Unformatted text preview: Figure 13: A ParallelPlate Capacitor
conductor is zero, the only ﬂux will be through the part of our Gaussian surface in between the
two plates, and 2) Since the ﬁeld is uniform, E · dA = E dA. Hence,
E · dA = q ⇒ 0 E= q
0A where A is the area of one of the plates. Now that we know the ﬁeld between the plates, we can
calculate the potential between them.
d d E · ds = V=
0 0 qd
q
ds =
0A
0A Now that we know the potential diﬀerence between the two plates, calculating the capacitance is
simple.
q
q
0A
C=
= qd =
V
d
A
0 Notice that, as stated before, the capacitance is simply a geometrical quantity. In the case of
a parallelplat capacitor, it is proportional to the area of each plate and inversely proportional to
the distance between the two plates.
6.2.2 Cylindrical Capacitor Figure 14: A Cylindrical Capacitor
We begin by assuming that there is a hypothetical charge q placed on the inner and outer
conductor. Because of symmetry, the charge on the inner conductor will distribute itself uniformly
on the outside edge of the cylinder. If we draw our Gaussian surface in the usual way and make
the usual symmetry arguments, then the electric ﬁeld between the plates is
q
E=
r
ˆ
2πrL
0
Hence, the potential diﬀerence between the plates is
b V=
a q
dr =
2πrL
0
19 q
ln
2πL
0 b
a ...
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 Spring '08
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 Physics, Energy, Electric charge, Inductor

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