W9_parallel_resonance.eps

350 Week 10: Maxwell’s Equations and Light

Let us now guess that this invariant current is thecorrectone to use in Ampere’s Law,

and see if it gives us the right answer in at least one problem where weknowthe answer

and Ampere’s Law as it was before got it wrong. That is, suppose Ampere’s Law is

really:

∮

C

B~·d~ℓ=μ 0

{∫

S/C

J~·nˆdA+ǫ 0 d

dt

∫

S/C

E~·nˆdA

}

(824)

Note well the location of the brackets: theμ 0 isoutsideof them, and everything inside

of them has the units of current.

If we use this expression to computeIin our capacitor problem above, when we compute

theinvariantcurrent throughS 1 we still getI(because the field due to the capacitor is

confined to live in between the plates of the capacitor and doesn’t pass throughS 1. If

we apply it to the surfaceS 2 , no physical current gets through, but thefield inside the

capacitoris (recall):

E=

Q

ǫ 0 A

(825)

whereAis the area of the capacitor. The integral:

∫

S 2 /C

E~·ˆndA=EA=Q

ǫ 0

(826)

and hence

Iinv=ǫ 0

d

dt

∫

S 2 /C

E~·ˆndA=EA=dQ

dt

=I (827)

becauseIis the rate at which the capacitor is charging! We get the sameI for both

surface! We therefore get the same (correct) magnetic field aroundCfrom both surfaces.

The extra term we have added to the physical current was originallyadded by James

Clerk Maxwell, and the implications of this term were so profound, so overwhelming,

that the entire set of equations (and the term itself) were named inhis honor. It is

called theMaxwell Displacement Current:

IMDC=ǫ 0 d

dt

∫

S 2 /C

E~·ˆndA (828)

From now on we will assume that equation 824 is the actual, correct form for Ampere’s

Law, the one that will always give the right answer, the law of nature. As we’ve seen,

for many “static” problems where there is no time-varying electric field we can use the

old form without error, but it won’t work when charge is building up andthe electric

field is varying.

In fact, there is one very important place where it fails. It fails to describe the magnetic

fieldinsidethe parallel plate capacitor. Let’s work that out as an example.

Example 10.0.1: The Magnetic Field Inside a Parallel Plate Ca-

pacitor

In figure 140 we see a parallel plate capacitor with cylindrical symmetry being charged by

a (momentarily) steady currentI. As charge flows onto the capacitor, the field (assumed

as usual to be strictly confined to be between the two plates, ignoring the fringe) increases

uniformly. This increasing field creates an increasing flux through cylindrically symmetric

Amperian loops of radiusrin between the plates, generating a magnetic field there. Our

job is to evaluate this field, both between the plates and in free space outside of the

plates (but in the plane that separates them).