Re: Capacitor and Force


"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote in message
news:685Pi.1534$wF3.246@xxxxxxxxxxxxxxxxxxxxxxx
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


The formula I have is,
Newtons attract = [8.856e-12 x relative permeability x Area in mtr^2 x
Voltage^2] / [2 x gap in mtrs]
Would seem my 10n 400V multilayer poly cap has 1/4 ounce squeezing it
together at max voltage.
Maybe this is the source of dielectric absorbsion (sp?). E.g. the more rigid
dielectrics will 'spring' together under a voltage strain and relax at
leisure.
Suggests a dielectric with similar cold flow properties to putty would be
ideal. I.e once the plates have been squeezed together a few times then no
more compaction can take place.



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