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lec06

# lec06 - Scott Hughes Massachusetts Institute of Technology...

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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 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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lec06 - Scott Hughes Massachusetts Institute of Technology...

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