Re: Capacitor and Force


"Jim Thompson" <To-Email-Use-The-Envelope-Icon@xxxxxxxxxxxxxxx> wrote in
message news:7bppg397rjiopa21p4agncupihm89j1moq@xxxxxxxxxx
On Wed, 10 Oct 2007 09:15:30 -0500, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:

I'm trying out a little experiment but its not working ;/

Since

Q = C*V

and

F = k*Q1*Q2/r^2,

The force one parallel plate capacitor due to the charge is

F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N. (assuming r ~=
1mm)

I've tried this with two pennies and a piece of paper(as the
dielectric/insulator) which gives about 20pF and about 1nF(theoretically
and
to small for me to measure).

I grinded down one side of 40nF ceramic plate capacitor and a penny and it
has a total of 40nF but I experience no force when moving the plates close
together(which is surely < 1mm). But why? Surely since they act as a 40nF
capacitor and there is 20V across is then there should be a significant
force between the two. A large enough force to feel at when trying to
seperate the two?

But this isn't the case so I must be wrong either in theory or
application.
The theory is pretty straight forward and even if my calulations are off
by
1000 I probably should feel some force between the two but as far as I can
tell there is nothing. Not sure about the application either as it also
is
pretty straight forward(the main problem is getting a large enough
capacitance in a simple way).

Any ideas where I went wrong?

Thanks,
Jon



Trying to remember some fundamentals...

Isn't Force = dE/dx

change of energy with spacing?



lol. No. F = q*E. E = grad(V) which is probably what you were thinking.


.

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