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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
Whatever the potential difference V ≡ φ 2 − φ 1 = −
∫ (^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,
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.
6.1.2 Units of capacitance
In SI units, we measure charge in Coulombs and potential in Volts, so the unit of capacitance is the Coulomb/Volt. This combination is given a special name: the Farad, or F. In cgs units, we measure charge in esu and potential in esu/cm. The unit of capacitance is thus just the centimeter! This is actually kind of cool: objects that are big (have lots of
6.2.2 Parallel plates
Consider a pair of plates, each with area A and separated by a distance s.
Area A
s
Charge −Q
Charge +Q
If the plates are close together (s ø
A), we can approximate the field between these plates as those coming from a pair of infinite planes:
positive negative
2πσ away from 2πσ towards
plate plate
4πσ between plates
0 outside plates
Summing an infinite plane with charge density σ = +Q/A and σ = −Q/A we find
E = 4πσ = 4πQ/A
between the plates, and zero outside of the plates. The field points from the positive plate to the negative plate, so the potential difference from the negative to the positive plate is
V = −
∫ (^) pos
neg
E^ ~ · d~s = 4πσ
∫ (^) s
0
ds =
4 πQs A
Rewriting this in the form Q = CV , we read off the capacitance:
Cplates =
A
4 πs
This is an area divided by a length, so it has the correct units. It’s pretty important to know about capacitance in SI units, since most circuit elements are discussed using things like Volts and Farads rather than statvolts and centimeters. Con- verting is easy: we just need to remember that the electric field in general has a factor of k (the constant from Coulomb’s law) attached to it. Hence, the voltage V likewise picks up this factor; rearranging to solve for the capacitance, we find
Cplates =
k
A
4 πs
In cgs units, k = 1, so this rather trivially reduces to what we had before. In SI units, k = 1/ 4 π≤ 0 , and we find
Cplates =
≤ 0 A
s
Whenever you work out a formula for capacitance in cgs units, you can easily convert to SI by multiplying by 1/k = 4π≤ 0.
6.2.3 Nested shells
Return to the example discussed last lecture: a pair of nested spherical shells,
R
R
Q
Q 1
1
2
2
The potential difference we found in this case was
∆φ = V =
Q 2
R 2
Q 2
R 1
Set Q 2 = Q = −Q 1 :
∆φ = V = Q
R 2
R 1
) .
Rearranging into the form Q = CV , we find
C =
R 1 R 2
R 1 − R 2
6.4 Dielectrics
The capacitors we’ve discussed here are somewhat artificial: the space between their plates is taken to be empty. In the real world, the normal situation is that this space is filled with something — air, plastic, paper, compressed leprechauns, whatever. The “stuff” that fills this space interacts with the electric field, and can have an important impact on the the properties of the capacitor. The basic idea can be understood fairly simply. The key point is that many materials are polarizable: they are made out of molecules that, though neutral overall, have a net excess of + charge at one end and an excess of − charge at the other. This means that each molecule looks like a little electric “dipole” — a distribution whose total charge is zero, but that shows non-trivial charge separation. Normally, the polar character of these molecules is unimportant — the random, mostly thermal motion of the molecules means that at any moment there is no net polarization to a big mass of them. If we take a block of such material and slide it between two parallel plates, it looks (roughly) like this:
+
+
+
+
+
+
+
+ − −
−
− −
−
−
−
Suppose we now charge up these plates, so that there is an electric field between them. We leave these plates with fixed charge. All of these little dipoles will now want to line up — the + ends will align with the negatively charged plate, and vice versa: By lining up in this way, they will tend to reduce the electric field between the plates — the field due to the charges on the plates themselves is superposed upon a field pointing in the opposite direction due to the alignment of the dipoles. For an enormous number of molecules, the reduction in the applied electric field can be summarized using a single number:
Fixed charge situation: Ewith dielectric^ = Ewithout dielectric/K.
The number K is called the dielectric constant; it’s something we can measure and catalog for various substances. Table 10.1 of Purcell lists a whole bunch; the values range from K ' 1 for air to K ' 80 for water. (Water has a huge dielectric constant because of the charge distribution associated with its bent molecules.) For us, the value of K is always greater than or equal to one.
+ +
+
+
+ +
+
+ −
−
−
− −
−
− − −
−
−
−
−
−
+ + + + + +
Since the electric field is reduced by K, the voltage between the plates is likewise reduced. For the parallel plate capacitor and in cgs units, we have
V with dielectric^ =
4 πQs KA
Invoking the definition of the capacitance, we find
Cwith dielectric^ =
KA
4 πs
= KCwithout dielectric^.
It’s worth thinking for a moment about what the dielectric does in some specific circum- stances. Suppose I have a capacitor C whose plates have been charged up to some fixed charge Q. This is easily done by hooking a capacitor up to a voltage source V (e.g., a bat- tery) and then removing the source — the capacitor will then hold the charge Q = CV. If I insert a dielectric between the plates, the capacitance increases; since the charge is fixed, the voltage must decrease:
Fixed charge: V new^ =
Q
Cnew^
Q
KCorig^
V orig K
Suppose instead I take the capacitor and hold it at fixed voltage. This is also easy to do — we hook up our capacitor to a voltage source and just leave it there. The capacitance again increases, which means the charge must increase:
Fixed voltage: Qnew^ = CnewV = KCorigV = KQorig^.
The “extra” charge is actually pulled from the voltage source. Note that in this circumstance the electric field between the plates does not decrease when we insert the dielectric! However, it takes more charge to hold this field than it does without the dielectric — the dielectric “wants” to reduce the electric field; the voltage source needs to supply more charge to “fight” this tendency.
