Re: Capacitor and Force
- From: Tom Bruhns <k7itm@xxxxxxx>
- Date: Fri, 12 Oct 2007 21:22:06 -0700
On Oct 12, 8:22 pm, The Phantom <phan...@xxxxxxx> wrote:
On Fri, 12 Oct 2007 22:48:59 GMT, "Jon Slaughter"
<Jon_Slaugh...@xxxxxxxxxxx> wrote:
Heres the problem, due to John Perry.
My assumption about ceramic caps was wrong. I didn't know it was layered.
Since one has layers the charge is distributed between each layer. Suppose
there are 2n interleaved layers and charge Q. Then each layer has charge
Q/n. Every layer experiences a force between every other but almost all
cancel out.
By assuming all the charge Q was on just one plate seperated by a distance r
was wrong(Which is how jus 2 plates would actually work).
Some initial investigation shows that the layers have a significant effect
in reducing the force. No only is the charge on each plate reduced by Q/n
but the distance between the first and last is increased because of all the
layers inbetween. So treating a ceramic cap as a parallel plate cap is wrong
as I did it. (using CV to get the charge and assuming it was on each plate
because it should be Q/n but I have no idea what n is).
So there are forces there but they are much smaller than what I was
thinking. For a very large parallel plate cap one has C = e*A/d, Q = C*V, F
= k*(Q/d)^2 = k*(e*A/d*V/d)^2 = k*(e*A*V/d^2)^2.
You seem particularly resistant to the notion that you are using the
wrong formula to compute the force between the plates of a parallel plate
capacitor. The formula for the force between point charges won't give the
correct result. You have to divide the plates up into little dxdy pieces,
with each having the appropriate little bit of the total charge on a plate,
and integrate the force between each little bit of charge on one plate and
all the little bits on the other plate. See:
http://web.utk.edu/~kamyshko/P231/Problem_HW_Chapter24_Force_Between_...
and:
http://mysite.du.edu/~jcalvert/phys/caps.htm
The second reference goes through the integration necessary, and even gives
an example of plates of dimensions 10 cm X 10 cm and 1 mm spacing charged
to 300 volts. Their calculated force is .00398 newtons (398 dynes).
Using your formula k*(e*A*V/d^2)^2 and plugging in their numbers:
k = 8.988E9
e = 8.854E-12
A = 10cm * 10cm = .01 m^2
V = 300
d = 1mm = .001 m
we get 6.34 newtons, considerably more than the .00398 newtons they got.
For A = 1/4 m^2, V = 10V, d = 1mm, one gets F = 4N. This is without a
dielectric in a vacuum. I imagine that that if one inserts a
dielectric(which they will probably have to if they want to insulate it)
that it will cut the charge in half because the dielectric sorta acts as a
layer itself. In any case it should be an order of magnitude approximation.
Note that increasing d = 2mm gives F = 0.2N so the F is heavily dependent on
d. (obviously it is d^(-4)). Changing d by a factor of 10 changes F by a
factor of 10000.
Jon
Now there, finally, is a reasonable resolution to the problem, and an
explanation of just why Jon's result was incorrect. As a check, I did
a d(energy)/dx calculation for the 10cm x 10cm, 1mm spacing plates
charged to 300V. My capacitor calculator tells me it will be about
88.54pF, and if the plates are spaced by x, then d(energy)/dx =
C*(V^2)/(2*x) -- which in this case evaluates to 3.984e-6 joules/mm,
or 3.984e-3 joules/meter.
Thanks, "Phantom."
Cheers,
Tom
.
- References:
- Capacitor and Force
- From: Jon Slaughter
- Re: Capacitor and Force
- From: Jon Slaughter
- Re: Capacitor and Force
- From: The Phantom
- Capacitor and Force
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