= a
0
2π
0
VL(r, ϕ)J
ρjr a
cosϕ sinϕ
rdrdϕ≡ j (5–96) Performing the integration over ϕand using the orthogonality of the trigono- metric functions to eliminate all terms for which m= (for = 0, replace π by 2πfor the cosine term in all the following formulas), we find
i
Ci Di
π
a 0
J
ρir a
J
ρjr a
rdrdϕ=j (5–97) and using (E–20)
a
0
J ρir
a
J ρjr
a
rdr= 12a2J2+1(ρj)δij (5–98) we perform the remaining integration overr to obtain
Cj
Dj
1
2πa2J2+1(ρj) = a
0
2π
0
VL(r, ϕ)J
ρjr a
cosϕ sinϕ
rdrdϕ (5–99) The original expansion coefficients are now easily found:
Aj Bj
= 2
πa2J2+1(ρj) sinhρjL
a
a
0
2π
0
VL(r, ϕ)J ρjr
a
sinϕ cosϕ
rdrdϕ (5–100) The complete solution satisfying both boundary conditions is then the sum of (5–90) with coefficients given by (5–92) and of (5–94) with coefficients given by (5–100).
with no restrictions on yet. Our rather peculiar choice of writing the separation constant anticipates a well-behaved solution. The radial equation
r2 d2
dr2(rR) =(+ 1) (rR) (5–104)
is easily solved to giverR =Cr+1+Dr−, whence we conclude R =Cr+ D
r+1 (5–105)
The remaining equation for Θ, 1
sinθ d dθ
sinθdΘ dθ
+
(+ 1)− m2 sin2θ
Θ = 0 (5–106)
may be converted to the associated Legendre equation by setting x = cosθ ⇒ d/dθ=−sinθ·d/dx, which results in
d dx
(1−x2)dΘ dx
+
(+ 1)− m2 1−x2
Θ = 0 (5–107)
The solutions to (5–107) are the associated Legendre functions Pm (x) and Qm (x). Qm (±1) = ±∞ and Pm (x) diverges as x → ±1 unless is an integer.
P0(x) ≡ P(x) are the Legendre Polynomials encountered in section 5.2.3 and earlier.
Recapitulating, we have found the well-behaved solutions to∇2V = 0 to be of the form
V(r, θ, ϕ) =
,m
(Amsinmϕ+Bmcosmϕ)
Cr+ D r+1
Pm (cosθ) (5–108) or, more briefly, the entire solution may be written in terms of spherical harmonics,
V(r, θ, ϕ) =
,m
A,mr+B,m
r+1
Ym (θ, ϕ) (5–109) whereA,m andB,m are complex constants.
Example 5.15: Find the potential inside a hollow sphere of radius a given that the potential on the surface isVa(θ, ϕ).
Solution: Since V is finite at the origin, B,m of equation (5–109) must vanish;
hence the general solution in the region of interest is of the form V(r, θ, ϕ) =
,m
A,mrYm (θ, ϕ) (Ex 5.15.1) Atr =a, this specializes to
Va(θ, ϕ) =
,m
A,maYm (θ, ϕ) (Ex 5.15.2)
Multiplying both sides by Y∗m(θ, ϕ), the complex conjugate of Ym(θ, ϕ), and integrating over the entire solid angle, we have with the aid of the orthonormality (F–43) of the spherical harmonics
2π
0
π
0
Va(θ, ϕ)Y∗m (θ, ϕ) sinθdθdϕ=A,ma (Ex 5.15.3) The coefficientsA,m of the expansion follow immediately.
Example 5.16: A sphere of radius a centered on the origin has potential V = V0sin2θcos 2ϕon its surface. Find the potential both inside and outside the sphere.
Solution: Inside the sphere the potential must take the form V(r < a, θ, ϕ) =
,m
A,mrYm(θ, ϕ) (Ex 5.16.1)
Whereas we could in principle use the prescription in the previous problem to de- termine the coefficients it is easier to write the potential on the surface in spherical harmonics and then depend on the linear independence of the spherical harmonics to find the coefficients. Therefore, using (F–42), we write the boundary condition for the problem as
Va(θ, ϕ) = 12V0sin2θ
e2iϕ+e−2iϕ
=
&
8π 15
Y22(θ, ϕ) + Y−22 (θ, ϕ)
(Ex 5.16.2) Equating the general potential (Ex 5.16.1) to (Ex 5.16.2) atr=a, we have
,m
A,maYm (θ, ϕ) =
&
8π 15V0
Y22(θ, ϕ) + Y−22 (θ, ϕ)
(Ex 5.16.3)
We conclude immediately
A2,2=A2,−2=V0 a2
&
8π
15 (Ex5.16.4)
We can, of course express the potential in terms of trigonometric functions by substituting the trigonometric expressions for Y±22 (θ, ϕ)
V(r < a, θ, ϕ) =V0r2 a2
&
8π 15
&
15
32πsin2θ
e2iϕ+e−2iϕ
=V0r2
a2 sin2θcos 2ϕ (Ex 5.16.5) The exterior solution is obtained in much the same way. The solution must take the form:
V(r > a, θ, ϕ) =
,m
B,m
r+1 Ym (θ, ϕ) (Ex 5.16.6)
We again equate this potential atato that on the sphere to obtain
,m
B,m
a+1 Ym (θ, ϕ) =
&
8π 15V0
Y22(θ, ϕ) + Y−22 (θ, ϕ)
(Ex 5.16.7) leading to the conclusion that the only nonvanishing coefficients are
B2,±2=a3V0
&
8π
15 (Ex 5.16.8)
and the potential forr > ais V(r > a, θ, ϕ) =a3V0
r3
&
8π 15
Y22(θ, ϕ) + Y−22 (θ, ϕ)
=a3V0
r3 sin2θcos 2ϕ (Ex 5.16.9) It is evident in both cases that the solution satisfies the boundary conditions.
Example 5.17: Find the potential inside and outside a sphere of radius R that has surface charge with densityσ=σ0sin2θcos 2ϕdistributed on its surface.
Solution: The presence of a surface charge means the electric field is discontinuous byσ/ε0 across the surface of the sphere. The second boundary condition that the tangential component ofE is continuous across the surface means that both∂V /∂θ and ∂V /∂ϕ are continuous which in turn implies that the potential is continuous across the surface. The form of the potential is
V(r < R, θ, ϕ) =
,m
A,mrYm (θ, ϕ) (Ex 5.17.1)
V(r > R, θ, ϕ) =
,m
B,mr−(+1)Ym(θ, ϕ) (Ex 5.17.2) We apply the radial boundary condition to obtain
∂V
∂r
R−
− ∂V
∂r
R+
= σ
ε0 (Ex 5.17.3)
or
,m
A,mR−1+(+ 1)B,m R+2
Ym(θ, ϕ) =σ0 0
&
8π 15
Y22(θ, ϕ) + Y−22 (θ, ϕ)
(Ex 5.17.4) The linear independence of the spherical harmonics means that for (, m)= (2,±2)
A,m=−(+ 1)B,m
R2+1 (Ex 5.17.5)
while for (, m) = (2,±2) we have
2A2,±2R+3B2,±2 R4 =σ0
ε0
&
8π
15 (Ex 5.17.6)
The continuity of the potential across the boundary gives for all (, m) A,m= B,m
R2+1 (Ex 5.17.7)
(Ex 5.17.7) and (Ex 5.17.5) together yield only trivial solutions for A,m andB,m, leaving us to solve only (Ex 5.17.7) and (Ex 5.17.6) for the (, m) = (2,±2) coeffi- cients. Solving these two equations simultaneously yields
A2,±2= σ0 5ε0R
&
8π
15 and B2,±2= σ0R4 5ε0
&
8π
15 (Ex 5.17.8)
We conclude the potentials are V(r < R, θ, ϕ) = σ0r2
5ε0R
&
8π 15
Y2−2(θ, ϕ) + Y22(θ, ϕ)
(Ex 5.17.9) and
V(r > R, θ, ϕ) = σ0R4 5ε0r3
&
8π 15
Y−22 (θ, ϕ) + Y22(θ, ϕ)
(Ex 5.18.10)
5.3.3 Oblate Ellipsoidal Coordinates
As a last example of separation of variables we consider a problem in oblate (pancake shaped) ellipsoidal coordinates. This by no means exhausts the separable coordi- nate systems: prolate ellipsoidal (cigar-shaped), bispherical, conical, and toroidal systems (not strictly speaking separable, see Appendix B) come to mind immedi- ately.16 Rather than obtain the general solution we will solve what is really a one variable problem, the potential due to a charged, constant potential oblate ellipsoid of revolution. We will specialize to find the potential around a charged, conducting, flat circular plate. We begin by constructing the Laplacian in oblate ellipsoidal coordinates ρ,α, andϕdefined by
x=aρsinαcosϕ y=aρsinαsinϕ
4
s≡
x2+y2=aρsinα z=a
ρ2−1 cosα ρ≥1, 0≤α≤π, 0≤ϕ ≤2π
(5–110)
where, for convenience we have defined the cylindrical radius s as ρ has already been used as one of the oblate elliptical coordinates. Surfaces of constant ρobey the equation
s2
a2ρ2 + z2
a2(ρ2−1) = 1 (5–111)
16Toroidal coordinates are discussed in Appendix B. The metric tensor, curl, grad, div, and the Laplacian for a large number of orthogonal coordinate systems may be found in Parry Moon and Domina E. Spencer. (1961)Field Theory Handbook, Springer-Verlag, Berlin. The separation of the Laplace equation and the Helmholtz equation is effected for each of the coordinate systems in this small volume, and the general solutions are given.
demonstrating that the constantρsurface is an ellipsoid of revolution with major radius aρ in the x-y plane and minor radius a(ρ2−1)1/2 along the z axis. The surface described by ρ= 1 has z≡0 ands lies between 0 and a on this surface.
It is a (two-sided) circular flat disk of radiusa.
Although there are other methods, we construct the Laplacian using the methods of Appendix B, by means of the metric tensor whose nonzero elements are
gρρ= ∂x
∂ρ 2
+ ∂y
∂ρ 2
+ ∂z
∂ρ 2
= a2(ρ2−sin2α) ρ2−1 gαα=
∂x
∂α 2
+ ∂y
∂α 2
+ ∂z
∂α 2
=a2(ρ2−sin2α)
gϕϕ= ∂x
∂ϕ 2
+ ∂y
∂ϕ 2
+ ∂z
∂ϕ 2
=a2ρ2sin2α (5–112) The Laplacian may now be constructed:
∇2V =
ρ2−1 a2ρ2(ρ2−sin2α)
∂
∂ρ
ρ
ρ2−1∂V
∂ρ
+ 1
a2sinα(ρ2−sin2α)
∂
∂α
sinα∂V
∂α
+ 1
a2ρ2sin2α
∂2V
∂ϕ2 (5–113) For the particular problem of the (constant potential) metal circular disk, the boundary conditions are independent of ϕ or α (as ρ → ∞, α → θ), leading us to seek a solution depending only onρ. V must then satisfy
∂
∂ρ
ρ
ρ2−1∂V
∂ρ
= 0 (5–114)
Integrating twice we obtain
∂V
∂ρ = k ρ
ρ2−1 (5–115)
followed by
V(ρ) =−ksin−1 1
ρ
+C (5–116)
wherek andC are constants of integration. At very large values ofρ, (whereaρis indistinguishable fromr) we expect the potential to vanish. The value ofC follows, V(∞) = 0 =−ksin−10 +C⇒C= 0. Setting the potential on the surface of the ellipsoid defined by ρ=ρ0 toV0, we findV0=−ksin−1(1/ρ0). [For the flat plate defined byρ0= 1,V0=−ksin−1(1) =−12kπ]. The potential at any pointρis then
V(ρ) = V0
sin−1(1/ρ0)sin−1 1
ρ
(5–117)
To relateV0 to the total charge on the ellipsoid, we first compute the electric field E =−∇V =− 1
√gρρ
∂V
∂ρρˆ (5–118)
where ˆρis a unit vector in theρdirection. The electric field is then E = V0
sin−1(1/ρ0)
ˆ ρ aρ
ρ2−sin2α (5–119)
To obtain the charge on the ellipsoid, we integrateE over any constantρ=ρ0 surface containing the charged ellipsoid and equate the surface integral to Q/ε0 (Gauss’ law).
Q
ε0 = E ·dS= Eρ√
gααgϕϕdαdϕ
= V0
sin−1(1/ρ0) 2π
0
π 0
a2ρ0sinα
ρ20−sin2α ρ0a
ρ20−sin2α
dαdϕ
= V0
sin−1(1/ρ0) 2π
0
π
0
asinα dαdϕ= 4πV0a
sin−1(1/ρ0) (5–120) The potentialV0on the conductor is then related to the charge by
V0= Qsin−1(1/ρ0)
4πε0a (5–121)
giving
V(ρ) = Q
4πε0asin−1 1
ρ
(5–122) and
E = Q 4πε0a
ˆ ρ aρ
ρ2−sin2α (5–123)
Specializing now to the flat circular plate of radiusa corresponding to ρ= 1, we have
E(ρ = 1) = Qˆρ 4πε0a√
a2−s2 (5–124)
with ˆρ=±ˆk.
The surface charge density on the plate is given by σ=ε0E⊥= Q
4πa√
a2−s2 (5–125)
(At s = 0, the center of the plate, this is just half of the charge density it would have if the charge were uniformly distributed over both sides.)
Exercises and Problems
Figure 5.19: The rectangular pipe has three sides grounded, while the fourth has uniform potentialV0.
Figure 5.20: The pipe has potential difference V0applied between the upper and lower halves.
5-1 Show that the potential at the cen- ter of a charge-free sphere is precisely the average of the potential on the surface of the sphere.
5-2Two large flat conducting plates are placed so that they form a wedge of angle α < π/2. The plates are insulated from each other and have potentials 0 on one and V =V0 on the other. Find the po- tential between the plates.
5-3 Two conducting coaxial cones with vertex at the origin and apex anglesα1 andα2are isolated from each other. The inner and outer cones have potentialV1 and V2, respectively. Find the potential between the cones.
5-4 A rectangular pipe (Figure 5.19) of dimensionsa = 10 cm andb= 8 cm has three of its sides maintained at zero po- tential while the remaining side is insu- lated and maintained atV0. Use separa- tion of variables to evaluate the potential along they =b/2 line atx= 1 cm, 5 cm, and 9 cm to three significant figures.
5-5 A cylindrical pipe (Figure 5.20) of radius a is sawn lengthwise into two
equal halves. A battery connected be- tween the two halves establishes a po- tential difference ofV0 between the two halves. Use separation of variables to find the potential inside and outside the pipe.
5-6A conducting sphere of radiusawith its center at the origin is cut into two halves at thex-y plane. The two halves are separated slightly, and the top half is charged toV0while the bottom half is charged to−V0. Find the potential both interior to the sphere and exterior to the sphere.
5-7 Find the scalar magnetic potential in the vicinity of the center of a solenoid of lengthL withN turns, each carrying currentI.
5-8Use the mapping f = ln
a+z a−z
to map the infinite parallel plates at v=±π/2 to the cross section of the split cylinder of Figure 5.20. (Hint: Express [a+z]/[a−z] as Reiα, giving u+iv =
lnR+iαorv=α.) Express the potential in the pipe in closed form, and compare this to the result obtained by summing the first five nonzero terms of the series of exercise 5-5 for the pointy= 0.2aand x = 0.
5-9Obtain the mapping that carries the +u axis in thef plane to the positivex axis and the−uaxis to an axis inclined atαto thex axis.
5-10 Obtain the charge density on the constant potential right angled plate of Figure 5.9 (page 116).
5-11 Verify that the mapping z= 2a
π ln coshf
carries the infinite lines at v = π/2, 0, and −π/2 to two infinite lines and one half-infinite line. Use this mapping to find the capacitance of a metal plate in- serted in the middle between two larger grounded plates.
5-12 Calculate the capacitance of two parallel flat circular conducting plates with a space d between them and a ra- dius R d. How good an approxima- tion is your result?
5-13 Draw the image in the z plane of thev = 1 and v= 2 (potential) lines as well as the u = 14b and 12b (field) lines under the mapping (Ex 5.13.3).
5-14Find the image of the real axis un- der the mappings and determine what happens off the real axis
z=i−f
i+f and z= i+f i−f 5-15Find the image of the real axis un- der the mappings
z=
√f −1
√f + 1 and z= √
f −1
√f + 1 1/n
5-16Find the image of the real axis un- der the mapping
z=d(f1/2−f−1/2)
5-17Find the image of the real axis un- der the mapping
z={2(f+1)1/2−2 ln[(f+1)1/2+1]+lnf}
5-18 Find the image of the upper half plane under the mappingz=a
f2−1.
5-19 A right circular cylinder of radius ahas a potential
V =V0
1− r2 a2
on its top surface and zero on all its other surfaces. Find the potential anywhere inside the cylinder. Note that
d
dz(zνJν(z)) =zνJν−1(z) implies that
xnJn−1(x)dx =xnJn(x), allowingxJ0(x) to be integrated directly while x3J0(x) may be integrated by parts.
5-20 A hollow spherical shell has sur- face charge σ0cosθ distributed on its surface. Find the electro-static potential and electric field due to this distribution both inside and outside the sphere.
5-21 A sphere of radiusa has potential sin 2θcosϕ on its surface. Find the po- tential at all points outside the sphere.
5-22 A hollow spherical shell carries charge density σ(θ, ϕ) = σ0sinθsinϕ.
Calculate the potential both inside and outside the sphere
5-23 Use prolate ellipsoidal coordinates x=a
ρ2−1 sinαsinϕ y=a
ρ2−1 sinαcosϕ z=aρcosα
to compute the potential in the vicin- ity of a thin conducting needle carry- ing charge Q. Show that the charge dis- tributes itself uniformly along the needle.
5-24 Bispherical coordinates defined by x= asinαcosϕ
coshρ−cosα y= asinαsinϕ
coshρ−cosα z= asinhρ
coshρ−cosα
may be used to obtain the potential in the vicinity of two identical conducting spheres of radius 1 m with centers sepa- rated by 10 m charged to potentialsVa andVb, respectively. The relevant coor- dinate surface satisfies
x2+y2+ (z−acothρ)2= a2 sinh2ρ Sketch the method of obtaining the so- lution. (Hint: Use the substitution
V = (coshρ−cosα)1/2U to separate the variables.)