment is equal to zero. The electrostatic potential of this charge distribution is dominated by the quadrupole term.

The electrostatic potential at large distance from the rod will be equal to

Example: Problem 3.27
Four particles (one of charge q, one of charge 3q, and two of charge -2q) are placed as shown in Figure 3.12, each a distance d from the origin. Find a simple approximate formula for the electrostatic potential, valid at a point P far from the origin.

The total charge of the system is equal to zero and therefore the monopole term in the multipole expansion is equal to zero. The dipole moment of this charge distribution is equal to

The Cartesian coordinates of P are

The scalar product between and is therefore

The electrostatic potential at P is therefore equal to

Example: Problem 3.38
A charge Q is distributed uniformly along the z axis from z = -a to z =a. Show that the electric potential at a point is given by

for r > a.

The charge density along this segment of the z axis is equal to

Therefore, the nth moment of the charge distribution is equal to

This equation immediately shows that

The electrostatic potential at P is therefore equal to