W9_parallel_resonance.eps

134 Week 4: Capacitance

whereV is the potential difference across the arrangement as a function of the common chargeQ
used to create it. In the case of our example, the capacitance of an isolated conducting sphere is:

C= 4πǫ 0 R (217)

In general theSI unitsof capacitance are easily remembered (as always) from the defining
relation:
1 Farad =1Coulomb1Volt

which we shouldalsorecognize as being the natural units ofǫ 0 (or 1/ke) times alength.

Although we might have occasion to refer to the capacitance of an isolated conductor used (for
example) as the storage ball on a VandeGraff generator, we willalmost alwaysuse capacitance in
the context ofspecific arrangementsoftwo conductorsthat are designed and intendedjustto store
charge in this way. Those three arrangements are:

• Aparallel platecapacitor. This is our template model, and you should thoroughly learn it
  as it is quite simple and informative.


• Acylindrical shellcapacitor.


• Aspherical shellcapacitor.

The latter two are primarily useful as teaching models, as you know everything you need to know
in order to compute their capacitance from Gauss’s Law and the definition of potential difference.
Let’s examine these three cases in some detail.

Example 4.1.1: Parallel Plate Capacitor

+Q

−Q

d

A

Figure 39: An “ideal” parallel plate capacitor of cross-sectional areaAand plate separationd.

In figure 39 you can see the archetype for all capacitor problems.Two parallel conducting plates
are arranged so that they are separated by asmallinsulating gapd(which may or may not be
filled with a dielectric material, see section on dielectrics below). A metaphorical “blue devil” armed
with a metaphorical micro-pitchfork (that is, a still undefined process we will discuss later) forks
up charge from one plate and shoves it, working against an ever increasing electric field, over to
the other plate, eventually creating (after doing an amount of work that we will of course calculate
shortly) the situation portrayed, with a charge +Qon the lower plate and−Qon the upper plate.
We will invariably assume that a charged capacitor has thesame magnitudeof opposing charges on
the two plates – in the static limit this is an exact result^43.

We wish to compute the capacitance, showingall the steps. We proceed as follows:

(^43) Why? Consider the properties of a conductor in electrostatic equilibrium, which requires perfect cancellation of
the fields inside the conductors just inside the opposing surfaces...