W9_parallel_resonance.eps

150 Week 4: Capacitance

Now just lay down one sheet of foil on the table. Cover it (symmetrically) with the paper. Top
it with the second piece of foil. Tape the foil to the paper on both sides. Congratulations! You’ve
made a capacitor! When the foil is pressed tight to the paper, the gapdis roughly 0.1 mm (a ream
of 500 sheets of printer paper is roughly 5 cm = 50 mm thick) and has an areaA= 0. 12 = 0. 01
square meters. The paper prevents the paper from touching andis more resistant to arcing than 0.1
mm = 10−^4 meters of air!

To compute the capacitance, we have to solve the parallel plate capacitor problem all over again.
Suppose you put a charge±Qon your capacitor (e.g. moving a net chargeQfrom one plate and
putting it on the other). This charge isfree charge, unbalanced charge that distributes itself on
the conducting plates of the capacitor, so perhaps we should refer to it asQfto cleanly differentiate
it from bound charge on the surface of the dielectric paper.

The capacitor plates have an areaA, so the magnitudeσ=Qf/Aand Gauss’s Law tells you that
the magnitude of the field in between the plates if there werenopaper there would be:

E 0 = 4πǫ 0 σf=

σf

ǫ 0

(258)

However, now thereisa dielectric in that space. The field is modified to become:

E=

E 0

ǫr

=

σf

ǫrǫ 0

=

σf

ǫ

(259)

Next, we compute as usual the potential difference:

V=−

∫ 0

d

Qf

Aǫrǫ 0

dz=

Qfd

Aǫrǫ 0

(260)

and the capacitance:

C=

Qf

V

=ǫr

ǫ 0 A

d

=ǫrC 0 (261)

where (note well!) the definition of capacitance involves only thefreecharge on the plates, as that
is the charge we actually moved around charging it and whereC 0 is the capacitance of the same
plate geometrywithoutthe dielectric!

Recall thatǫr>1. We see that the presence of a dielectric between the platesincreases the
capacitancecompare to a vacuum, or air, between the plates,in additionto mechanically separating
the strongly attracting plates and prevenint dielectric breakdown. So what (approximately) is the
capacitance of our homemade capacitor?

That’s left as an exercise, a few seconds work with a calculator. To save you a bit of time for
thisestimate, you can assume that

ǫ 0 = 8. 85 × 10 −^12 ≈ 10 −^11

farads

meter

(262)

and now you can probably do the estimatewithouta calculator!

Before we move on, we need to do one final thing: relate thefreesurface charge that we put on
the actual conducting plates of our parallel plate capacitor with a dielectric to theboundsurface
charge that appears on the polarized dielectric in the resulting field. We can easily do this with
Gauss’s Law or equivalently with our knowledge of the free field and the reaction field in terms of
the surface charges.

In figure 48 we can write the field in the dielectric in two ways:

E=

E 0

ǫr =E^0 −Er (263)