Scott Hughes
17 February 2005
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2005
Lecture 6:
Capacitance
6.1
Capacitance
Suppose that I have two chunks of metal. A charge +
Q
is on one of these chunks,

Q
is
on the other (so that the system is neutral overall). Each chunk will be at some constant
potential. What is the potential difference between the two chunks of metal?
+Q
Q
2
1
+
+
+
+
+
+






Whatever the potential difference
V
≡
φ
2

φ
1
=

R
2
1
~
E
·
d~s
turns out to be, it must of
course turn out to be independent of the integration path. In fact, as we can show with a
little thought, the potential difference
V
must be proportional to the geometry. This means
that the potential difference (or “voltage”) between the two chunks must take the form
V
= (Horribly messy constant depending on geometry)
×
Q .
The horrible mess that appears in this proportionality law depends only on geometry. In
other words, it will depend on the size and shape of the two metal chunks, their relative
orientation, and their separation. It does
not
depend on their charge.
This constant is defined as 1
/C
, where
C
is the
Capacitance
of this system. A system like
this is then called a
capacitor
. The 1
/
may seem a bit wierd; the point is that one usually
writes the capacitance formula in a different way: we put
Q
=
CV .
A key thing to bear in mind when using this formula is that
Q
refers to the charge
separation
of the system. In other words, when I say that a capacitor is charged up to some level
Q
,
54
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Figure 1:
Q
=
CV
.
I mean that I have put +
Q
on one part of the capacitor,

Q
on the other.
The capacitor
as a whole remains neutral!!
I emphasize this now because I often find students are some
what confused by this point, particularly when we start thinking about circuits that have
capacitors in them.
6.1.1
Why are the charge and voltage proportional?
The real concern here is, how do we know that when we say double the charge on the
capacitor, the charge doesn’t move around and cause the dependence to be more complicated?
The simple answer is the principle of superposition:
whenever we double the amount of
charge, the fields that these charges create simply double. Hence, the potential difference
(which is just the line integral of the field) must also double.
There’s an implied assumption here, though: we are assuming that there is enough charge
smeared out on the “plates” of the capacitor to uniformly cover them. Imagine we just had
one electron on the minus plate, and one electron “hole” on the plus plate. If we just doubled
the charge, the electron on the  plate indeed might move somewhere else. This situation
could be messy!
Fortunately, in every situation that is of practical interest, the assumption that the plates
are evenly coated with a charge excess is a good one. This is because the elementary charge
is so small — even if there’s just a tiny tiny amount of charge — say 10

12
Coulombs —
that means we’ve got something like 6 million excess electrons.
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 Spring '10
 ScottHughes
 Charge, Potential difference, Electric charge, Farad

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