Re: Capacitor and Force

On Oct 13, 5:19 pm, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx> wrote:
"The Phantom" <phan...@xxxxxxx> wrote in message

news:nji2h3t085k1vn56l672m34qh80ov8b994@xxxxxxxxxx



On Sat, 13 Oct 2007 17:15:03 -0500, "Jon Slaughter"
<Jon_Slaugh...@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."

Its nice how you take it out of context. BTW, theres nothing in the quote
that says volume integerals are wrong or anything.

It says distributions don't matter and they don't because all are done the
same way... by a volume integeral. Any one is conceptually the same as any
other.

Now maybe I said that when you reduce the plates to a point that it doesn't
matter and I guess it shouldn't have said that... but for an approximation
it shouldn't as you can assume point charges and get a resonable
approximation in most cases(if the two ojbects are far enough apart then it
will work just fine).

Take two plates 1mx1m and put them 100000m apart. Don't you think that for
all practical purposes each plate looks like a point charge to the other?
Why? Because the vector between any two "charges" is approximately if the
plates were pointed...

i.e., 1/100000 is very small. In the case of two plates very close one has
L/r where L is very large compared to r so the ratio is large and the
vectors have a wide range of angles.

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.

Physics is used to describe reality.... everything is a proximation and
there are no mathematical exacts. Hence sometimes getting an order of
magnitude appromixation is good enough.





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."

duh. I don't know what your caught up so much about volume integrals.
Obviously you have to do that in any case. everything involving volume in
volume integeral. thats what the hell volume integerals are for... to do
calculations over volumes...

But let me see you do even a simple volume integeral that is not a special
case found in a calculus book by hand. Now do two nested volume integerals
and lets see how far you get.

The fact of the matter is, you can say what you want but everything in
physics is a approximation... and guess what? All those volume integerals
your caught up on... they were derived by approximations(that is how you
prove integeration in the first placed and that is how all
integerals(lebesque, stieljies, reimann, gauge, etc..) are all defined.

When the is a high degree of symmetry then one usually uses that to simplify
the problem if doing it by hand.

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.

Well, your hung up on volume integerals likes something special. Its basic
calculus. But you have an easy time saying that one should use them... I
challenge you to compute the total force between two parallel plates using
the the full volume integerals(or even surface integerals if you want). You
cannot use the approximation that the site you gave uses where the plates
are infinite, i.e., where E is constant between the plates and there is no
fringe effect.

I'll be waiting for your solution, in full, to the problem. Lets see how far
you really get.
Do you even happen to know what an elliptical integeral is? if not then your
going to have a lot of fun.





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.

Nope, it was actually only off by a factor of 2. I made a mistake in the
logic which is why it was so wrong... that is not hte fault of the
approximation by my fault for using bad logic.

Guess what? even the site you used to prove me wrong, which I have no
problem with that specically as it showed me that I was in error, is using
an approximation.

The problem I have wiht you is that you keep bring up volume integerals like
its a big deal. Its not. Its basic calculus. Almost all volume integerals
are intractable and initially that is what I actually did but couldn't
compete the integeral... even had maple try and it came up with nothing...

but since your a genius with volume integerals I want to see you do it.
(what it sounds like is you like tossing the word around but you have no
experience with them.)



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....

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

Actually I have enough books... but when you can show me your work for the
parallel plates

Heres the integeral you have to compute, I'll write it as a surface
integeral since that is more appropriate for this case.

We will assume the surface charge density is constant to make your life much
easier(of course your not even going to attempt to do them... doubt you ever
actually did a surface integeral in your life)

k*(Q/A)^2*(int(int((dS1 - dS2)/|dS1 - dS2|^(3/2)))

where, of course, the integeral is the differential form of coulumbs law,
i.e., dF = k*dq1*dq2*r/|r|^(3/2).

It sounds to me like you don't know much about what your talking about. True
you see see my problem after you found a site that did a...

read more »

Gee, Phantom hands you the correct answer to your rather gross
misconception on a silver platter, and that's the thanks he gets?
Wow, aren't YOU special!

.

Relevant Pages

  • Re: Capacitor and Force
    ... But if I bring them together, so the like charges are at ... that says volume integerals are wrong or anything. ... approximation in most cases(if the two ojbects are far enough apart then it ... Take two plates 1mx1m and put them 100000m apart. ...
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  • Re: Capacitor and Force
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    ... expression for the force between parallel plates, ... said to my attention so I could correct the approximation but I hope you ... that your volume integerals are not as ubiquiteous as you think. ... You can do the same thing with the integrals but then IT IS ...
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    ... of you tossing the term volume integerals around like it means something. ... said to my attention so I could correct the approximation but I hope you see ... You can do the same thing with the integrals but then IT IS ... but I made a mistake in my formula... ...
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