Static Potentials in Vacuum – Laplace’s Equation
5.1 Laplace’s Equation
5.2.6 Capacitance
Solution: For this mapping we need two vertices (the vertices at±∞are irrelevant), one to split the line and one to fold the upper ‘electrode’ back onto itself in order to have it terminate. Let the image of the vertex producing the 2πfold lie ataand the image of displacement vertex with 0 vertex angle atb on the real axis in the f plane. The mapping then obeys
dz
df =A(f−a)(f−b)−1 (Ex 5.14.1) Although we have three constants to determine, there is only one significant con- stant in the geometry, namelyd. Therefore any two of the constants may be given convenient values with the third left to fit the correct boundary. For convenience then, we chooseb= 0 and a=−1, resulting in
dz df =A
f+ 1 f
(Ex 5.14.2) which, when integrated gives
z=A(f+ lnf) +B (Ex 5.14.3)
For large positive real f, z is real so that we conclude B is real. For negative real f, z=A(u+iπ+ ln|u|) +B so that we conclude Aπ =d. Letting our fold point u=−1 correspond tox= 0 we obtainB=d/π. The required mapping is then
z= d
π(1 +f+ lnf) (Ex 5.14.4)
Before leaving the example above entirely, we note that the mapping used earlier mapped the semi-infinite plate not from the real axis but from a line above the real axis. Substituting for f the results from example 5.13 will complete the transfor- mation. In particular, letting lnf =wwith w=u+iv we can recast (Ex 5.14.4) as
z= d
π(1 +w+ew)
which mapsw=u+iπonto the top ‘plate’ andw=uonto the x axis. Cascading maps in this fashion is frequently useful.
Figure 5.15:The charge on the lower plate betweenx1, y1andx2, y2is just ε0(U1−U2).
The complex mapping Φ of the potential, as we have seen, has a real partU, perpendicular to the lines of constant potential, parallel to the electric field lines.
The relationship between lines of constantU andE is in fact more complete. In the case of electric field lines, we interpret the density of lines as field strength. Under this interpretation, the amount of charge contained on a conductor bounded by two given field lines is just proportional to the number of lines between the boundaries.
Not surprisingly the density of constant U lines, may also be interpreted as field strength. We therefore surmise that it may be possible to evaluate the charge on a conductor betweenU =U1andU =U2 from the differenceU2−U1.
Let us consider two segments of the equipotential surfaces havingV =V1 and V = V2 (formed perhaps by two plates of a capacitor) bounded by U = U1 and U =U2, as shown in Figure 5.15, extending a distancedz into thez direction (out of the page). (We abandon the use of z as a complex variable using it instead as the third spatial dimension in order that we have surfaces to deal with when we employ Gauss’ law.) With the aid of Gauss’ law, the charge on the lower surface is given by
Q=ε0 x2,y2
x1,y1
E ·dS (5–67)
Writing the electric field asE =−∇V , we express the chargeQ as Q=−ε0
2
1
∂V
∂xdSx+∂V
∂ydSy (5–68)
where we have abbreviated the limits of integration (x1, y1) as 1 and (x2, y2) as 2.
We use the Cauchy-Riemann equations
∂U
∂x = ∂V
∂y and ∂U
∂y =−∂V
∂x (5–69)
to recast (5–68) in terms ofU: Q=ε0
2
1
∂U
∂ydSx−∂U
∂x dSy (5–70)
Figure 5.16: dSx=−dy dzanddSy=dx dz. The decomposition ofdSinto components is not to scale but is intended only to show directions.
With reference to Figure 5.16, dSx =−dydz and dSy =dxdz. (Had we taken an downward sloping segment, we would have had dSx = dydz and dSy = −dxdz.) Substituting fordSxanddSy in (5–70), we have
dQ dz =−ε0
2
1
∂U
∂ydy+∂U
∂xdx
=−ε0
2
1
dU=−ε0(U2−U1) (5–71) Recognizing that the sign is inconsequential, the capacitance per unit length is now easily obtained as
C1−2=ε0U2−U1
V2−V1 (5–72)
We use this result to find the capacitance of a large widthw of the semi-infinite capacitor considered on page 119. The deviation from the infinite capacitor result will yield the contribution of the fringing field to the capacitance. We begin by writing Φ in thef plane, Φ =V0f /2π=V0(u+iv)/2π=U +iV. Equation (5–72) is therefore equivalent to
C=ε0u2−u1
v2−v1 (5–73)
We note from Figure 5.12, any pointx on the plate has two corresponding pointsu, one on the outside surface withu >0, and one on the inside withu <0. To obtain the capacitance of a strip of widthw to the edge of the plate, we chooseu1andu2 on one of the two plates, say the upper plate, corresponding tox=−w. In other words,u1 andu2 are the roots of (a/2π)(1 +u−eu) =−w. When|w/a| 1, the approximate roots areu1∼=−2πwaandu2∼= ln(2πw/a). The capacitance per unit length of a width w (including the edge) of a long parallel plate capacitor is now easily obtained:
C=ε0 w
a + 1
2πln2πw a
(5–74) The term ε0w/a is just the capacitance per unit length of a segment widthw of an unbounded parallel plate capacitor. The remaining term reflects the effect of
the edge on the capacitance. It is now a simple matter to deduce the capacitance per unit length of a finite width (widthspacing) strip bounded on both sides to be
C=ε0 w
a +1 πln2πw
a
(5–75) Moreover, the capacitance of a strip of metal above a parallel ground plane satisfies exactly the same boundary conditions so that its capacitance is also given by (5–75).
Finally, we can get a good approximation of the capacitance of a pair of rectangular plates of dimensionsw×b (eachthe spacinga) to be
C=ε0 wb
a + 1 π
bln2πw
a +wln2πb a
(5–76) The corners, of course, make this only an approximate solution becoming less valid as the plates get smaller.