Ch3. 特殊技巧 (Special Techniques). 3.1 拉普拉斯方程式 (Laplace’s Equation) 3.2 映像方法 (The Method of Images) 3.3 變數分離 (Separation of Variables) 3.4 多極展開 (Multipole Expansion). 3.1.1.1 拉普拉斯方程式 (Laplace’s Equation).
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3.1 拉普拉斯方程式(Laplace’s Equation)
3.2 映像方法(The Method of Images)
3.3 變數分離(Separation of Variables)
3.4 多極展開(Multipole Expansion)
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 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 the Poisson's equation :
Very often we only want to determine the potential in a region where = 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.
In one dimension the electrostatic potential V depends on only one variable x. The electrostatic potential V(x) is a solution of the onedimensional Laplace equation :
The general solution of this equation is V(x) = mx + b, where m and b are arbitrary constants. These constants are fixed when the value of the potential
is specified at two different positions.
Property 1:
It is a consequence of the Mean Value Theorem.
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 onedimensional Laplace equation does. However, the properties of solutions of the onedimensional Laplace equation are also valid for solutions of the twodimensional Laplace equation:
Property 1:
The value of V at a point (x, y) is equal to the average value of V around this point. If you draw a circle of any radius R about the point (x,y), the average value of V on the circle is equal to the value at the center:
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 onedimensional Laplace equation does. However, the properties of solutions of the onedimensional Laplace equation are also valid for solutions of the threedimensional Laplace equation:
Property 1:
The value of V at a point is equal to the average value of V over a spherical surface of radius R at .
Property 2:
V has no local maxima or minima; all extremes occur at the boundaries.
Proof of property 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 .
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 VPacross 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.
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
Therefore, Laplace's equation can be rewritten as
The solution V of this secondorder differential equation must satisfy the following firstorder differential equation:
This differential equation can be rewritten as
The general solution of this firstorder 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
Therefore, Laplace's equation can be rewritten as
The solution V of this secondorder differential equation must satisfy the following firstorder differential equation:
This differential equation can be rewritten as
The general solution of this firstorder differential equation is
where b is a constant. The constants a and b are determined by the boundary conditions.
Consider a volume within which the charge density is equal to zero. Suppose that the value of the electrostatic potential is specified at every pointon the surface of
this volume.
The first uniqueness theorem states that in this case the solution of Laplace's equation is uniquely defined.
Proof
We will consider what happens when there are two solutions V1 and V2 of Laplace's equation in the volume. Since V1 and V2 are solutions of Laplace's equation,we know that
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.
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 many other electrostatic problems we do not know the potential at the boundaries of the system.
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.
Proof
Suppose that there are two fields and that are solutions of Poisson's equation in the region between the conductors. Thus
where 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 qi, are equal to qi/0 .
The difference between and , , satisfies the following equations:
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, V3 =V1 V2 will also be constant on the surface of each conductor. The surface integral of V3 over the surface of conductor i can be written as
Since the surface integral of V3 over the surface of conductor i is equal to zero, the surface integral of V3 over all conductor surfaces will also be equal to zero.
The surface integral of V3 over the outer surface will also be equal to zero since V3 = 0 on this surface. Thus
Invoking product rule number (5) :
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 = 0 everywhere. This implies immediately that E1 = E2 everywhere, and proves the second uniqueness theorem.
Consider a point charge q held as a distance d above an infinite grounded conducting plane.
The electrostatic potential of this system must satisfy the following two boundary conditions:
when
1.
2.
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.
Consider a second system, consisting of two point charges with charges +q and q, located at z = d and z = d, respectively.
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 satisfies the same boundary conditions.
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 last page. It is equal to
Since this solution satisfies the boundary conditions, it must be the correct solution in the region z > 0. 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 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
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 :
where r 2 = x2 + y2. Substituting the expression for in the integral we obtain
As a result of the induced surface charge on the conductor, the point charge q will beattracted towards the conductor.
Since the electrostatic potential generated by the charge image charge system is the same as the chargeconductor 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 imagecharge system and the real system. This difference is the total electrostatic energy of the system.
The electric field in the imagecharge 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 E2, the electrostatic energy of the real system will be 1/2 of the electrostatic energy of the imagecharge system (only 1/2 of the total volume has a nonzero electric field in the real system). The electrostatic energy of the imagecharge 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 3.2
Solution :
Imagecharge system.
Consider a system consisting of two charges q and q', located on the z axis at z = s and z = z', respectively. 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).
The electrostatic potential at P is equal to
Theequation above can be rewritten as
The electrostatic potential at Q is equal to
Theequation above can be rewritten as
Combining the two expression for q' we obtain
Theequation above 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 we can express d and d' in terms of R, s, and :
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 chargesphere system.
Consider an arbitrary point (r ,). 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 using the cosine rule:
The electrostatic potential at (r ,) will therefore be equal to
Answer of (a)
(b)
The surface charge density 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.
The equation above 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 over the surface of the sphere.
Answer of (b)
(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
Answer of (c)
Example 3.3 :
Two infinite, grounded, metal plates lie parallel to the xz plane, one at y = 0, the other at y = a (see Figure below). The left end, at x = 0, is closed off with aninfinite stripinsulated from the two plates and maintained at a specified potential V0(y). Find the potential inside this "slot".
a
V0(y)
The electrostatic potential in the slot must satisfy the threedimensional Laplace equation. However, since V does not have a z dependence, the threedimensional Laplace equation reduces to the twodimensional 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 = a) = 0 (grounded top plate).
3. V(x = 0, y) = V0(y) (plate at x = 0).
4. When .
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.
Consider solutions of the following form:
If this is a solution of the twodimensional Laplace equation then we must require that
This equation can be rewritten as
The first term of the lefthand side of this equation depends only on x while the second term depends only on y.
If the above equation must hold for all x and y in the slot we must require that
The differential equation for X can be rewritten as
If C1 is a negative number then this equation can be rewritten as
where
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 C1can notbe a negative number. If C1 is a positive number then the differential equation for X can be written as
where
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 the above equation is
Therefore, the general solution for the electrostatic potential V(x,y) is equal
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 = a) = 0:
which requires that sin ka = 0. This condition limits the possible values of k :
where n = 1, 2, 3, …
Note: negative values of k are not allowed since exp(kx) approaches zero at infinity only if k >0.
However, since n 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
To satisfy boundary condition # 3 we must require that
Multiplying both sides by sin(n’y/a) and integrating each side between y = 0 and y = a we obtain
The integral on the lefthand side of this equation is equal to zero for all values of n except n = n’. Thus
where
The coefficients Dn’can thus be calculated easily:
The coefficients Dnare called the Fourier coefficientsof V0(y)
The solution of Laplace's equation in the slot is therefore equal to
where
Problem 3.12 :
Find the potential in the infinite slot of Example 3.3 if the boundary at x = 0 consists to two metal stripes: one, from y = 0 to y = a/2, is held at constant potential V0 , and the other, from y = a/2 to y = a is at potential V0 .
The boundary condition at x = 0 is
The Fourier coefficients of the function V0(y) are equal to
The values for the first four G coefficients are
It is easy to see that Gn + 4 = Gnand therefore we conclude that
The Fourier coefficients Dnare 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 ( ∂V / ∂ = 0) Laplace's equation reads
Consider the possibility that the general solution of this equation is the product of a function R(r), which depends only on the distance r, and a function ( ), which depends only on the angle :
Substituting this "solution" into Laplace's equation we obtain
The first term in this expression depends only on the distance r while the second term depends only on the angle . This equation can only be true for all r and if
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
This equation gives us the following expression for k
The general solution for R(r) 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 polynomialP (cos).
P (cos)is most conveniently defined by the Rodrigues formula :
Combining the solutions for R(r) and ( ) we obtain the most general solution of Laplace's equation in a spherical symmetric system with azimuthal symmetry:
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 regionsince otherwise V(r,) would blow up at r = 0. Thus
The potential at r = R is therefore equal to
Using trigonometric relations we can rewrite
Substituting the above expression in the equation for V(R,) we obtain
This equation immediately shows that A = 0 unless = 1 or = 3.
If = 1 or = 3,
The electrostatic potential inside the sphere is therefore equal to
Now consider the region outside the sphere (r > R). In this region since otherwise V(r,) 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
This equation immediately shows that B = 0 unless = 1 or = 3.
If = 1 or = 3,
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 lefthand 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 V0() 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
Solution:
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 Pn(cos)sin and integrating with respect to between = 0 and = :
Remember that
Thus
In the region outside the sphere the electrostatic potential is given by
and the boundary condition is
The coefficients can be determined by multiplying both sides of this equation by Pn(cos)sin and integrating with respect to between = 0 and = :
Thus
The charge density () on the surface of the sphere is equal to
Differentiating V(r,) with respect to r in the region r > R we obtain
Differentiating V(r,) with respect to r in the region r < R 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.
For a system with cylindrical symmetry the electrostatic potential does not depend on z. This immediately implies that ∂V / ∂z = 0. 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 R(r)() we obtain
The first term in this equation depends only on r while the second term in this equation depends only on . This equation can therefore be only valid for every r and every if each term is equal to a constant. Thus we require that
First consider the case in which = m2 < 0. 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 is equal to the solution for +2. Obviously this condition is not satisfied for this solution, and we conclude that = m2 ≥ 0.
The differential equation for () can be rewritten as
The most general solution of this differential solution is
The condition that m() = m(+2) requires that m is an integer.
Now consider the radial function R(r). We will first consider the case in which = m2 > 0. Consider the following solution for R(r):
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 R(r) under the assumption that m2 > 0 is therefore
Now consider the solutions for R(r) when m2 = 0. In this case we require that
If a0 = 0 then the solution of this differential equation is
If a0 0 then the solution of this differential equation is
Combining the solutions obtained for m2 = 0 with the solutions obtained for m2 > 0 we conclude that the most general solution for R(r) 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 () = a sin(5) (where a is a constant) is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside the cylinder.
In the region inside the cylinder the coefficient Bmmust be equal to zero since otherwise V(r,) would blow up at r = 0. For the same reason a0 = 0. Thus
In the region outside the cylinder the coefficients Ammust be equal to zero since otherwise V(r, ) would blow up at infinity. For the same reason a0 = 0. Thus
Since V(r,) must approach 0 when r approaches infinity, we must also require that bout, 0 is equal to 0.
The charge density on the surface of the cylinder is equal to
Differentiating V(r,) in the region r > R and setting r = R we obtain
Differentiating V(r,) 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 sin(5) we can conclude immediately that Cin,m=Cout,m= 0 for all m and that Din,m= Dout ,m= 0 for all m except m = 5.
Therefore
This requires that
A second relation between Din, 5 and Dout, 5 can be obtained using the condition that the electrostatic potential is continuous at any boundary. This requires that
Thus
We now have two equations with two unknown, Din, 5 and Dout, 5 , which can be solved with the following result:
The electrostatic potential inside the cylinder is thus equal to
The electrostatic potential outside the cylinder is thus equal to
Consider a given charge distribution . The potential at a point P (see Figure below) is equal to
where d is the distance between P and a infinitesimal segment of the charge distribution. d can be written as a function of r, r' and :
so
At large distances from the charge distribution r >> r’ and consequently r’/r <<1. Using the following expansion for :
we can rewrite 1/d as
Using the expansion of 1/d we can rewrite the electrostatic potential at P as
This expression is valid for all r (not only r >> r’). However, if r >> r' then the potential at P will be dominated by the first nonzero term in this expansion. This expansion is known as the multipole expansion.
In the limit of r >> r’ 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 V(P):
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 is the angle between r and r ' we can rewrite r’cos as
The electrostatic potential at P can therefore be rewritten as
In this expression p 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 V(P):
Consider a system of two point charges shown in Figure below. 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
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 . 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:
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 Qtot 0, 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 nonzero). 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 below, 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.
d
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