Re: Capacitor and Force


"The Phantom" <phantom@xxxxxxx> wrote in message
news:36l0h39u2uugnt6c8v2v90atpoe4r0v9v0@xxxxxxxxxx
On Fri, 12 Oct 2007 22:07:15 GMT, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:


"Tom Bruhns" <k7itm@xxxxxxx> wrote in message
news:1192219843.504190.120500@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Oct 12, 12:48 am, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx>
wrote:
Agreed. Getting the right idea about what's going on usually means
eliminating a lot of wrong - though superficially plausible - ideas.
Nobody talks much about the process of getting rid of the bad ideas,
but it always takes a while.

I think John Perry hit the nail on the head. I don't think anything is
wrong
with my analysis except I misunderstood how a ceramic capacitor worked.
I'm
going to try it with just two large plates and see what happens and if
it
doens't work then I'll go back and look at my calculations. (I suppose
I
should have tried that in the first places but large plates are harder
to
manage)

Thanks,
Jon

As a check on your calculations, you might try calculating it in a
different way. For example, if you move a plate distance dx, and the
capacitance changes by dC in the process, and you still have the same
charge on the capacitor, you can (1) calculate the voltage change,
then (2) calculate the energy change, then (3) knowing the distance
and the energy change, calculate the force. Of course, the answer
should come out the same as your calculation concerning the force
attracting charges. Often, a good way to discover "wrong" ideas is to
try to arrive at the answer by as different a path as you reasonably
can.

well, all your doing is taking the derivatives then plugging them in. Its
a
different path to another calculation that should work out too but if my
formulas are wrong in the first place or my logic is flawed then chances
are
I would either get the right answer but still be wrong or get the wrong
answer and still be wrong ;)

I see what your saying though but it is a elementary problem and I
seriously
doubt my calculations are wrong(I use maple to do them and I've done them
enough that any chance of it being wrong is pretty low). Now I might be
misapplying the formula but again, its pretty basic. If you have charges
then you have forces. Some thing that the distribution matters but it
doesn't. It is very similar to gravity(in fact it is equivalent except for
scale).

It isn't equivalent. For one thing, there is no such thing as shielding
with respect to gravitation. The presence of a *** of lead between the
earth and moon would have no effect on the forces the earth exerts on the
moon; the forces exerted by the *** of lead would be additive to the
earth's forces. But, if the moon and earth were oppositely charged, a
conducting shield between them would alter the electrostatic forces the
earth would exert on the moon.


No, no one is talking about shielding... and if there was anti-gravity then
you could have shielding and pretty much equivalency between the two.

F = G*m1*m2/r^2
F = k*q1*q2/r^2

that is why the forces are similar and all calculations involved are
similar. The only difference is that the m's have to be positive(unless
there is anti-gravity) while the q's can be possitive or negative. My point
is not the effect of that difference but the mathematics is the same.



Consider further that there is no such thing as a Faraday shield in
gravitation. In electrostatics, there is no electric field inside a
hollow
conducting container. Even with a tiny hole in it to give access to the
inside, this is essentially true, but such an effect does not exist for
gravity. It's true that inside a hollow sphere made of some isotropic
material, there is no gravitational field, but this is only true for a
sphere. Inside a hollow non-spherical isotropic object, there is a
gravitational field. But in the world of electrostatics, it doesn't
matter
what shape a hollow container has, there's no electric field inside. The
distribution of charge on the outside matters.


The reason is beause charge is much more mobile then gravity and because
there is polarity with charge. This doesn't mean that the calculations are
fundamentally different and this is what I was getting at when doing the
calculations. The superposition principle holes for both. Newton's law
holds, differential calculus holds... and there all the "same" because
functionally one has C/r^2 and thats whats important. C is different for
the two cases but its also different for different systems but all the
calculations will be functionally identical and if you can do one you can do
the other.



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