Re: Capacitor and Force

On Sat, 13 Oct 2007 17:15:03 -0500, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:


<SNIP>

Jon, please consider this simple example; perhaps then you will see
why Phantom's explanation is right on. Consider two point charges,
let's say equal magnitude and opposite polarity, separated by a
distance x. They are attracted with force F. If I have another
identical pair, they will also be attracted by F. So if the two pairs
are at a distance from each other much greater than x, the total force
will be 2*f. But if I bring them together, so the like charges are at
the same points, the force will be 2^2 or 4 times as much. OK? It
really DOES matter how in space the charges are distributed.


I never said it didn't matter.

Cut and pasted right out of your post of 10/12/07 5:18 AM with an obvious
single letter typo fixed:

"If you have charges then you have forces. Some think that the distribution
matters but it doesn't."

volume integerals are not exact.

They are exact if the object being analyzed is a mathematical construct.
If the object is real, like the earth, then we can't do a volume integral
because we don't have a mathematical description of the density at every
point in the interior.

Its a matter of how accurate one wants and
I was just trying to estimate the result. The problem is that I was
calcuating the force between all charge "pairs" which is Q^2 but this is
wrong because its more like Q^2/A. So my estimation was flawed and not the
fact that I can't assume they are concentrated at a point. Its just that
when I do that I should have realized what I was doing which was essentially
making the interacting force the same for all charged particle pairs(this is
not true for plates... just take the opposite corner of the opposite plate).

1 2
* *
* *
* *
* *
* *
* 3

if 1 and 2 are the particles that we are trying to find the force on then
its simply Coulumbs force. But with 1 and 3 its also coulumbs force but the
component in orthogonal to the plate is very small... I did not take this
into account in my approximation. Essentially what I did was treat the force
between 1 and 3 as the same as 1 and 2... hence the grossly over
approxmation. Bu only looking at the forces between directly opposite
charges(like 1 and 2) then its much closer and should be a lower bound.

e.g.,

1 2
3 4
5 6
7 8

and compute the forces between 1 2, 3 4, 5 6, etc.. and add them up. In this
case since they are all the same(because of the approximation) its really
just as if they were point charges but we have k*(Q/A)^2/r^2 = force between
1 and 2. Now adding all the forces on one plate is same as multiplying by A
so that its k*Q^2/A/r^2.

This of course assumes all forces that are at angles, like 1 4, 1 6, 1 8,
etc.. are 0. This is not the case but for small d most of the orthogonal
components are very small.


As Phantom says, the SAME is true for gravitational attraction; two
massive plates separated by x where x is small compared with the
extent of the plates will not be attracted with the same force as two
point masses separated by x.


Hmm, so what your saying is that the force of gravity due to the earth,
i.e., m*g, is wrong?

What your saying is that any time someone wanted to calculate there weight
they would have to compute a volume integeral?

This is simply not true. We assume the earth as a point mass and compute its
force.


What I said was:

"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."

Of course, in the case of a real object such as the earth, we can't do a
volume integral because we don't have a mathematical description of the
density at every point in the extended body; we must simply measure the
force. Having done so, we can treat the problem as though the entire mass
were concentrated at the center of gravity, as I never denied, but plainly
asserted.


e.g.,

F = G*mE*m/r^2 = m*a ==> g = G*mE/r^2


which works

mE = 5.97*10^24 kg,
G = 6.67*10^(-11)
r = 6.356*10^(6)

and gives g = 9.8 m/s^2


So assuming points isn't a wrong idea as you and phantom are trying to make
it out to be. The problem is that it needs to be applied correctly which I
didn't do. Theres nothing wrong with assuming point charges for a first
approximation though... don't know why you guys are trying to make it out to
be bad.

Because it gives answers that are wrong by orders of magnitude? That's
hardly what I'd call even a first approximation.


Again though, if you think computing two nested volume integrals is fun then
by all means go ahead and do it. I'll be waiting for your result.

It's not a matter of fun. It's a matter of getting the correct result.
And you can find a lot of this sort of thing worked out in Roark's
Formulas:

http://www.amazon.com/gp/offer-listing/0071210598/ref=pd_bbs_sr_olp_1/102-8481842-1092946?ie=UTF8&s=books&qid=1192316491&sr=1-1

if you don't want to do it yourself, or if you want to check your own
results.




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