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Re: [Phys-l] Physics Challenge for Teachers and Students (Feb08)

W/ out doing the work of comparison -- I wonder if using a cell instead of an infinite capacitance caspacitor is not the same. In similar problems Harnwell writes "... the batteries supply twice the change in electrostatic energy." He develops the maths techniques for these problems w/ out directly considering the capacitances, instead only the fields' energies. In his general method he deals w/ various dielectrics including the work and forces relating to their movement in addition to the forces and movements of the conductors.

bc fifty years ago.


On 2008, Mar 15, , at 23:49, John Denker wrote:

On 03/15/2008 09:49 PM, carmelo@pacific.net.sg wrote:

Perhaps most of you agree with ...
http://scitation.aip.org/journals/doc/PHTEAH-home/challenges/ feb2008.pdf

I hope not.

However, I have had an interesting discussion ....

The "problem" lies with differentiating "1/2 CV squared" with respect
to x and claiming this to be the electric force. The "problem" is that
V also depends on x. In order to keep V constant while we change x
(and hence change the capacitance), we will have to change the amount
of charge stored in the capacitor, ....

The energy change in changing x does not completely go towards moving
the plates. It also goes partly to moving the charges. So
"differentiating "1/2 CV squared" with respect to x" is NOT equal to
the electric force between the plates.

<facetious>
This illustrates the principle that when doing a long calculation,
the last mistake should be just big enough to cancel out all the
previous mistakes.

Mindlessly plugging .5 C V^2 into PVW (principle of virtual work)
is not just wrong in principle, it gives the wrong answer. In
particular it differs by a minus sign from the correct answer.
But no problem, if we mindlessly blow off the minus sign we get
the desired answer.
</facetious>

If you want to do real physics, the simplest approach is to consider
a model system. The standard advice applies: draw the diagram already.


----------------------------------------
| |
| |
| |
------- ---------------------------------------
cx C
------- ---------------------------------------
| |
| |
| |
| |
----------------------------------------


The movable capacitor is cx, so called because its capacitance is a
function of x, the separation. The other capacitor, C, is a huge
buffer, a huge reservoir of charge. If C is large enough, the details
of how it works don't matter; we only need to understand C to first
order.

Write down the energy of this circuit as a whole, impose conservation
of charge, and differentiate w.r.t x. Actually if you're clever,
differentiate w.r.t cx to obtain a more general result.

You will find that the contribution from the reservoir is twice the
size of the contribution from cx, and opposite in sign, so that when
you add the two contributions you get the right answer, with the
right sign and everything.

============================================

You might think that it would be fortuitous that the wrong answer would
be equal to the right answer except for the minus sign. But in fact
this relationship is relatively robust.

In particular, it survives even if C is a nonlinear capacitor, i.e. if
it has a nonlinear V-versus-Q relationship. Try it with V(Q) = a + Q/C
for some nonzero intercept (a). Lots of terms cancel, reproducing the
same result as above.

If you want to understand at a deeper level why this should be so,
here's a hint as to one way of understanding it: The minus sign that
we're messing with here is basically the same minus sign that shows up
in the formula for integration by parts.

There are probably other ways of understanding it.

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