Re: Capacitor and Force

On Sat, 13 Oct 2007 12:12:33 GMT, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:


"The Phantom" <phantom@xxxxxxx> wrote in message
news:pad0h3df4mvk2fdl348gaf36g70e7slcma@xxxxxxxxxx
On Fri, 12 Oct 2007 22:48:59 GMT, "Jon Slaughter"
<Jon_Slaughter@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:


Yes I am resistant because the formula applies to this situation. I might be
making some mistake in applying it but that is a totally different story
from using the wrong formula.

I know you can use calculus for the general case and that can be done but it
shouldn't result in any significant different in this case(because of the
large degree of symmetry) and actually my result *should* be a lower bound
since I don't take into account any forces acting at an angle.

http://web.utk.edu/~kamyshko/P231/Problem_HW_Chapter24_Force_Between_Plates.pdf

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.



I do not know why the results are so significantly different. Obviously the
formula I wrote above doesn't come out with any reasonable answer and is
several orders of magnitude off. (it should actually be lower) I either
there is a factor wrong or something else is wrong. The logic itself should
work just fine and its exactly the logic that site uses except I do not
worry about charges that are not directly across from any other.

I think that is my problem though. I'm computing Q^2 which gives the
interaction force between every pair of charges but doens't take into
account the angle at which the forces act so its treating it as a if the
force is always constant for every particle pair. So actually I'm guessing
what I should have is k*(Q/A)*(Q/A)/d^2*A (as an approximation.. dimensions
are wrong of course)

in this case I get about 0.006N for the example above. So the real problem,
like a few have mentioned is that its wrong to assume that all the charge is
concentrated at a point because when I do this I'm basically saying the
force is the same over a all charge pairs when its definitely not. By
realizing that the force is really only that strong for one pair and then
gets drastically weaker(which for my approximation I just say its 0) I get a
much more reasonble approximation. But in this case it should be lower and
its still higher ;/

I'll work on it some more and see if I can still apply the coulumbs law or
not. I do think it can be used as a first approximation but it needs to be
used properly instead of blindly like I did.

Of course you can apply Coulomb's law; in fact, you MUST use Coulomb's
law. And, using it properly means integrating over the entire charge
distribution.

Likewise, in the case of gravitational attraction, you MUST start with
the expression for the force between two point masses. It's true that in
the gravitational case the concept of a center of gravity can be used, but
how do you think the location of the center of gravity is found? In
extended bodies with a high degree of symmetry, such as a sphere, it's
obvious where it is. But in the general case, you must integrate over the
volume of the object.

I also think you're making ambiguous use of the phrase "apply a formula".
It's true that in calculating the force between objects, gravitational or
electrostatic, you must start with the appropriate formula for forces
between point masses or charges. But, for example, suppose you wanted the
gravitational force between two frustums of a cone (somewhat asymmetric
bodies). They each have a center of gravity, and once found, the force
between them can be accurately represented by using the distance between
the centers of gravity in the point to point formula, F = G*m1*m2/r^2. But
what I would call a "formula" for the force between these two particular
objects would involve the dimensions of each object; the formula must be
found by carrying out the volume integrals. So, it's true, the point
formula "applies" (in carrying out the appropriate integral), but THE
(final?) formula for the force between two extended bodies is something
more than just F = G*m1*m2/r^2 in the (possibly asymmetrical) gravitational
case or F = k*q1*q2/r^2 in the electrostatic case.

(Actually my result works well
if the distance between the plates is much larger than the size instead of
vice versa).

Thanks,
Jon




.

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