Static Potentials in Vacuum – Laplace’s Equation
5.1 Laplace’s Equation
5.2.4 Conformal Mappings
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.
Taking ∆zalong the real axis results in df
dz = lim
∆x→0
u(x+ ∆x, y)−u(x, y) +i
v(x+ ∆x, y)−v(x, y)
∆x
= ∂u
∂x+i∂v
∂x (5–43)
whereas taking the derivative in the imaginary direction results in df
dz = lim
∆y→0
u(x, y+ ∆y)−u(x, y) +i
v(x, y+ ∆y)−v(x, y) i∆y
=−i∂u
∂y +∂v
∂y (5–44)
Comparing the two expressions (5–41) and (5–42), we find that the real and imaginary parts of an analytic function f are not arbitrary but must be must be related by
∂u
∂x = ∂v
∂y and ∂u
∂y =−∂v
∂x (5–45)
These two equalities (5–45) are known as theCauchy-Riemann equations. The va- lidity of the Cauchy-Riemann equations is both a necessary and sufficient condition forf to be analytic.
To illustrate these ideas, we consider the function f(z) = z2 = (x+iy)2 = (x2−y2) + 2ixy. Then u = x2 −y2 and v = 2xy. The derivatives are easily obtained
∂u
∂x = 2x, ∂v
∂y = 2x; ∂u
∂y =−2y, ∂v
∂x = 2y in accordance with the Cauchy-Riemann equations.
The Cauchy-Riemann equations can be differentiated to get
∂2u
∂x2 = ∂2v
∂x∂y and ∂2u
∂y2 =− ∂2v
∂y∂x (5–46)
Since the order of differentiation should be immaterial, these can be added to give
∂2u
∂x2 +∂2u
∂y2 =∇2u= 0 (5–47)
In similar fashion
∂2v
∂y2 = ∂2u
∂y∂x and ∂2v
∂x2 =− ∂2u
∂x∂y ⇒ ∇2v= 0 (5–48)
The functionsu andv each solve Laplace’s equation. Any analytic function there- fore supplies two solutions to Laplace’s equation, suggesting that we might look for the solutions of static potential problems among analytic functions. The functions u and v are known as conjugate harmonic functions. It is easily verified that the curvesu = constant andv = constant are perpendicular to one another, suggesting
Figure 5. 9: The image of the rectangular grid in thef plane under the mappingz =
f is shown on the right in thez plane.
that if either u or v were to represent potential, then the other would represent electric field.
To return now to the notion of conformal mappings,14 let us consider the map- ping produced by the analytic function f(z) =z2, (Figure 5.7). The inverse map- ping, z = √
f, is analytic everywhere except at z = 0. The image of the line v
= 0 in the z plane is easily found: for u > 0, z = √
u produces an image line along the positive x axis, while foru <0,z=√
u=i√
−uproduces an image line along the positive y axis of the z plane. The image of v = 1 is easily found from (u+iv) = x2−y2+ 2ixy, implying that 2xy = 1. Similarly, the line u =c has imagex2−y2=c.
Let us now consider the potential above an infinite flat conducting plate with potential V =V0, lying along theu axis. Above the plate, the potential will be of the form V =V0−av. (A second plate at a different potential, parallel to the first would be required to determine the constant a. In a more general problem, V would be a function of bothu andv. V may generally be considered to be the imaginary (or alternatively, real) part of an analytic function
Φ(u, v) =U(u, v) +iV(u, v) = Φ(f) (5–49) For this example, taking V to be the imaginary part of an analytic function Φ, Φ(u, v) = iV0−af = −au+i(V0 −av). The corresponding electric field has components Ev=aandEu= 0.
14The mappings are called conformal because they preserve the angles between intersecting lines except when singular or having zero derivative. To see this, we consider two adjacent pointsz0
andz on a line segment that makes angleϕwith thex axis atz0. Thenz−z0 may be written in polar form as reiϕ. The image of the segment ∆f =f(z)−f(z0) can be expressed to first order asf(z0)(z−z0) =aeiα(z−z0). If we write each term in polar form, ∆f takes the form Reiθ=aeiαreiϕ. Thus the image line segment running fromf(z0) tof(z) makes angleθ=α+ϕ with theu axis. This means that any line passing throughz0 is rotated through angleαin the mapping to thef plane providing that the derivativef(z0) exists and that a unique polar angle αcan be assigned to it. Whenfis zero as it is forf(z) =z2atz= 0, conformality fails.
Figure 5.10: The figure on the left is the mapping of the rectangular grid on the right produced by the functionz= sinf.
Now the mapping that takes the grid lines from thef plane to thez plane also takes Φ to thez plane. Expressed in terms of x andy
Φ (f(z)) =iV0−az2=a(y2−x2) +i(V0−2axy) (5–50) Thus, in thez plane,U(x, y) =a(y2−x2) andV(x, y) =V0−2axy. The latter is the potential produced by two conducting plates at potentialV0intersecting at right angles. The field lines produced byv= constant in theu-v plane are now obtained from u =y2−x2= constant, while the constant potential lines are produced by v
= 2xy = constant. Figure 5.9 illustrates this mapping as it takes line segments and areas from the f plane to thez plane
A second example is offered by the mappingz =sinf, which maps the entire u axis onto a finite line segment of length 2in thez plane. We expressz in terms ofu andv by expanding sinf as
sinf =eif −e−if
2i (5–51)
to obtain
sinf = ei(u+iv)−e−i(u+iv)
2i =e−veiu−eve−iu 2i
= e−v(cosu+isinu)−ev(cosu−isinu) 2i
=isinhv·cosu+ coshv·sinu (5–52) Thusx=coshv·sinuandy=sinhv·cosu.
We again take for the potential above thev = 0 “plane” the imaginary part of Φ(u, v) =V0−af. It is somewhat awkward, however, to find Φ in terms ofx andy directly by substituting forf in the expression for Φ. Fortunately this is not really necessary. For a given point (u, v) in thef plane (whereV is assumed to be known), the corresponding (x, y) value is readily found. In particular, the equipotentials at v = constant are easily obtained, for
x2
2cosh2v + y2
2sinh2v = 1 (5–53)
In other words, the equipotentials around the flat strip are ellipses of major axis coshv and minor axis sinhv. At large distances coshv and sinhv both tend to
1
2ev, meaning the equipotentials tend to circles. The potential and field lines are shown in Figure 5.10. (This problem can also be solved using separation of variables in elliptical coordinates.)
For a last example of somewhat more interest, let us consider the transformation z= a
2π
1 +f+ef
(5–54) We begin by considering some of the properties of this mapping. If f is real, then z is also real, and, asu varies from −∞to +∞, x will take on values from
−∞to +∞. The real axis maps (albeit non-linearly) onto itself. Next consider the horizontal line f =u+iπ. Substituting this into the expression forz gives
Figure 5.11: The infinite “planes” at±iπin thef plane map to the semi- infinite “planes” in thez plane aty =±a/2 on the right.
z= a 2π
1 +u+iπ+eu+iπ
= a
2π(1 +u+iπ−eu) (5–55) We conclude that
x= a
2π (1 +u−eu) and y= a
2 (5–56)
For u large and negative, x au/2π increases to 0 as u goes to zero. As u passes zero, eu it becomes larger than 1 +uand x retraces its path at 12a above thex axis from 0 to −∞. The line in thef plane folds back on itself atu = 0 on being mapped to thez plane. Similarly, the line atv=−πmaps to the half-infinite line at y=−a/2 with negative x. If we let the lines at ±iπ in thef plane be the constant potential plates of the infinite parallel plate capacitor of Figure 5.9, the mapping carries this potential to that of the semi-infinite capacitor on the right.
We can, with this mapping, find the fringing fields and potentials near the edge of a finite parallel plate capacitor. As we will see in the next section, we will also be able find the correction to the capacitance due to the fringing fields.
Letting the plates of the infinite capacitor in thef plane be at potential±V0/2, with the top plate positive, the potential at any point in between the plates becomes V = V0v/2π. The analytic function of which V is the imaginary part is easily obtained as Φ(f) = V0f /2π. Again it is somewhat awkward to substitute for f in
terms of x andy. Instead, we separate the equation forz into real and imaginary components to get
x= a
2π (1 +u+eucosv) (5–57)
and y= a
2π(v+eusinv) (5–58)
which we can evaluate parametrically by varyingu to obtain the constantv (con- sequently also the constant V =V0v/2π) curves. Similarly, the electric field lines are obtained in thex-y plane by varyingv. The field and potential lines are plotted in Figure 5.12.
Figure 5.12:The equipotential and field lines near the edge of a semi-infinite plate capacitor.
Verifying that a given mapping does indeed map a particular surface onto an- other is fairly straightforward. It is not, however, obvious how mappings for given boundaries are to be obtained, other than perhaps by trial and error. For polygonal boundaries, a general method of constructing the mapping is offered by Schwarz- Christoffel transformations. For nonpolygonal boundaries, one must rely on “dic- tionaries” of mappings.15