Static Potentials in Vacuum – Laplace’s Equation
5.1 Laplace’s Equation
5.2.3 Spherical Polar Coordinates with Axial Symmetry
Applying the first of these conditions (Ex 5.5.8) to the series forV, we obtain (Asinϕ−Bcosϕ) =
(Csinϕ−Dcosϕ) (Ex 5.5.10) Grouping terms, we have
(A−C) sinϕ+
(B−D) cosϕ= 0 (Ex 5.5.11) The linear independence of the trigonometric functions then requires that the co- efficients of each term in the sum above equal zero; therefore A =C and B = D.
The second boundary condition (Ex 5.5.9) applied to the series gives
a(Ccosϕ+Dsinϕ) =−
a(Acosϕ+Bsinϕ) +σ0 ε0cosϕ
(Ex 5.5.12) We again equate the coefficient of each linearly independent function to zero, giving for the cosϕterm
C1+A1−σ0a ε0
cosϕ= 0 (Ex 5.5.13)
while the remaining terms give B +D = 0 and A+C = 0. Together with the equations from (Ex 5.5.11) these latter imply that except for A1 and C1, the coefficients vanish. SubstitutingA1 =C1 in the equation above yieldsA1=C1 = σ0a/2ε0. Finally, then,
V(r < a) = σ0r
2ε0 cosϕ (Ex 5.5.14)
and
V(r > a) =σ0a2
2ε0rcosϕ (Ex 5.5.15)
It might be noted that Exercise 5.5 is essentially a Neumann problem in that the derivatives of the potential are specified, albeit indirectly, on the boundary. It should be obvious that an arbitrary constant may be added toV without changing the fields.
We assume a separable solution of the form V = R(r)Θ(θ), then r2∇2V /V = 0 becomes
r2 rR
d2
dr2(rR) =− 1 Θ sinθ
d dθ
sinθdΘ dθ
=(+ 1) (5–26)
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) (5–27)
is readily solved to giverR =Ar+1+Br−, whence we conclude R(r) =Ar+ B
r+1 (5–28)
The remaining equation for Θ, 1 sinθ
d dθ
sinθdΘ dθ
+(+ 1)Θ = 0 (5–29)
is may be converted to the Legendre equation (see Appendix F) by setting x = cosθ⇒d/dθ=−sinθ·d/dx, which results in
d dx
(1−x2)dΘ dx
+(+ 1)Θ = 0 (5–30)
The solutions to (5–30) are well known as the Legendre functions P(x) and Q(x). Q(±1) = ±∞and P(x) diverges as x → ±1 unless is an integer. The integralsolutions P(x) are the Legendre Polynomials encountered earlier.
Recapitulating, we have found the well-behaved solutions to∇2V = 0 to be of the form
V(r, θ) =
Ar+ B r+1
P(cosθ) (5–31)
and the general solution for problems includingθ= 0 andθ=πis V(r, θ) =
∞
=0
Ar+ B r+1
P(cosθ) (5–32)
Example 5.6: A grounded conducting sphere of radius a is placed in an initially uniform electric fieldE =E0kˆ. Find the resulting potential and electric field.
Solution: Besides the obvious boundary condition V(a, θ, ϕ) = 0, we have also V(r→ ∞) =−E0rcosθ. In addition we expect the solution to be symmetric about thez axis . We therefore rewrite
V(r, θ) =
Ar+ B
r+1
P(cosθ) (Ex 5.6.1)
The boundary condition at ∞ implies that the potential can have no terms growing faster thanr: V(r)→ −E0rcosθ=A1rP1(cosθ) as r→ ∞then givesA1
=−E0, andA= 1 = 0. With this restriction the solution atr =a reduces to 0 =V(a, θ) =−E0acosθ+
∞
=0
B
a+1P(cosθ) (Ex 5.6.2)
= B0 a +
B1 a2 −E0a
cosθ+
∞
=2
B
a+1P(cosθ) (Ex 5.6.3) As the Legendre polynomials are linearly independent functions, they can add to zero only if the coefficient of each vanishes. We conclude then that
B0
a = 0 ⇒ B0= 0 B1
a2 −E0a= 0 ⇒ B1=E0a3 B
a+1 = 0 ⇒ B=1= 0
(Ex 5.6.4)
Recapitulating, we have found that the potential is V(r, θ) =
a3 r2 −r
E0cosθ (Ex 5.6.5)
from which we easily deduce the electric fieldE =−∇V : E =
1 + 2a3
r3
E0rˆcosθ+ a3
r3 −1
E0θˆsinθ (Ex 5.6.6) It is worth noting that as ˆk= ˆrcosθ−θˆsinθ,
E =E0kˆ+E0a3 r3
3ˆrcosθ+ ˆθsinθ−ˆrcosθ
(Ex 5.6.7)
=E0ˆk+E0a3 r3
$
3(ˆr·ˆk)ˆr−ˆk
%
(Ex 5.6.8) In other words, the field of the charge distribution on the sphere induced by the initially uniform external fieldE0 is just that of a dipole of strength 4πε0a3E0k.ˆ
As a somewhat more powerful use of the spherical polar solution of Laplace’s equation, we can use as boundary condition the known (by other means) potential along a symmetry axis. The solution thus obtained will give us the potential at any point in space.
Example 5.7: Obtain the magnetic scalar potential of a circular current loop of radius aat a point in the vicinity of the center (r < a).
Solution: The general solution forVm that does not diverge asr→0 is Vm(r, θ) =
ArP(cosθ) (Ex 5.7.1) Forr along thez axis θ= 0, allowing us to replace P(cosθ) by P(1) = 1. Along thez axis,Vmbecomes
Vm(z) =
Az (Ex 5.7.2)
The magnetic scalar potential along the central axis of a circular current loop was found in (Ex 1.15.3) to be
Vm(z) =−I 2
√ z
z2+a2 (Ex 5.7.3)
which can be expanded by the binomial theorem as a power series in (z/a):
Vm=−Iz 2a
1 + z2
a2 −1/2
(Ex 5.7.4)
=−Iz 2a
1−1
2 z2 a2 +3
2 1 2
1 2!
z4 a4 −5
2 3 2 1 2
1 3!
z6 a6 +. . .
(Ex 5.7.5)
=−I 2az+ I
4a3z3− 3I
16a5z5+ 15
96a7z7−. . . (Ex 5.7.6) A comparison of coefficients yields
A1=−I
2a, A3= I
4a3, A5=− 3I
16a5, A7= 15I 96a7, . . .
The general result for the magnetic scalar potential of a circular current loop is then Vm=−I
2 r
aP1(cosθ)− r3
2a3P3(cosθ) + 3r5
8a5P5(cosθ)−15r7
48a7P7(cosθ) +. . . (Ex 5.7.7) The magnetic induction field is now easily obtained asB =−µ0∇Vm.
The magnetic induction field of a plane coil is sufficiently important that we might try expressing the result above (Ex 5.8.5) in cylindrical coordinates in some- what more elementary terms. Writing P1(cosθ) = cosθ=z/r, we expand
P3(cosθ) = 1 2
5
z r
3
−3 z
r
(5–33)
P5(cosθ) =1 8
63
z r
5
−70 z
r 3
+ 15 z
r
(5–34)
Then, using the cylindrical coordinate ρ for distance from the axis,r2 =ρ2+z2 and we obtain
r
aP1(cosθ) = r a·z
r =z
a (5–35)
r a
3
P3(cosθ) =1 2
5z3
a3−3 r2z a3
=1 2
5z3
a3−3(z2+ρ2)z a3
= z3 a3−3
2 ρ2z
a3 (5–36) r
a 5
P5(cosθ) = 1 8
63z5
a5−70z5
a5 −70ρ2z3
a5 + 15(ρ2+z2)2z a5
= 1 8a5
8z5−40z3ρ2+ 15zρ4
(5–37) Collecting terms, we obtain finally forVm(ρ, z)
Vm(ρ, z) =−I 2
z a− 1
2a3
z3−3 2ρ2z
+ 3
64a5
8z5−40ρ2z3+ 15ρ4z
· · ·
(5–38) We see that centrally, the magnetic induction field takes the value Bz =µ0I/2a, decreases quadratically along thez axis and increases quadratically as the wire is approached.
Example 5.8: Obtain the magnetic scalar potential of a circular current loop of radius aat a pointr > a.
Solution: The general expression forVm that does not diverge asr→ ∞is Vm(r, θ) =A0+
B
r+1P(cosθ) (Ex 5.8.1) giving along thez axis Vm(z) =A0+
B
z+1 (Ex 5.8.2)
The previously found expression for the potential along the z axis can again be expanded in a power series, but this time in (a/z) to ensure convergence.
Vm(z) =−I 2
√ z
z2+a2 =−I 2
1 +
a z
2−12
=−I 2
1−1
2 a
z 2
+1 2 3 2
1 2!
a z
4
−1 2 3 2 5 2
1 3!
a z
6 +· · ·
(Ex 5.8.3) Comparing terms, we have B1 = 14Ia2, B3 =−163Ia4, B6 = 325Ia6, and so forth.
The expansion for arbitraryr, r > a, is then Vm(r, θ) = I
2
−1 + 1 2
a2
r2P1(cosθ)−3 8
a4
r4P3(cosθ) + 5 16
a6
r6P5(cosθ) +· · · (Ex 5.8.4)
As a final example of the utility of this technique we compute the noncentral field of a pair of coils with parallel planes, separated by their radius. Such pairs of coils are known as Helmholtz coils and are useful for producing uniform magnetic induction fields over relatively large, open volumes. The spacing of the coils is exactly that required for the cubic terms of the scalar potential to cancel one- another.
Example 5.9: Find the scalar magnetic potential and magnetic induction field at points near the axis of a pair of Helmholtz coils each of which has radiusaand has one turn carrying currentI.
Solution: The magnetic scalar potential of two coils centered atz=±12ais easily written as
Vm(0,0, z) =−I 2
z−12a
(z−12a)2+a2
+ z+12a
(z+12a)2+a2
(Ex 5.9.1)
We begin by expanding this axial potential as a power series inz/a. Some care is required in carrying all terms to order (z/a)6. The result is
Vm(0,0, z) = −8I 53/2
z a−144
54 z
a 5
+O z
a 7
(Ex 5.9.2) The general solution in spherical polar coordinates (assuming the obvious az- imuthal symmetry) may be written
Vm(r, θ) =
A r
a
P(cosθ) (Ex 5.9.3)
forr < a. Along thezaxis the general solution specializes to Vm(z) =
A z
a
(Ex 5.9.4) Comparison of series leads to A1 = −(8I/5)3/2, A2 = A3 = A4 = 0, A5 =
−(8I/5)3/2(−144/54),A6= 0,· · · ; therefore Vm(r, θ) =−8I
53/2 r
a
P1(cosθ)−144 54
r a
5
P5(cosθ) +· · ·
(Ex 5.9.5) where P5(x) = 18(63x5−70x3+ 15x).
In cylindrical polar coordinates with z = rcosθ and r2 =ρ2+z2, the scalar potential of the Helmholtz pair becomes
Vm(r, θ) =− 8I 53/2
z a−144
625 z5
a5 +15 8
zρ4
a5 −5z3ρ2 a5
+· · ·
(5–39)
The components of the magnetic induction field are now easily found, giving Bz(ρ, z) = 8µ0I
53/2a
1−144 125
z a
4 +3
8 ρ
a 4
−3z2ρ2 a4
+· · ·
0
(5–40) and
Bρ(ρ, z) =8µ0I 53/2 · 144
125
4ρz3−3zρ3 2a5
+· · · (5–41) It will be noted that the central field is very homogeneous, a 10% of the radius displacement in any direction leads to only about a 10−4deviation from the field’s central value.
As mentioned before, when charges or currents reside entirely on the boundary, these may frequently be taken into account by the boundary conditions. In the next example we consider how we may account for a current on the bounding surface.
Example 5.10: Obtain the magnetic induction field both inside and outside a uni- formly charged rotating spherical shell.
Solution: The magnetic scalar potential will be of the form Vm(r) =
∞
=0
ArP(cosθ) forr < a (Ex 5.10.1) and
Vm(r) = ∞
=0
B
r+1P(cosθ) forr > a (Ex 5.10.2) The boundary conditions are obtained by integrating Maxwell’s equations at the surface of the sphere. Integrating∇ · B = 0 over the thin volume of Figure 5.7, we have with the aid of the divergence theorem
0 =
τ
∇ · Bd 3r=
S
B ·S = ¯BrextS1−B¯rintS2+ 2πrBt (Ex 5.10.3)
Figure 5.7: A thin pillbox shaped volume whose top and bottom surfaces are, respectively outside and inside the spherical boundary is used to obtain the boundary condition onBr.
where Bt is the mean value of the component of tangential field perpendicular to curved side of the pillbox. Letting the thickness approach zero, the last term vanishes, and S1 approaches S2. As the expression above is independent of the surface area S of the pillbox, we conclude Brext = Brint. The perpendicular com- ponent of the induction field is continuous across the sphere.
Figure 5.8: The loop lies in a plane perpendicular to the current and is traversed by currentlσωsinθ.
Integrating∇ × B =µ0Jover the area Σ illustrated in Figure 5.8 gives
Σ
(∇ × B) ·dS=
µ0J·dS (Ex 5.10.4)
or, with the aid of Stokes’ theorem (dS=−ϕdS),ˆ B ·d=−µ0
JϕdS (Ex 5.10.5)
Following the boundary in the direction indicated in Figure 5.16 we break the contour into four segments to get,
B ·d=lB¯θint−lB¯θext+ 2B¯r (Ex 5.10.6) The bar indicates a mean value of the quantity under it. As we lettend to zero the last term vanishes. On the surface of the sphere, the current density isJ=σδ(r−a)v
=σδ(r−a)rωsinθϕ. The current passing through the thin rectangular area is thenˆ
µ0Jϕ·dS =µ0
a+/2
a−/2
rθ0+l
rθ0
ωσδ(r−a)rsinθ drd
=µ0σaωlsinθ (Ex 5.10.7)
Again, the relation must hold independent ofl, meaning that
Bθext−Bθint=µ0ωσasinθ (Ex 5.10.8)
These boundary conditions are easily translated to boundary conditions on Vm. SinceB =−µ0∇V m, these conditions reduce to
∂Vm
∂r
r=a+
= ∂Vm
∂r
r=a−
(Ex 5.10.9) and
1 a
∂Vm
∂θ
r=a+
− 1 a
∂Vm
∂θ
r=a−
=−ωσasinθ (Ex 5.10.10)
Using the general interior and exterior forms of the potential to evaluate the derivatives ata− anda+, respectively, we write for the latter equation (Ex 5.10.10)
∞
=0
B
a+2 −Aa−1
P(cosθ) sinθ=aσωsinθ (Ex 5.10.11) which requires that
B
a+2 =Aa−1 (Ex 5.10.12)
except when P(cosθ) is a constant in which case= 1 and P1 = 1. Hence for =
1, we obtain
B1 a3 −A1
=aσω (Ex 5.10.13)
The boundary condition involving the radial derivatives (Ex 5.10.9) gives
−(+ 1)B
a+2 P(cosθ) =Aa−1P(cosθ) (Ex 5.10.14) which reduces to
−(+ 1)B
a+2 =Aa−1 (Ex 5.10.15)
Comparison of this term with the one above for which, for = 1, shows that the two can be compatible only if equals−(+ 1). No integer solutions exist leading us to conclude thatA=B= 0 for= 1.
When= 1, the relation (Ex 5.10.15) leads to
−2B1
a3 =A1 (Ex 5.10.16)
Substituting forA1 in the equation (Ex 5.10.13) above, we conclude B1
a3 +2B1 a3
=aσω ⇒ B1= a4σω
3 andA1=−2aσω
3 (Ex 5.10.17) The potentials are then
Vm(r > a) = a4σω
3r2 cosθ (Ex 5.10.18)
and
Vm(r < a) =−2σaω
3 rcosθ=−2σaωz
3 (Ex 5.10.19)
The magnetic induction fieldB is now readily evaluated asB =−µ0∇Vm,
B(r < a) = 23µ0aσωˆk (Ex 5.10.20) B(r > a) = 23µ0a4σω
r3 cosθˆr+13µ0a4σω
r3 sinθθˆ (Ex 5.10.21)
The magnetic induction field inside the sphere above is uniform throughout the entire sphere. While it is not practical to construct an apparatus inside a charged, rotating sphere, the same current density and hence the same magnetic field are achieved by a coil wound around the sphere with constant (angular) spacing between the windings.