136 Week 4: Capacitance
At the edges of the plate, the field “bulges” out from between the plates and forms curved field
lines that resemble those of an electric dipole (because after all, theplatesdoform a sort of dipole).
This “fringing field” rapidly falls off in magnitude compared to its strength between the plates, and
in this course we willalwaysidealize this by asserting that the field “vanishes” at and outside of the
edges of the plates and is perfectly uniform in between, even though this isn’t precisely true. This
situation is portrayed in figure 40
With the fields in hand, it is but the work of a moment to compute the potential difference of
the upper plate relative to the lower (or vice versa):
V= ∆V=−∫d0Ezdz=− 4 πkeσd=−Qd
ǫ 0 A(220)
Note that the integral we computed isnegative, which simply means that the upper plate is at a
lower potential than the lower plate (consistent with the field pointing from the lower to the upper
plate).
We are ready to form the capacitance. Our potential difference is negative, but when we form the
capacitance we by convention make it a positive number – obviously the capacitance is symmetric
and we can charge the plates in either direction, so there is no point ingiving it a sign. We
correspondingly form:
C=|Q|
|V|= QdQ
ǫ 0 A=ǫ^0 A
d(221)
Note well the dependence of thisarchtypicalcapacitance on the dimensions of the capacitor. The
dielectric permittivity of free spaceǫ 0 appears on top and clearly has SI units (above others) of farads
per meter. The capacitance varieswiththe cross-sectional area of the facing plates andinversely
withtheir separation. Bigger plates (more area) means bigger capacitance; closer plates (smaller
separation) also means bigger capacitance.
This is an important enough result that you should probably try to remember itas wellas being
able to derive it in detail, following all three steps outlined above. Notethat this is agreatproblem
to practice because thisoneproblem requires you to use Gauss’s Law for the electric field, the
superposition principle, the definition of potential (difference) in terms of an integral of the field,
the definition of capacitance, and a certain amount of common sense as far as idealization of the
plate fields and the self-consistent distribution of charge in static equilibrium.
We’ll now quickly indicate the key step for cylindrical and spherical capacitors, but without
presentingallof the steps. Your very first homework problem is to fill in the missingstepsyourself,
creating “perfect” derivations of the capacitance for conducting plates with all three Gauss’s Law
geometries. Don’t forget to draw your own figures!
Example 4.1.2: Cylindrical Capacitor
Given two concentric cylindrical conducting shells of lengthLand radiiaandbsuch thatδ=b−a≪
L, find their capacitance.
As before, assume that they are charged up to +Qon the inner and−Qon the outer by means
of our little blue devil dude and his charged-particle pitchfork. This puts a charge per unit length
of±λ=±Q/Lon the inner and outer shell, respectively. From Gauss’s Law it is easyto show that:
Er=^2 keλ
ra < r < bandEr= 0 otherwise (idealizing by neglecting the fringing fiends that might exist at the ends of
the cylinders). Then:
V= ∆V=−∫baErdr=− 2 keλln(
b
a)
=−
1
2 πǫ 0Q
Lln(
b
a