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Re: There's work, and then there's work



I'm starting to miss the point in this thread so I'm going to
butt in and ask some stupid questions again...

Suppose you had two like charges, A and B, of equal charge q
that are a distance L apart.

Hold charge A still and bring charge B toward
charge A until the charges are a distance L-x apart.

What was the total work done _on the charges_?

When both charges are released at once, the charges separate.
When they reach a distance L apart, what is the kinetic
energy of each charge?

I would think that the kinetic energy equals the total work
done on the charges but maybe I'm missing something.

If two people have a different answer for the total work done,
that seems to mean that they would have different answers for
the total kinetic energy, which should be easily testable, no?

If this has nothing to do with the discussion, please
tell me why.

Thank you for any guidance you can provide in this area.
____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

-----Original Message-----
From: David Rutherford [mailto:drutherford@SOFTCOM.NET]
Sent: Thursday, January 23, 2003 8:57 PM
To: PHYS-L@lists.nau.edu
Subject: Re: There's work, and then there's work


John Mallinckrodt wrote:

David Rutherford wrote:

Answer my first question, first:

"How do you propose to assemble a configuration, in the
first place, if
the assembled charges keep flying off every time you bring in a new
one?"

I'd hold them so they don't move.

Why? In the conventional way of calculating the energy of a
distribution
of charges, you don't even consider the field of the particle you are
bringing in, so you don't have to worry about it. In the case of a two
particle distribution, the field of the second particle is not even
included in the calculation of the energy of the distribution. So by
calculating the energy the conventional way, you get only
_half_ of the
_actual_ energy of the distribution (for two identical particles). Go
figure!?#@*

Then please show me how you determined that my result
predicts that the
energy stored by a capacitor is CV^2. Show your work.

Fine. You double count the contribution of each charge to the total
energy. Therefore, since the correct answer is 1/2 CV^2, you get
CV^2.

No. Think of a parallel plate capacitor as two oppositely charged
particles flattened into two oppositely charged parallel plates. Using
the same (conventional) method that was used to find the energy of the
two charge distribution, above, you can see that the total
field of the
(assembled) capacitor only includes the field of _one_ of the plates.
That means that, using this method, you get 1/4 CV^2, not 1/2
CV^2. But
using my method, in which the fields of _both_ plates are considered,
you get 1/2 CV^2, which is the _correct_ result.

I'm assuming that, since you used this example, there is experimental
evidence corroborating this. In that case, the evidence verifies my
result for the energy (density) of a charge distribution (and
a parallel
plate capacitor), and proves that the conventional result for
the energy
(density) of a charge distribution is incorrect.

If you have any other "relevant experimental result[s]" that my theory
"violates", let's see them. It'll be a pleasure to shoot down yet
another conventional physics Dodo Bird.