W9_parallel_resonance.eps

Week 4: Capacitance 137

This is negative because we integrated from inside out (in the direction of the field). We could just
as easily have integrated from outside in and gotten a positive potential difference. As always, the
only thing that matters is that the potential must decrease when moving in the direction of the field.

The capacitance is now easy:

C=

Q

V

=

2 πǫ 0 L

ln

(b

a

) (223)

which has the right units –ǫ 0 times a length. Still, it isn’t at all obvious that this has the limiting
form ofǫ 0 A/d. You are asked to show that it does, after all, have this form for homework. You
might want to remember that ln(1 +x)≈xforx≪1 is the limiting form of the power series
expansion for the natural log function when you get to this part ofthe first problem.

Example 4.1.3: Spherical Capacitor

Similarly, we can do two concentric spherical conducting shells of radiusaandb, charged to±Qon
inner and outer shell respectively by our intrepid devil. From Gauss’sLaw:

Er=

keQ

r^2

a < r < b

andEr= 0 otherwise, withnoidealization or fringing fields. From this we trivially find:

V = ∆V=−

∫a

b

Erdr

= keQ

{

1

a

−

1

b

}

= keQ

{

b−a

ab

}

=

1

4 πǫ 0

Q

{

b−a

ab

}

(224)

This time I cleverly integrated from the outside in,recognizingthat this would give me a positive
potential difference as I integrateagainstthe direction of the field. Now finding the capacitance is
easy:

C=ǫ 0

4 πab

b−a

(225)

where I’ve deliberately arranged it this way as a hint as to how to proceed to answer the “limiting
form” part of the first homework problem.

4.2: Energy of a Charged Capacitor

It’s time to compute how much work our little devil dude does shovellingcharge from one plate over
to the other. Imagine that he starts with the plates uncharged. The first pitchfork full of charge
∆Qthat he moves over is “free”. There is no field to push against yet. The second one, however,
he must push against the field of the first one. The third one he mustpush against the field of the
total charge of the first two. And so on.

Suppose he has been shovelling for a while on a capacitorC(where the particular geometry of
the capacitordoes not matteras long as we know the capacitance) and at this moment the total
charge on capacitor plates is±Q, so that:

V=

Q

C

(226)