y.m. hu, associate professor, department of applied physics, national university of kaohsiung ch3....

Y.M. Hu, Associate Professor, Department of Applied Physics, National University

of Kaohsiung

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)


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3.1.1.1 拉普拉斯方程式 (Laplace’s Equation)

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

)'r(

'dr)r(

)'r(

4

1)r(E 2

0

where r'rr

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

'dr

)'r(

4

1)r(V

0

since the integrand of the integral is a scalar.

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

E

VE


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3.1.1.2 拉普拉斯方程式 (Laplace’s Equation)

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).

)r(

A better approach to determine the electrostatic potential is to start with the Poisson's equation :

0

2VV

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 :

0V2

There are an infinite number of functions that satisfy Laplace's equation and the appropriate solution is selected by specifying the appropriate boundary conditions.


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3.1.2.1 一維的拉普拉斯方程式 (Laplace’s Equation in One Dimension)

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 :

0dx

Vd2

2

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 potentialis specified at two different positions.

It is a consequence of the Mean Value Theorem.

Property 1:


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3.1.2.2 一維的拉普拉斯方程式 (Laplace’s Equation in One Dimension)

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.


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3.1.3 二維的拉普拉斯方程式 (Laplace’s Equation in Two Dimension)In two dimensions the electrostatic potential depends on two variables x and y. Laplace's equation now becomes

0y

V

x

V2

2

2

2

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. 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:

circle

VdlR2

1)y,x(V

Property 2:

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


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3.1.4.1 三維的拉普拉斯方程式 (Laplace’s Equation in Three Dimension)In three dimensions the electrostatic potential depends on three variables x , y , and z. Laplace's equation now becomes

0z

V

y

V

x

V2

2

2

2

2

2

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 is equal to the average value of V over a spherical surface of radius R at .

sphere

2 VdaR4

1)r(V

Property 2:

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

r

r


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3.1.4.2 三維的拉普拉斯方程式 (Laplace’s Equation in Three Dimension)

Proof of property 1. The potential at P, generated by charge q, is equal to

d

q

4

1V

0p

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

cosrR2Rrd 222

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

cosrR2Rr

q

4

1V

220

p

The average potential on the surface of the sphere can be obtained by integrating VP across the surface of the sphere.


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The average potential is equal to

sphere22

0

sphere

2p2average

dsin2cosrR2Rr

q

4

1

4

1

ddsinRVR4

1V

r4

q)

rR

Rr

rR

Rr(

8

q

rR

cosrR2Rr

8

q

000

22

0

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.


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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

0]V

sin

1)

V(sin)

r

Vr(

r[sin

sinr

12

22

2

If V is only a function of r, then

0V

0V

Therefore, Laplace's equation can be rewritten as

0)dr

dVr(

dr

d

r

1 22


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

constantadr

dVr2

This differential equation can be rewritten as2r

a

dr

dV

The general solution of this first-order differential equation is br

a)r(V

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

r

a)r(V


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Laplace's equation in cylindrical coordinates is

0z

VV

r

1)

r

Vr(

rr

12

2

2

2

2

If V is only a function of r, then

0V

0z

V

Therefore, Laplace's equation can be rewritten as 0)dr

dVr(

dr

d

r

1

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

constantadr

dVr


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This differential equation can be rewritten as

r

a

dr

dV

The general solution of this first-order differential equation is

b)rln(a)r(V

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


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3.1.5.1 邊界條件與唯一定理 (Boundary Conditions and Uniqueness Theorems)

Consider a volume 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 ofthis 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

0VV 22

12

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.


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Now consider a third function V3, which is the difference between V1 and V2 :

213 VVV The function V3 is also a solution of Laplace's equation. This can be demonstrated easily: 0VVV 2

21

23

2 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 mustbe equal to zero everywhere. This immediately implies that

21 VV 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.


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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 theconductors is known.

Suppose that there are two fields and that are solutions of Poisson's equation in the region between the conductors. Thus

Proof

1E

2E

01E

02E

where is the charge density at the point where the electric field is evaluated.


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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 .

1E

2E

0

i

iconductorsurface

1

qadE

0

i

iconductorsurface

2

qadE

The difference between and , , satisfies the following equations:

1E

2E

213 EEE

0EEE00

213

0qq

adEadEadE0

i

0

i

iconductorsurface

2

iconductorsurface

1

iconductorsurface

3


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Since the potential on the surface of any conductor is constant, theelectrostatic 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

1E

2E

3E

0adEVadEV

iconductorsurface

33

iconductorsurface

33

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

3E

3E

3E

0adEVsurfaceAll

33


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Invoking product rule number (5) :

)f(A)A(f)Af(

23333333 )E()V(E)E(V)EV(

V

2333

VsurfaceAll

33 d)E(d)EV(adEV0

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.

23E 2

3E23E


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3.2.1.1 典型的映像問題 (The Classic Image Problem)

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:

0)0,y,x(V 0)z,y,x(V when

z

y

x

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.


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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

0]dyx

q

dyx

q[

4

1)0,y,x(V

2222220

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.


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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

])dz(yx

q

)dz(yx

q[

4

1)z,y,x(V

2222220

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.


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3.2.2.1 感應的表面電荷 (Induced Surface Charge)

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

E

nE0

outside

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

n

0z0z0 z

VE


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3.2.2.2 感應的表面電荷 (Induced Surface Charge)

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 :

2

0 0Surface

total rdrd)r(daQ

where r 2 = x2 + y2. Substituting the expression for in the integral we obtain

q]d

10[qd

dr

qdrdr

)dr(

1qdQ

022

02/322total

Substituting the solution for V in this equation we find

2/32220z

2/32222/3222 )dyx(

d

2

q}

])dz(yx[

)dz(

])dz(yx[

)dz({

4

q


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3.2.3.1 力與能量 (Force and Energy)

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

n)d2(

q

4

1F 2

2

0


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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 E2, 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

d2

q

4

1W

2

0image

The electrostatic energy of the real system is therefore equal to

d4

q

4

1W

2

1W

2

0imagereal


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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

d4

q

4

1

z4

q

4

1dz

z4

q

4

1W

2

0

d2

0

d

2

2

0real

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


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3.2.4.1 其他的映像問題 (Other Image Problem)

Example 3.2

A point charge q is situated a distance s from the center of a grounded conducting sphere of radius R.

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.


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Solution : Image-charge 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

'zR

'q

4

1

Rs

q

4

10V

00p


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)'zR(Rs

q'q

The equation above can be rewritten as

The electrostatic potential at Q is equal to

'zR

'q

4

1

Rs

q

4

10V

00Q


)'zR(Rs

q'q

Combining the two expression for q' we obtain

)'zR(Rs

q)'zR(

Rs

q


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)'zR)(Rs()'zR)(Rs(


)Rs(R)Rs(R)Rs('z)Rs('z

2R2'sz2

The position of the image charge is equal to

s

R'z

2

The value of the image charge is equal to

s

Rq)

s

RR(

Rs

q)'zR(

Rs

q'q

2


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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 :

cosRs2sRd 22

coss

RR2

s

RRcos'Rz2'zR'd

2

2

4222

The electrostatic potential at P' is equal to

0]cosRs2sR

q

cosRs2sR

q[

4

1

]

coss

RR2

sR

R

sR

q

cosRs2sR

q[

4

1]'d

'q

d

q[

4

1V

22220

2

2

42

2200

'p


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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.

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:

cosrs2srd 22

coss

Rr2

s

Rrcos'rz2'zr'd

2

2

4222


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The electrostatic potential at (r ,) will therefore be equal to

]

cosrs2R)Rrs

(

q

cosrs2sr

q[

4

1

]

coss

Rr2

sR

r

sR

q

cosrs2sr

q[

4

1]'d

'q

d

q[

4

1),r(V

2222

0

2

2

42

2200

Answer of (a)

(b)

The surface charge density on the sphere can be obtained from the boundary conditions of E

rnEEE00

outsideinsideoutside

where we have used the fact that the electric field inside the sphere is zero.


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r

VE 0r0

Substituting the general expression for V into this equation we obtain

Rr

2/322

2

2

2/322 }]cosrs2R)

Rrs

[(

cossRrs

)cosrs2sr(

cossr{

4

q

}]cosRs2Rs[

cossRs

)cosRs2sR(

cossR{

4

q2/322

2

2/322

2/322

22

)cosRs2sR(

Rs

R4

q


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

0

2/322222

)cosRs2sR(

dsin)Rs(R

2

qddsinRQ

s

Rq]

sR

1

sR

1[

s

Rs

2

q]

cosRs2sRsR

1

[)Rs(R2

q 22

0

22

22

(c)

Answer of (b)

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).


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This force is equal to

k)Rs(

sRq

4

1k

)s

Rs(

)qsR

(q

4

1k

)'zs(

'qq

4

1F 222

2

022

02

0'qq

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'qqF

22

2

0

s

22

2

0

s

222

2

0

s

'qq Rs

Rq

8

1

)Rz(2

Rq

4

1dz

)Rz(

zRq

4

1ldFW

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'.

Answer of (c)


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3.3.1.1 分離變數法 :直角座標 (Separation of Variables : Cartesian Coordinates)

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 an infinite strip insulated from the two plates and maintained at a specified potential V0(y). Find the potential inside this "slot".

aV0(y)


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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:

0y

V

x

V2

2

2

2

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 .0V 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.


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Consider solutions of the following form:

)y(Y)x(X)y,x(V If this is a solution of the two-dimensional Laplace equation then we must require that

0dy

)y(Yd)x(X

dx

)x(Xd)y(Y)]y(Y)x(X[

y)]y(Y)x(X[

x 2

2

2

2

2

2

2

2

This equation can be rewritten as

0dy

)y(Yd

)y(Y

1

dx

)x(Xd

)x(X

12

2

2

2

The first term of the left-hand side of this equation depends only on x while the second term depends only on y.


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If the above equation must hold for all x and y in the slot we must require that

constant 12

2

Cdx

)x(Xd

)x(X

112

2

Cdy

)y(Yd

)y(Y

1

The differential equation for X can be rewritten as

0)x(XCdx

)x(Xd12

2

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

0)x(Xkdx

)x(Xd 22

2

where 21 kC

The most general solution of this equation is

)kxsin(B)kxcos(A)x(X However, this function is an oscillatory function and does not satisfy boundary condition # 4, which requires that V approaches zero when x approaches infinity.


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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

0)x(Xkdx

)x(Xd 22

2

where 21 kC

The most general solution of this equation is

kxkx BeAe)x(X

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

kxBe)x(X

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

0)y(Ykdy

)y(Yd 22

2


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The most general solution of the above equation is

)kysin(D)kycos(C)y(Y

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

)]kysin(D)kycos(C[e)y(Y)x(X)y,x(V kx

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:

0Ce)]0sin(D)0cos(C[e0)0,x(V kxkx

which requires that C = 0.


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The second boundary condition requires that V(x, y = a) = 0:

)kasin(De0)a,x(V kx

which requires that sin ka = 0. This condition limits the possible values of k :

a

nk

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

1n

xa

n

n )ya

nsin(eD)y,x(V


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To satisfy boundary condition # 3 we must require that

1n

n0 )ya

nsin(D)y(V)y,0x(V

Multiplying both sides by sin(n’y/a) and integrating each side between y = 0 and y = a we obtain

a

0

01n

a

0

n dy)y(V)ya

'nsin(dy)y

a

nsin()y

a

'nsin(D

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

'n1n

'nnn1n

a

0

n D2

a

2

aDdy)y

a

nsin()y

a

'nsin(D

where'nn

a

0 2

ady)y

a

nsin()y

a

'nsin(


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a

0

0'n dy)y(V)ya

'nsin(

a

2D

The coefficients Dn’ can thus be calculated easily:

a

0

0n dy)y(V)ya

nsin(

a

2D

The coefficients Dn are called the Fourier coefficients of V0(y)

The solution of Laplace's equation in the slot is therefore equal to

1n

xa

n

n )ya

nsin(eD)y,x(V

where a

0

0n dy)y(V)ya

nsin(

a

2D


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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

ay2

aforV

2

ay0forV

)y,0(V

0

0

The Fourier coefficients of the function V0(y) are equal to

a

2/a

0

2/a

0

0

a

0

0n dy)ya

nsin(V

a

2dy)y

a

nsin(V

a

2dy)y(V)y

a

nsin(

a

2D


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n00

n Gn

V2)]n

2

1cos(2ncos1[

n

V2D

The values for the first four G coefficients are

0G1 4G2 0G3 0G4

It is easy to see that Gn + 4 = Gn and therefore we conclude that

The Fourier coefficients Dn are thus equal to

otherwise0

,...10,6,2nforn

V8

D

0

n

The electrostatic potential is thus equal to

,..10,6,2n

xa

n0 )y

a

nsin(e

n

1V8)y,x(V


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3.3.2.1 分離變數法 :球狀座標 (Separation of Variables : Spherical Coordinates)

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


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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

constant )1(]dr

)r(dRr[

dr

d

)r(R

1 2

constant )1()d

d(sin

d

d

sin

1

Consider a solution for R of the following form:

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


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constant )1()1k(kr)1k(r

k)

dr

dArr(

dr

d

Ar

1 kk

k2

k

This equation gives us the following expression for k

)1(2

)21

(21

2

)1(411k

The general solution for R(r) is thus given by

1r

BAr)r(R

where A and B are arbitrary constants.


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The angle dependent part of the solution of Laplace's equation must satisfy the following equation

0

sin)1()d

d(sin

d

d

The solutions of this equation are known as the Legendre polynomial P (cos).

P (cos) is most conveniently defined by the Rodrigues formula :

)1x()dx

d(

!2

1)x(P 2

1)x(P0 x)x(P1 2/)1x3()x(P 22 2/)x3x5()x(P 3

3


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The Legendre polynomials have the following properties:

1. If = even,

2. If = odd,

3. for all

4. or .

)cos(P)(cosP

)cos(P)(cosP

1)1(P

'

1

1

' 12

2dx)x(P)x(P

'

0

' 12

2dsin)(cosP)(cosP

)(cosPcos 1

)(cosP3

1)(cosP

3

2)(cos 022

)(cosP5

3)(cosP

5

2)(cos 133


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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:

)(cosP)r

BrA(),r(V

01

Problem 3.18

The potential at the surface of a sphere is given by

)3cos(k)(V0

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.)


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

)(cosP)r

BrA(),r(V

01

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

0B

)(cosPrA),r(V0

The potential at r = R is therefore equal to

)3cos(k)(cosPRA),R(V0

Using trigonometric relations we can rewrite

)(cosP5

3)(cosP

5

8cos3cos4)3cos( 13

3


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Substituting the above expression in the equation for V(R,) we obtain

)(cosP5

k3)(cosP

5

k8)(cosPRA),R(V 13

0

This equation immediately shows that A = 0 unless = 1 or = 3.

If = 1 or = 3,

R5

k3A1 33 R5

k8A

The electrostatic potential inside the sphere is therefore equal to

)(cosPR

r

5

k3)(cosP

R

r

5

k8),r(V 133

3


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3.3.2.9 分離變數法 :球狀座標 (Separation of Variables : Spherical Coordinates)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

0A

)(cosPr

B),r(V

01

The potential at r = R is therefore equal to

)(cosP5

k3)(cosP

5

k8)(cosP

R

B),R(V 13

01

This equation immediately shows that B = 0 unless = 1 or = 3.

If = 1 or = 3, 2

1 R5

k3B 4

3 R5

k8B

The electrostatic potential outside the sphere is thus equal to

)(cosPr

R

5

k3)(cosP

r

R

5

k8),r(V 12

2

34

4


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

r)(

EE0

RrRr

Since this boundary condition can be rewritten asVE

0RrRr

)(

r

V

r

V

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

)](cosP32)(cosP6[R5

k)](cosP

r

R

5

k6)(cosP

r

R

5

k32[

r

V31

Rr

13

2

35

4

Rr


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In the same manner we obtain

)](cosP24)(cosP3[R5

k)](cosP

R

1

5

k3)(cosP

R

r

5

k24[

r

V31

Rr

133

2

Rr

Therefore,

031

RrRr

)()](cosP56)(cosP9[

R5

k

r

V

r

V

The charge density on the sphere is thus equal to

)](cosP56)(cosP9[R5

k)( 31

0


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3.3.2.12 分離變數法 :球狀座標 (Separation of Variables : Spherical Coordinates)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

)(cosPC)12(R2

)(0

20

0

0 dsin)(cosP)(VC

Solution:

First consider the electrostatic potential inside the sphere. The electrostatic potential in this region is given by

)(cosPrA),r(V0


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and the boundary condition is

)(V)(cosPRA),R(V 00

The coefficients can be determined by multiplying both sides of this equation by Pn(cos)sin and integrating with respect to between = 0 and = :

A

Remember that '

0

' 12

2dsin)(cosP)(cosP

nn

0 0

n

0

n0 RA1n2

2dsin)(cosP)(cosPRAdsin)(cosP)(V

Thus

0

n0nn dsin)(cosP)(VR2

1n2A


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

)(cosPr

B),r(V

01

and the boundary condition is

)(V)(cosPR

B),R(V 0

01

The coefficients can be determined by multiplying both sides of this equation by Pn(cos)sin and integrating with respect to between = 0 and = :

B

1nn

0 0

n10

n0 R

B

1n2

2dsin)(cosP)(cosP

R

Bdsin)(cosP)(V

Thus

0

n01n

n dsin)(cosP)(VR2

1n2B


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

)r

V

r

V()(

RrRr0

Differentiating V(r,) with respect to r in the region r > R we obtain

02

Rr

)(cosPR

B)1(

r

V

Differentiating V(r,) with respect to r in the region r < R we obtain

0

1

Rr

)(cosPRAr

V

})(cosP]RAR

B)1[({

})](cosPRA[)](cosPR

B)1[({)

r

V

r

V()(

0

120

0

1

020

RrRr0


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0

01 dsin)(cosP)(VR

2

12B

0

0 dsin)(cosP)(VR2

12A

})](cosPC)12[({R2

})(cosPC]RR2

)12(

R2

R)12()1[({)(

0

20

0

12

1

0

where

0

0 dsin)(cosP)(VC


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3.3.3.1 分離變數法 :圓柱座標 (Separation of Variables : Cylindrical Coordinates)

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:


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Substituting this solution into Laplace's equation we obtain

0d

)(d

r

)r(R)

dr

)r(dRr(

dr

d

r

)(2

2

2

Multiplying each term in this equation by r2 and dividing by R(r)() we obtain

0d

)(d

)(

1)

dr

)r(dRr(

dr

d

)r(R

r2

2

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

constant)dr

)r(dRr(

dr

d

)r(R

r

2

2

d

)(d

)(

1


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First consider the case in which = -m2 < 0. The differential equation for () can be rewritten as

0)(md

)(d 22

2

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

0)(md

)(d 22

2


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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

221k1k

21k

1k

k

k mkArAr

k)rkAr(

dr

d

Ar

1)

dr

dArr(

dr

d

Ar

r

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

mkmk


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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

0)dr

)r(dRr(

dr

d

0adr

)r(dRr constant r

a

dr

)r(dR 0

If a0 = 0 then the solution of this differential equation is

If a0 0 then the solution of this differential equation is


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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


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3.3.3.7 分離變數法 :圓柱座標 (Separation of Variables : Cylindrical Coordinates)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 Bm must 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 Am must 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.


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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


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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


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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


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3.4.1.1 多極展開 (Multipole Expansion)

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

d

d4

1)P(V

volume0

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 :

]cos)r

'r(2)

r

'r(1[rcos'rr2'rrd 22222

so

cos)r'r

(2)r'r

(1

1

r

1

d

1

2


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At large distances from the charge distribution r >> r’ and consequently r’/r <<1. Using the following expansion for :x1/1

....x16

5x

8

3x

2

11

x1

1 32

we can rewrite 1/d as

0nn

n22

22

2

)(cosP)r

'r(

r

1....})

2

1cos

2

3()

r

'r(cos)

r

'r(1{

r

1

.....}]cos2)r

'r[()

r

'r(

8

3]cos2)

r

'r)[(

r

'r(

2

11{

r

1

cos)r'r

(2)r'r

(1

1

r

1

d

1


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

0n Volume

nn

1n0

d)(cosP'r)'r(r

1

4

1)P(V

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 non-zero 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:

...}d)2

1cos

2

3('r

r

1dcos'r

r

1d

r

1{

4

1)P(V

Volume

223

Volume2

Volume0

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.


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3.4.2.1 單極項 (The Monopole Term)

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

r

Q

4

1d

r

1

4

1)P(V

0Volume0

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) :

rr

Q

4

1)

r

1(

4

Q)P(V)P(E 2

00


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3.4.2.2 偶極項 (The Dipole Term)

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.

Volume

20

dcos'rr

1

4

1)P(V

Since is the angle between r and r ' we can rewrite r’cos as

'rrcos'r

The electrostatic potential at P can therefore be rewritten as

20Volume

20 r

rp

4

1d'r

r

r

4

1)P(V

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

Volume

d'rp


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

20Volume

20 r

rp

4

1d'r

r

r

4

1)P(V

30

r r

1

4

cosp2

r

)P(V)P(E

30 r

1

4

sinp)P(V

r

1)P(E

0)P(V

sinr

1)P(E

)ˆsinrcos2(r4

p),r(E 3

0dip


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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

sq)rr(qr)q(r)q(p

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

Volume

SS drp


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

drr S'S

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

Qdpdddrdrp S

VolumeVolume

S

Volume

'S'S

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.


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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.


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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 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


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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

-


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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

kqd2k)d)(q3(j)d)(q2(k)d)(q(j)d)(q2(rqpi

ii


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The Cartesian coordinates of P are

The scalar product between and is thereforep

r

cosqd2rp

The electrostatic potential at P is therefore equal to

y.m. hu, associate professor, department of applied physics, national university of kaohsiung ch3....

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