W9_parallel_resonance.eps

Week 4: Capacitance 135

a) Compute the electric field at all points in space, but in particular in between the plates,

using a mix of Gauss’s Law and the superposition principle. The field will, of course, be

directly proportional toQ. We will idealize the field at the edges of the plates, something

that is permissible ifd≪

√

Aand that in any event will not substatively affect their potential

difference.

b) Compute the potential difference between the plates. Like the field, this will depend on the

chargeQtransferred from one plate to the other. Note well that we will always be computing

a potentialdifferencebut we will often be lazy and write it asV, not bothering to add the ∆

as in ∆V. It just makes the algebra a bit simpler, and keeps us from having todo the same

thing forQvs ∆Q.

c) Form the capacitance,C=Q/V. Note that theQwillalways cancel outand leave us with

something that depends onǫ 0 and thegeometricparameters of the plate. Pay close attention

to the dimensions and units, as you will need to be able to tell if your answers to problems

“make dimensional sense” on the fly!

So here are the steps. First we note that the charges distribute themselves (approximately)
uniformly on the facing surfaces of the two plates, getting as closetogether as they can. This forms
two equal and opposite sheets of charge with charge per unit area±σ=±Q/A. Applying Gauss’s
Law to either one of them, say the lower, we get:
∮

S

E~·nˆdA = 4πkeQinS

|Ez| 2 A =

σA

ǫ 0

Ez=

σ

2 ǫ 0

= 2πkeσ (218)

(pointing away from the sheet of charge above and below it). We getexactly the same for the upper
plate, except that the field pointstowardthe negative sheet of charge.

We then apply the superposition principle. Above and below both sheets, the fields produced by
the upper and lower chargescancel, as e.g. field from the upper one points down and the field from
the lower one points up, and the fields have equal magnitudes. In between the plates, the field from
the upper plate points up and so does the field from the lower one – the two fieldsadd. Thus we
obtain a total field of:
Ez= 4πkeσ=

σ

ǫ 0

(219)

directedupwardsbetween the plates, as drawn, andEz= 0 above and below the plates. Note well
that this field is automagicallyzeroinside the conducting metal of the plates themselves and in the
wires above and below the plates! Our assumption of charge distributing itself in two uniform sheets
isconsistentas it leads to the field vanishing inside the conductor, as we expect.

Actual Ideal

Figure 40: Fringe fields at the edge of an actual pair of parallel plates carrying opposite charge
compared to the idealized field that vanishes sharply at the edge andis uniform in between the
plates. Note that the field, and hence the potential difference, is almost identical in most of the
volume between the plates.