Theoretische Physik 2: Elektrodynamik Home assignment 4 · Home assignment 4 Problem4.1 Green’s...

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WiSe 2012 8.10.2012 Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Matthias Saba am Lehrstuhl für Theoretische Physik I Department für Physik Friedrich-Alexander-Universität Erlangen-Nürnberg Theoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4 Problem 4.1 Green’s reciprocation theorem Green’s reciprocation theorem states that if Φ is the potential due to a volume-charge density ρ within a volume V and a surface-charge density σ on the conducting surface S bounding the volume V , and Φ 0 is the potential due to another charge distribution ρ 0 and σ 0 , then Z V ρΦ 0 dV + Z S σΦ 0 dS = Z V ρ 0 ΦdV + Z S σ 0 ΦdS. a) Two infinite, grounded, parallel conducting planes are separated by a distance d. A point charge q is placed between the planes. Use the reciprocation theorem of Green to prove that the total induced charge on one of the planes is equal to (-q) times the fractional perpendicular distance of the point charge from the other plane. Hint: As your comparison electrostatic problem with the same surfaces choose one whose charge densities and potential are known and simple. b) Consider a potential problem in the half-space defined by z 0, with Dirichlet boundary conditions on the plane z =0 (and at infinity). (i) Write down the appropriate Green’s function G(~x,~x 0 ). (ii) Suppose that the potential on the plane z =0 is specified to be Φ= V inside a circle of radius a centered at the origin, and Φ=0 outside that circle. (1) Find an integral expression for the potential at the point P specified in terms of the cylindrical coordinates (ρ, φ, z ). (2) Show that, along the axis of the circle (ρ = 0), the potential is given by Φ= V 1 - z a 2 + z 2 (3) Show that at large distances (ρ 2 + z 2 a 2 ) the potential can be expanded in a power series in (ρ 2 + z 2 ) -1 , and that the leading terms are Φ= Va 2 2 z (ρ 2 + z 2 ) 3/2 1 - 3a 2 4(ρ 2 + z 2 ) + 5(3ρ 2 a 2 + a 4 ) 8(ρ 2 + z 2 ) 2 + ... . Verify that the results of parts (2) and (4) are consistent with each other in their common range of validity.

Transcript of Theoretische Physik 2: Elektrodynamik Home assignment 4 · Home assignment 4 Problem4.1 Green’s...

WiSe 2012 8.10.2012

Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl für Theoretische Physik IDepartment für PhysikFriedrich-Alexander-UniversitätErlangen-Nürnberg

Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

Home assignment 4

Problem4.1 Green’s reciprocation theorem

Green’s reciprocation theorem states that if Φ is the potential due to a volume-charge density ρ withina volume V and a surface-charge density σ on the conducting surface S bounding the volume V , andΦ′ is the potential due to another charge distribution ρ′ and σ′, then∫

VρΦ′ dV +

∫SσΦ′ dS =

∫Vρ′Φ dV +

∫Sσ′Φ dS .

a) Two infinite, grounded, parallel conducting planes are separated by a distance d. A point chargeq is placed between the planes. Use the reciprocation theorem of Green to prove that the totalinduced charge on one of the planes is equal to (−q) times the fractional perpendicular distanceof the point charge from the other plane. Hint: As your comparison electrostatic problem withthe same surfaces choose one whose charge densities and potential are known and simple.

b) Consider a potential problem in the half-space defined by z ≥ 0, with Dirichlet boundaryconditions on the plane z = 0 (and at infinity).(i) Write down the appropriate Green’s function G(~x, ~x′).(ii) Suppose that the potential on the plane z = 0 is specified to be Φ = V inside a circle

of radius a centered at the origin, and Φ = 0 outside that circle. (1) Find an integralexpression for the potential at the point P specified in terms of the cylindrical coordinates(ρ, φ, z). (2) Show that, along the axis of the circle (ρ = 0), the potential is given by

Φ = V

(1− z√

a2 + z2

)(3) Show that at large distances (ρ2 + z2 � a2) the potential can be expanded in a powerseries in (ρ2 + z2)−1, and that the leading terms are

Φ =V a2

2

z

(ρ2 + z2)3/2

[1− 3a2

4(ρ2 + z2)+

5(3ρ2a2 + a4)

8(ρ2 + z2)2+ ...

].

Verify that the results of parts (2) and (4) are consistent with each other in their commonrange of validity.

Problem4.2 Method of images

a) Using the method of images, discuss the problem of a point charge q inside a hollow, grounded,conducting sphere of inner radius a. Find(i) the potential inside the sphere;(ii) the induced surface-charge density;(iii) the magnitude and direction of the force acting on q. Is there any change in the solution

if the sphere is kept at a fixed potential V ? If the sphere has a total charge Q on its innerand outer surfaces?

b) A conducting sphere is placed in a uniform external field. Calculate the potential Φ and thesurface charge density σ.

Problem4.3 Hollow cube

A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and x = a, y = a,z = a. The walls z = 0 and z = a are held at a constant potential V . The other four sides are at zeropotential.

a) Find the potential Φ(x, y, z) at any point inside the cube.b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures.

How many terms in the series is it necessary to keep in order to attain this accuracy?c) Find the surface-charge density on the surface z = a.

Due date: Tuesday, 15.11.12

WiSe 2012 8.10.2012

Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl für Theoretische Physik IDepartment für PhysikFriedrich-Alexander-UniversitätErlangen-Nürnberg

Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

Solutions to Home assignment 4

Solution of Problem4.1 Green’s reciprocation theorem

a)

Fig. 1: A charge q between conducting planes.

Let us place the two conducting planes perpendicu-larly to the z-axis, one through the origin, and theother through the point z = d, while the charge q is atthe distance a from the left plane (see Figure 1). Theplanes are grounded

Φ∣∣∣z=0

= 0

Φ∣∣∣z=d

= 0 ,

while the point charge density is

ρ(~x) = qδ(x)δ(y)δ(z − a) .

On the planes, the surface charge densities σ1, σ2 arebeing induced.Consider the following problem: the charges q and −q are uniformly distributed on the two infinitelylong, conducting planes. This configuration describes a parallel plate capacitor, for which we know thepotential at a general point to be (Figure 1)

Φ′ = −Vd

(z − d) (whereV is the potential difference between the plates)

Φ′∣∣∣z=0

= V

Φ′∣∣∣z=d

= 0 ,

We now use Green’s reciprocation theorem. Here, the unprimed variables represent the quantities inthe case of there being a charge between the two planes, and the primed variables represent those inthe case of the parallel plate capacitor configuration, without a charge present in between. We get∫

VρΦ′dV +

∮SσΦ′dS =

∫Vρ′ΦdV +

∮Sσ′ΦdS

Now we calculate the different terms in the above equation.∫VρΦ′dV = −

∫Vqδ(x)δ(y)δ(z − a)

V

d(z − d) dV = −V q

d(a− d)

∮SσΦ′dS =

∫z=0

σ1Φ′dS +

∫z=d

σ2Φ′dS︸ ︷︷ ︸

=0

= V

∫z=0

σ1dS = V Q

where Q is the total charge induced on the z = 0 plane, which we want to find.∫Vρ′ΦdV = 0 (as ρ′ = 0)

∮Sσ′ΦdS = 0 (as the planes are grounded).

Putting the result for these four terms back into the Green’s reciprocity relation, we get

Q = −qd

(a− d) ,

where Q is the charge on the plane z = 0.

b) (i)

Fig. 2: Unit charge near a conducting, groundedplane.

The Green’s function for the specified region is equalto the solution for the potential in the problem of theunit charge near an infinite grounded conducting plane.If the unit charge is at the point with coordinates(x′, y′, z′), and its image is at the point (x′, y′,−z′),then the Green’s function reads

G(~x, ~x′) =1√

(x− x′)2 + (y − y′)2 + (z − z′)2

− 1√(x− x′)2 + (y − y′)2 + (z + z′)2

. (1)

(ii) (1)

Rewrite the Green’s function in cylindrical coordinates

x = ρ cosφ

y = ρ sinφ

z = z

(x− x′)2 + (y − y′)2 + (z − z′)2 = (ρ cosφ− ρ′ cosφ′)2 + (ρ sinφ− ρ′ sinφ′)2 + (z − z′)2

= ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + (z − z′)2

G(~x, ~x′) =1√

ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + (z − z′)2− 1√

ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + (z + z′)2.

The general solution of the Dirichlet problem without the space distribution of a charge ρ(x) reads

Φ(~x) = − 1

∮S

Φ(~x′)∂G(~x, ~x′)

∂n′dS′ .

2

The normal is directed outside of the volume z ≥ 0 and therefore ~n′ = −~ez (see fig. 3). The derivationover the normal at the plane z′ = 0 is

∂G(~x, ~x′)

∂n′

∣∣∣z′=0

= ~n′ · ∇′G(~x, ~x′)∣∣∣z′=0

= −∂G(~x, ~x′)

∂z′

∣∣∣z′=0

= −

{(−1

2) 2 (z − z′)(−1)

[ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + (z − z′)2]3/2

−(−1

2) 2 (z + z′)

[ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + (z + z′)2]3/2

}∣∣∣∣∣z′=0

= − 2z

[ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + z2]3/2

At the surface at infinity the potential is zero. The final solution is then

Φ(ρ, φ, z) =V z

∫ 2π

0dφ′∫ a

0

ρ′dρ′

[ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′) + z2]3/2. (2)

Note that the potential distribution is azimuthally symmetric on the edge and hence the potential isindependent of the coordinate φ. So, in the above solution we can put φ = 0.(2)For ρ = 0, the solution in (2) becomes

Φ(z) =V z

∫ 2π

0dφ′∫ a

0

ρ′dρ′

[ρ′2 + z2]3/2= V

(1− z√

a2 + z2

).

(3)

Fig. 3: Area of the disc shape at the potential Von the conducting, grounded plane.

We will expand the solution (2) into its Taylor seriesin powers of ρ2 + z2. For ρ2 + z2 the expansion willconverge fast.

ρ′

[ρ2 + ρ′2 − 2ρρ′ cosφ′ + z2]3/2

=ρ′

[ρ2 + z2]3/21[

1 + ρ′2−2ρρ′ cosφ′ρ2+z2

]3/2=

ρ′

[ρ2 + z2]3/2

{1− 3

2

ρ′2 − 2ρρ′ cosφ′

ρ2 + z2

+3× 5

2× 4

(ρ′2 − 2ρρ′ cosφ′

ρ2 + z2

)2

+ ...

}.

Integration of the first term gives∫ 2π

0dφ′∫ a

0

ρ′dρ′

[ρ2 + z2]3/2=

πa2

[ρ2 + z2]3/2(3)

Integration of the second term gives

− 3

2(ρ2 + z2)[ρ2 + z2]3/2

∫ 2π

0dφ′∫ a

0ρ′(ρ′2 − 2ρρ′ cosφ′)dρ′ = − 3πa4

4(ρ2 + z2)[ρ2 + z2]3/2(4)

Integration of the third term gives

15

8(ρ2 + z2)2[ρ2 + z2]3/2

∫ 2π

0dφ′∫ a

0ρ′(ρ′2 − 2ρρ′ cosφ′)2dρ′

=15

8(ρ2 + z2)2[ρ2 + z2]3/2

(π3a6 + πa4ρ2

)(5)

3

utting in the results from (3), (4) and (5) into (2), we get

Φ(ρ, z) =V a2

2

z

(ρ2 + z2)3/2

[1− 3a2

4(ρ2 + z2)+

5(3ρ2a2 + a4)

8(ρ2 + z2)2+ ...

]. (6)

Verification. Does the solution (6) for ρ = 0 match with the one we got in part (2) for z � a? From(6),

Φ(ρ = 0, z) =V a2

2

1

z2

[1− 3

4

(az

)2+

5

8

(az

)4+ ...

].

The solution from part (2) for z � a becomesz√

a2 + z2=

1√1 +

(az

)2 = 1− 1

2

(az

)2+

3

8

(az

)4+ ...

Φ(z) = V

[1−

(1− 1

2

(az

)2+

3

8

(az

)4− 5

16

(az

)6+ ...

)]=V a2

2

1

z2

[1− 3

4

(az

)2+

5

8

(az

)4+ ...

].

Hence we have shown that the solutions match in the specified range.

Solution of Problem4.2 Method of images

a) (i) The solution for the potential inside a spherical cavity is of the form (Fig. 4.)

Φ(~x) =q√

r2 + d2 − 2rd cos θ+

q′√r2 + d′2 − 2rd′ cos θ

,

where q′ is the value and d′ is the position of the image charge. Both q′ and d′ have to beevaluated from the boundary condition

Φ(~x)|r=a = 0 .

Using the assumed form of the potential with the boundary condition,

Φ(~x)|r=a =q√

a2 + d2 − 2ad cos θ+

q′√a2 + d2 − 2ad′ cos θ

= 0

=⇒ q√a2 + d′2 − 2ad′ cos θ = −q′

√a2 + d2 − 2ad cos θ

=⇒[q2(a2 + d′2)− q′2(a2 + d2)

]+ 2a cos θ

[dq′2 − d′q2

]= 0

z d

0d’

n q q’

Fig. 4: Point charge inside of the sphericalcavity in the conductor.

Functions 1 and cos θ are linearly independent,thus the coefficients multiplying them have tobe both 0 in order for the upper equality to besatisfied. From the two obtained equations wenow find q′ and d′:

q′ = − qad

d′ =a2

d.

The solution for the potential inside the sphereis then

Φ(~x) =q√

r2 + d2 − 2rd cos θ− a

d

q√r2 +

a4

d2− 2r

a2

dcos θ

.

4

(ii) We are going to evaluate the induced surface charge density using the boundary conditionfor the normal component of the electric field. Here we also have to take into account thedirection of the normal (Fig. 4.)

σ =1

4π~E · ~n

∣∣∣∣r=a

(~n = −~er)

σ =1

∂Φ

∂r

∣∣∣∣r=a

=q

−1

2

2r − 2d cos θ

(r2 + d2 − 2rd cos θ)3/2+

a

2d32rd2 − 2da2 cos θ(

r2 +a4

d2− 2r

a2

dcos θ

)3/2

∣∣∣∣∣∣∣∣∣r=a

= − q

4πa2

(ad

) [(ad

)2− 1

][1 +

(ad

)2− 2

a

dcos θ

]3/2 , d < a .

As a check, we can calculate the induced charge on the inner surface:

qind =

∮Sphere

σ(θ)dS =

∫ 2π

0dφ

∫ π

0dθ sin θa2σ(θ)

= − q

2

(ad

)[(ad

)2− 1

] ∫ π

0

sin θdθ[1 +

(ad

)2− 2

a

dcos θ

]3/2

= − q

2

(ad

)[(ad

)2− 1

](d

a

)(−1)√

1 +(ad

)2− 2

a

dcos θ

∣∣∣∣∣∣∣∣π

0

=q

2

[(ad

)2− 1

](d

a+ d− d

a− d

)=qd

2

[(ad

)2− 1

]−2d

a2 − d2= −q .

We could have come to the same conclusion using Gauss’ law and by inspecting a surfacethat lies completely inside the conductor and surrounds the spherical cavity. The electricfield inside the conductor must equal 0, which means that the total charge inside thatsurface is 0. The induced charge is then equal to

qind = −q .

(iii) The force on the charge q is calculated using the image charge method.

F =qq′

(d′ − d)2= −q

2

a2

(ad

)3(1− a2

d2

)−2.

The direction of the force is parallel to the direction of the unit vector ~ez. The existence ofan induced charge with a sign opposite to that of the charge q on the inner surface impliesthat the force is attractive.

(iv) If the sphere is kept at a fixed potential V , the solution for the potential changes to thefollowing form:

Φ =⇒ Φ′ = Φ + V .

5

This solution satisfies the boundary condition

Φ′|r=a = V .

Thus, the surface charge density and the force on the charge q remain the same.If there is a charge q inside the spherical cavity and the sphere is insulated, then one canobtain the solution in the following way. First we ground the sphere. The solution of thatproblem is already known: a charge −q is induced on the inner surface, while on the outersurface the charge equals 0. Now we insulate the sphere and we add charge Q+q to it. Thetotal charge on the inner and outer surfaces of the conductor is now Q. The added chargeQ+ q is distributed uniformly on the outer surface of the spherical conductor. The solutioninside the sphere is now changed to

Φ =⇒ Φ′ = Φ +Q+ q

b,

where b is the outer radius of the spherical conductor. This change in the potential doesnot have any influence on the force exerted on a charge q and on the surface charge densityof the inner surface. Note that the charge Q + q cannot distribute on the inner surface,otherwise the Gauss law for the surface that lies inside the conductor would not hold.

b) Suppose that two opposite charges ±Q are located on the z-axis at z = ±R. Within a sphereof radius much smaller than R, the field is approximately constant. If we now take the limits

R→∞ and Q→∞ such that2kQ

R2is held constant at E0, we obtain a uniform field E0 at the

origin.

z=R

a −Q +Q

z=−R z

r

E0 θ

Fig. 5: Configuration of the system.

Now we let the conducting sphere be placed at theorigin, and allow R to be finite. In this case, bothQ and −Q have image charges inside the sphere:

image of Q =⇒ q1 = −Q a

Rat z = −a

2

R

image of −Q =⇒ q2 = Qa

Rat z =

a2

R

The potential is the superposition of the potentialsdue to all the charges

Φ(~r) = k

[Q∣∣~r − ~r(Q)

∣∣ +q1∣∣~r − ~r(q1)∣∣ +

q1∣∣~r − ~r(q2)∣∣ − Q∣∣~r − ~r(−Q)

∣∣]

Thus, it follows that

Φ(~r)

k=

Q

(r2 +R2 + 2rR cos θ)1/2− Q

(r2 +R2 − 2rR cos θ)1/2

+−Q a

R(r2 +

a4

R2+

2a2r

Rcos θ

)1/2+

Qa

R(r2 +

a4

R2− 2a2r

Rcos θ

)1/2

Since r � R:

(r2 +R2 + 2rR cos θ

)−1/2=R−1

(1 +

r2

R2+ 2

r

Rcos θ

)−1/2= R−1

(1− r2

2R2− r

Rcos θ + ...

)(r2 +

a4

R2+

2a2r

Rcos θ

)−1/2= r−1

(1− a4

2r2R2− a2

rRcos θ + ...

)

6

The potential now becomes

Φ(~r) = kQ

R

[1− r2

2R2− r

Rcos θ − 1 +

r2

2R2− r

Rcos θ

]+ ...

−kQarR

[1− a4

2r2R2− a2

rRcos θ − 1 +

a4

2r2R2− a2

rRcos θ

]+ ...

= − 2kQr

R2cos θ + 2k

Qa

rR

a2

rRcos θ + ... = −2k

Q

R2

(r − a3

r2

)cos θ

(all

Q

Rnterms, where n > 2,

are dropped)

In the limit R→∞, Q→∞

Φ(~r) = −E0

(r − a3

r2

)cos θ ,

where −E0r cos θ = −E0z, which is equal to the potential of the uniform field, and where

E0a3

r2cos θ is the potential related to the charge induced on the surface.

σ = − 1

4πk

∂φ

∂r= − 1

4πk

∂φ

∂r

∣∣∣∣r=a

= − 1

4πk

[−E0

(1 +

2a3

a3

)cos θ

].

Thus,

σ =3

4πkE0 cos θ .

The surface potential of this charge density vanishes, so there is no difference between a groundedand an insulated sphere.

Solution of Problem4.3 Hollow cube

a)

Fig. 6:

We have to solve Laplace’s equation in rectangular coordi-nates,

∂2Φ

∂x2+∂2Φ

∂y2+∂2Φ

∂z2= 0, (7)

inside a cube. A solution of this partial differential equationcan be found in terms of three ordinary differential equations,all of the same form, by the assumption that the potential canbe represented by a product of three functions, one for eachcoordinate

Φ(x, y, z) = X(x)Y (y)Z(z) . (8)

Substitution of (8) into (7) and division by X(x)Y (y)Z(z) yields

1

X(x)

d2X

dx2+

1

Y (y)

d2Y

dy2+

1

Z(z)

d2Z

dz2= 0 .

7

If the above equation is to hold for arbitrary values of the independent coordinates, then eachof the three terms must be separately constant:

1X(x)

d2Xdx2

= −α2

1Y (y)

d2Ydy2

= −β21

Z(z)d2Zdz2

= γ2

(9)

where α, β and γ are complex numbers, such that α2 + β2 = γ2 . The solutions of the threedifferential equations (9) are exp(±iαx), exp(±iβy) and exp(±z

√α2 + β2). The potential (8)

can thus be built up from the product solutions

Φ = exp(±iαx) exp(±iβy) exp(±z√α2 + β2) .

The coefficients α and β can be determined from the boundary conditions on the potential. In theproblem we are given that the walls z = 0 and z = a are held at a constant potential V while theother four faces of the cube are at zero potential. We will split the problem into two superposableparts and first consider the case in which the face z = a is at the potential Φ(x, y, z)|z=a = V ,and all the other faces are at the potential zero. Starting from the requirement that Φ = 0 forx = 0, y = 0, z = 0 it is easy to see that the required forms of X,Y, Z are

X = sin(αx)Y = sin(βy)

Z = sinh(z√α2 + β2)

In order that Φ = 0 at x = a, y = a it is necessary that αa = mπ and βa = nπ. With thedefinitions

αm =mπ

a; βn =

a; γmn =

π

a

√m2 + n2 .

we can write the partial potential Φmn

Φmn = sin(αmx) sin(βny) sinh(γmnz) .

The potential can be expanded in terms of these Φmn with initially arbitrary coefficients (to bechosen to satisfy all boundary conditions)

Φ(x, y, z) =

∞∑n,m=1

Amn sin(αmx) sin(βny) sinh(γmnz) .

Similarly we find the expression for the case in which the side z = 0 is at the potential V . In

conclusion, the general solution for the interior of the cube in which the face z = 0 or a is at thepotential Φ(x, y, z)|z = V , and all the other faces are at the potential zero (see figure 6) reads

Φ(1)(x, y, z) =∞∑

n,m=1

Amn sin(αmx) sin(βny) sinh(γmnz)for z = a

⇒ V =∞∑

n,m=1

Amn sin(αmx) sin(βny) sinh(γmna) (10)

Φ(2)(x, y, z) =∞∑

n,m=1

Bmn sin(αmx) sin(βny) sinh(γmn(a− z))for z = 0

⇒ V =∞∑

n,m=1

Bmn sin(αmx) sin(βny) sinh(γmna) (11)

8

The equations (10) and (11) comprise of double Fourier series. From (10), the coefficientAmn sinh(γmnz) of the Fourier series is given by

Amn sinh(γmnz) =4V

a2 sinh(π√m2 + n2)

∫ a

0dx

∫ a

0dy sin

(mπax)

sin(nπay)

⇒ Amn =4V

a2 sinh(π√m2 + n2)

∫ a

0dx

∫ a

0dy sin

(mπax)

sin(nπay)

=4V

a2 sinh(π√m2 + n2)

( a

)(−1)[(−1)m − 1]

×( a

)(−1)[(−1)n − 1] .

⇒ Amn =

{0 ; if m or n even16V/[π2mn sinh(π

√m2 + n2)] ; if m and n both odd

Similarly,

Bmn =

{0 ; if m or n even16V/[π2mn sinh(π

√m2 + n2)] ; if m and n both odd

The total solution of the problem is obtained by the superposition of Φ(1)(x, y, z) and Φ(2)(x, y, z)

Φ(x, y, z) = Φ(1)(x, y, z) + Φ(2)(x, y, z)

=16V

π2

∞∑m,n odd

1

mnsin(αmx) sin(βny)

[sinh(γmnz) + sinh(γmn(a− z))]sinh(π

√m2 + n2)

. (12)

Using the trigonometric identities

sinh(x) + sinh(y) = 2 sinh(x+ y

2) cosh(

x− y2

)

sinh(2x) = 2 sinh(x) cosh(x)

we can write

sinh(γmnz) + sinh(γmn(a− z)) = 2 sinh[π

2

√m2 + n2

]cosh

[ π2a

√m2 + n2(2z − a)

]sinh(π

√m2 + n2) = 2 sinh

[π2

√m2 + n2

]cosh

[π2

√m2 + n2

].

This simplifies the fraction in (12). Also, since the sum in (12) is over only the odd positiveintegral values of m and n, we can make it run over all the positive integers by replacing m andn by 2m− 1 and 2n− 1, respectively. Therefore,

Φ(x, y, z) =16V

π2

∞∑m,n=1

1

(2m− 1)(2n− 1)

sin[(2m−1)π

a x]

sin[(2n−1)π

a y]

cosh[π2

√(2m− 1)2 + (2n− 1)2

]× cosh

[ π2a

√(2m− 1)2 + (2n− 1)2 (2z − a)

]}.

b) The coordinates of the cube’s center are (a/2, a/2, a/2), so the potential is

Φ(a

2,a

2,a

2

)=

16V

π2

∞∑m,n=1

1

(2m− 1)(2n− 1)

sin[(2m−1)π

2

]sin[(2n−1)π

2

]cosh

[π2

√(2m− 1)2 + (2n− 1)2

] . (13)

The numerical results of the summation of the first m×n terms of the expansion (13) are givenin Table 1 in units of V

9

Table 1: The potential at the center of the box in units of V .m,n 1 2 3 4∑

0.347546 0.332958 0.333345 0.333333We see that it is enough to consider the first 9 terms (m,n = 3) to achieve an accuracy up tothe third decimal place.

c) The surface charge density on the side z = a is

σ|z=a =1

∂Φ

∂z

∣∣∣∣z=a

=4V

π3

∞∑m,n=1

πa

√(2m− 1)2 + (2n− 1)2 sin

[(2m−1)π

a x]

sin[(2n−1)π

a y]

(2m− 1)(2n− 1) cosh[π2

√(2m− 1)2 + (2n− 1)2

]× sinh

[π2

√(2m− 1)2 + (2n− 1)2

]}=

4V

π2a

∞∑m,n=1

{√1

(2m− 1)2+

1

(2n− 1)2tanh

[π2

√(2m− 1)2 + (2n− 1)2

]× sin

[(2m− 1)π

ax

]sin

[(2n− 1)π

ay

]}.

10