Problems
Application 4.5 The Dielectric Polarization P(r)
5.4 Capacitance
5.4.3 The Capacitance Matrix
A modern integrated circuit contains thousands of metallic contacts. Acapacitance matrixCdescribes how the contacts influence one another electrostatically. The matrix elements Cij (which must be measured or calculated) generalize (5.36) by relating the conductor charges Qi to the conductor potentialsϕj for a collection ofNconductors:
Qi= N j=1
Cijϕj. (5.42)
To see that (5.42) is correct, Figure 5.8 shows a conductor with potentialϕ1in the presence ofN−1 other conductors held at zero potential by connection to ground. The charge on conductor 1 associated with ϕ1 induces each zero-potential conductor to acquire a net charge of its own drawn, up from ground.6Specifically,
Q(1)1 =C11ϕ1 Q(1)2 =C21ϕ1 · · · · Q(1)N =CN1ϕ1, (5.43)
6 Technically, we need a theorem from Section 7.3 to guarantee that the charge drawn up to each grounded conductor is uniquely determined by the potentialϕ1.
5.4 Capacitance 137
1 0
3 0
4 0
2 0
Figure 5.8: Sketch of four conductors. Conductor 1 is held at a non-zero potentialϕ1. The other three conductors are held at zero potential. A few representative electric field lines are indicated for the case whenϕ1>0.
where thecoefficients of capacitanceC11, C21. . . , CN1 depend only on the shapes and geometrical arrangement of the conductors. If the wires used to ground the conductors are sufficiently thin, the charges and potentials in (5.43) do not change if we remove these wires after static equilibrium has been reached.
Now consider a second situation where conductor 2 is held at potentialϕ2=0 while all the other conductors are held at zero potential by connection to ground. The same considerations apply as above (including the removal of the physical connections to ground at the end) and we infer that the charges on each conductor are
Q(2)1 =C12ϕ2 Q(2)2 =C22ϕ2 . . . . Q(2)N =CN2ϕ2. (5.44) Repeating this scenarioN−2 times, each time grounding only one conductor, producesN−2 more expressions like (5.43) and (5.44) for the charge on each conductor. The final step is to superpose all of the foregoing solutions to produce a situation where conductor 1 has potentialϕ1, conductor 2 has potentialϕ2, etc. We get the desired formula (5.42) by straightforward addition because the charge on thekth conductor is
Qk=Q(1)k +Q(2)k + · · · +Q(Nk ). (5.45) The diagonal elementsCkkof the capacitance matrix arenotsimply the self-capacitances discussed in Section 5.4. This can be seen from Figure 5.8 where the charge induced on the grounded conductors influences the potential of the non-grounded conductor. The off-diagonal elementsCkjare calledmutual orcross capacitances. The most important property of these quantities is that they are symmetric in their indices:
Cij =Cj i. (5.46)
We prove (5.46) using Green’s reciprocity relation (Section 3.5.2), which relates a charge distribution ρ(r) and its potentialϕ(r) to an entirely different charge distributionρ(r) and its potentialϕ(r):
d3r ϕ(r)ρ(r)=
d3r ϕ(r)ρ(r). (5.47)
For a collection ofNconductors, reciprocity relates a situation where the conductors possess charges Qiand potentialsϕito an entirely different situation where thesameset of conductors possess charges
CUUK1954-05 CUUK1954/Zangwill 978 0 521 89697 9 August 10, 2012 9:15
138 CONDUCTING MATTER: ELECTROSTATIC INDUCTION AND ITS CONSEQUENCES
Qiand potentialsϕi. This reduces (5.47) to N i=1
ϕiQi= N
i=1
ϕiQi. (5.48)
On the other hand, (5.42) applies to bothQiandQiin (5.48). Therefore, N
i=1
N j=1
ϕiCijϕj = N
i=1
N j=1
ϕiCijϕj. (5.49)
Exchanging the dummy indices on one side of (5.49) proves that (5.46) is correct.
The elements of the capacitance matrix satisfy theMaxwell inequalities:
Ckk>0 Ckj <0
!
j
Ckj ≥0.
(5.50)
These inequalities can be understood from a version of Figure 5.8 where thekth conductor is held at unit positive potential and all the other conductors are held at zero potential. For this situation, all the electric field lines must begin on conductorkand end on one of the other conductors (or at infinity).
In particular, no field lines directly connect any other pair of conductors. The inequalities (5.50) follow from the facts thatQm=Cmkand that the number of fields lines that leave each conductor is proportional to the total charge on that conductor (Gauss’ law). We get!
j Ckj =0 only when no field lines escape to infinity.
Example 5.3 Figure 5.9(a) shows a point chargeqa perpendicular distancez0 from one of two parallel and infinite conducting plates separated by a distanced. Use Green’s reciprocity to find the amount of charge drawn up from ground by each plate.
L 0 R 0
q
QL QR
d
z0
′ 0
L ′
R V
′ 0
q
′L
Q QR′
d
z0
(a) (b)
( )z0 ′(z0)
Figure 5.9: Application of Green’s reciprocity relation to find the chargesQLandQRinduced on two infinite, grounded plates by an interposed point chargeq.
Solution: Treat the point charge in Figure 5.9(a) as an infinitesimal spherical conductor placed at z=z0with potentialϕ(z0) and chargeq. The planes atz=0 andz=dhave chargesQLandQR
and potentialsϕLandϕR, respectively. If we use the same notation for the “primed” charges and
5.4 Capacitance 139
potentials, the reciprocity relation (5.48) withN=3 reads
ϕLQL+ϕRQR+ϕ(z0)q=ϕL QL+ϕR QR+ϕ(z0)q.
The trick to using reciprocity is to choose physically sensible values for the primed variables in order to isolate the desired quantities and eliminate the unwanted quantities from the preceding equation. Sinceϕ(z0) is unknown, we chooseq=0 and fix the potentials on the plates atϕL =0 andϕR =V [see Figure 5.9(b)]. SinceϕL=ϕR =0, these choices reduce the preceding equation to
0=V QR+ϕ(z0)q.
By translational invariance, the charge density is uniform on the primed plates. This means thatEis constant andϕ(z)=Az+Bbetween the plates. The boundary valuesϕL(0)=0 andϕR(d)=V fix the constants so ϕ(z0)=z0V /d. This determines QR. A second comparison system with ϕL =V andϕR=0 determinesQL. Altogether, we find
QL= −"
1−z0
d
#
q and QR= −z0
d q.
The charge induced on each plate is directly proportional to the distance betweenqand the other plate. The total charge drawn by both plates is−q. This implies that every electric field line that begins on the point charge ends on one of the plates. We will explore the generality of this result in Chapter 7.
Example 5.4 N+1 identical conductors labeledm=0,1, . . . , Nare arranged in a straight line.
The capacitance matrix is
Cij =
⎧⎨
⎩
C0 i=j,
−C |i−j| =1, 0 |i−j| >1.
Suppose conductorm=0 at the end of the line has chargeQand all the other conductors are uncharged. Find an approximate expression for the potentialϕmon every conductor in the limit CC0.
Solution: The charge-potential relations (5.42) are Q=C0ϕ0−Cϕ1
0=C0ϕm−C(ϕm+1+ϕm−1) m=0, N 0=C0ϕN−CϕN−1.
The last equation isϕN =(C C0)ϕN−1. Substituting this into the middle equation withm=N−1 and usingCC0produces
C0
C − C C0
ϕN−1=ϕN−2 or ϕN−1≈(C C0)ϕN−2.
Repeating this argument givesϕm=(C C0)ϕm−1. Therefore,ϕm=(C C0)mϕ0.ButQ=C0ϕ0
in the same limit, so
ϕm≈ Q C0
C C0
m
= Q C0
exp
−mln(C0 C) .
CUUK1954-05 CUUK1954/Zangwill 978 0 521 89697 9 August 10, 2012 9:15
140 CONDUCTING MATTER: ELECTROSTATIC INDUCTION AND ITS CONSEQUENCES
The Triode
Before semiconductor technology made low-power vacuum tubes obsolete, thetriodeperformed many of the same functions as the transistor. In this device, a current in the form of a stream of electrons flows through an enclosed vacuum between a heatedcathodeand ananode. The figure below is the circuit diagram symbol for a triode. The essential feature is that a voltage applied to a metalgridcontrols the magnitude of the current.
Cathode Grid Anode
−e
The operation of the triode can be understood qualitatively as follows. LetϕC, ϕG,andϕAbe the potential (voltage) of the cathode, grid, and anode, respectively. According to (5.42), the charge QCon the cathode is
QC=CCCϕC+CCGϕG+CCAϕA.
This equation shows that the grid voltage modulates the charge on the cathodeQCregardless of the potential difference between the cathode and the anode. From (5.14),QC is proportional to the field strengthECjust outside the cathode. This field, in turn, determines the acceleration (and hence the current) of electrons toward the anode. Note that the physical placement of the grid betweenthe cathode and the anode in the figure above is irrelevant to this argument.