W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 351

I

C

R

r

(r > R)

Figure 140: A capacitor made up of two circular disks is being chargedby a currentI. The
increasing electric field between the two plates becomes aMaxwell Displacement Currentthat creates
a magnetic field identical to the one that would exist inside a uniform conductor of the same radius
(assuming the conductor had a magnetic permeability and electric permittivity identical to the
vacuum value, not really a very good assumption).

This description is a perfect recipe for our algebraic work, yet another example of how a

verbalunderstanding of the physics plus knowledge of the laws and ability todo relatively

simple math suffices to enable one to solve problems that at first glance are quite difficult.

We imagine that at some timetthe capacitor has a total chargeQ(t) on it such that

I=dQ/dt.

Then (from Gauss’s Law):

E=

σ

ǫ 0 =

Q

ǫ 0 A r < R (829)

(from left to right – remember that the field is a vector) andE= 0 forr > R. This just

represents in an equation and a solution that by now should beveryfamiliar to you the

first step in the recipe above.

Second, we have have to evaluate the flux through the Amperian pathC(forr < R) in

figure 140:

φC=

∫

S/C

E~·nˆdA=EA=Qπr

2

ǫ 0 A

=

Qπr^2

ǫ 0 πR^2

(830)

(where we have usedA=πR^2 at the end).

Third, we have to write Ampere’s Law for this Amperian path:

∮

C

B~·d~ℓ=Bφ 2 πr=μ 0

(∫

S/C

J~·nˆdA+ǫ 0 d

dt

∫

S/C

E~·nˆdA

)

(831)

J~= 0 (no actual current flows through the insulating vacuum between the plates) and

the only thing that varies with time in the flux is the chargeQ, so this becomes:

Bφ 2 πr=μ 0

dQ

dtr

2

R^2

=

μ 0 Ir^2

R^2

(832)

We rearrange this to obtain half of our answer:

Bφ=

μ 0 Ir

2 πR^2

r < R (833)