Re: Capacitor and Force

On Oct 13, 9:54 pm, The Phantom <phan...@xxxxxxx> wrote:
On Sun, 14 Oct 2007 00:19:18 GMT, "Jon Slaughter"

<Jon_Slaugh...@xxxxxxxxxxx> wrote:

<SNIP>





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 similar example but
from your first post

"
F = k*(CV)^2/r^2

So for a ceramic capacitor of 20nF with V = 20V, F ~= 3.6N.

The (CV)^2 term alone has a value of 16*10^(-14), so the force can't be
anywhere near 3.6N"

is just utter nonsense and says you have no idea what your talking
about(because its basic algebra to see your wrong). ***, there are two
other factors involved there yet you completely ignore them like they have
no effect(and it really only takes r)

I already admitted that I did in fact fail to look closely at the
expression you had there. I mistakenly assumed that it was the correct
expression for the force between parallel plates, and I knew from previous
experience that the force couldn't be as large as you calculated. I knew
that the (CV)^2 term would dominate (in the correct expression) in spite of
the fact that there are other terms. I showed this in another post.





But even if I ignore that as a lapse of ignorance there still ist he problem
of you tossing the term volume integerals around like it means something.
Believe me, I do know that any time your doing with any type of object and
you are computing something over that object you must integrate(even if its
a point).

But since you throw it around willy nilly it sounds like you actually never
computed one or you would know that its not an easy thing and in general can
only be done numerically. Now this case we have a nested integeral and so
its going to be 10 orders of magnitude more difficult except in special
cases where there is a high degree of symmetry(the highest being a point).

Again though, since you seem to love doing volume integerals I'd like to be
shut up by having you compute the one above(for all I know you might be able
to do it but I seriously doubt it). Of course I'm sure if you do do it then
you will make some approximation somewhere(such as the normal force is
constant everywhere) and you might actually be able to do it. (but I want to
see it in the general case that I described above).

Its one thing to be able to toss these terms around but do you actually have
any practical experience with them?

Again, my first approximation, when corrected, is not bad at all. (it would
be worse for extremely small r of course). So I'm glad you brought what you
said to my attention so I could correct the approximation but I hope you see
that its not wrong(after all its an approximation). But I hope you realize
that your volume integerals are not as ubiquiteous as you think. Whats the
point of using the volume integerals if you cannot compute them. (sure you
can do it numerically but in this case it is actually somewhat difficult
because of the time complexity in the general case(since its 6 integerals))

Usually one starts with a first order approximation and moves on when they
need better approximations. I have mine with my corrected approximation.
Its good enough for what I am doing. Maybe later I'll try to find a better
one or use the ones that already exist for constant force. In any case I
want you do solve the surface integerals I have since you seem to think they
are easy.

When did I ever say that they are easy? Any particular one might be, or
it might not be, but I didn't say anything about that.

Anyway, the appropriate surface integral solution to the parallel plate
problem is given in:

http://mysite.du.edu/~jcalvert/phys/caps.htm

and there's no need for me to duplicate it.

I'm sorry to say I don't think I can help your further.



Of course you can't... Cause you probably don't even know what an
integral is. That site uses an approximation. It assumes that the
plates are infinite in size then chops them down. It does that to get
rid of the fringe effects. This is why it is much easier to do it
there way. You can do the same thing with the integrals but then IT IS
AN APPROXIMATION. EVERYTHING IS AN APPROXIMATION.

The fact of the matter is, that all you did was bring to my attention
that I made a mistake by showing me someone else's work that but you
have not actually done any mathematical work yourself and have made
several blunders that make me think you even have problems with simple
algebra.

I figured you wouldn't attempt the integrals... even writing it as a
surface integral has an impossible solution(well, AFAIK)... but you
have tried to make it sound so easy to do with all your talk of volume
integrals. My approximation and thought process was actually correct
but I made a mistake in my formula... once that mistake was corrected
it is actually a very decent approximation. All you did was bring
that to my attention, which I'm thankful... but you have also tried to
make yourself out to be someone that actually knows whats going on. I
have as to yet see you do any work except search the internet and
found an exercise that someone else did to demonstrate my initial
formulation was in error and claim a bunch of crap that you most
likely have no clue about.

There were two others that actually formulated the approximation from
that site before you but I did not spend enough time to think about
why my formulation was in error. There was also the problem of using
ceramic caps which I was in error also. But fixing the error and
taking into account how ceramic caps work fixes my formula for a rough
approximation.

In any case it doesn't matter much any more... this thread isn't going
anywhere anymore.

.