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Re: [Phys-l] [Electric potential of charged spheres



Sorry, John. I had not intended to mock your answer. Your explanation here is very cogent.

Paul


On Nov 30, 2011, at 2:30 PM, John Denker wrote:

On 11/29/2011 03:08 PM, Paul Nord wrote:
Well, John argued that the total capacitance of the two spheres in
contact (or near each other) would be less than the sum of the two
spheres individually. I presumed that it might actually be greater
because the two-sphere system "looks like" a much larger single
sphere (to a very rough approximation). If John is correct, the
voltage is between 5V and 10V. If I'm correct, the voltage is less
than 5V.

OK.

John used larger words than I did. I'll go with his answer.

1) I assume that's a joke. Size of words is a perfectly terrible
reason for agreeing with me.

2) I apologize for the big words. One definition of "nerd" is
somebody who never uses a small word when a big word will do.

But I might need to test this one.

We should be able to work out the theory, in qualitative terms,
using only small words and well-understood physics principles.

1) Start with the original situation, v=1V and v=9V, spheres far apart.

2) Bring them together long enough to symmetrize the charge, then
move them back to their original positions. v=5V and v=5V now.

Note: In this step (and all other steps), total charge remains
constant. In all steps /except/ this one, the charge on each sphere
remains constant.

3) Bring them together again. In so doing, you do work against
their mutual electrostatic repulsion. That is to say, the energy
of the system goes up.

4) Voltage is energy per unit charge. The energy went up, so the
voltage went up. The new voltage /must/ be greater than 5V.

==================

Here's a second argument that comes to the same conclusion.

it might actually be greater because the two-sphere system "looks
like" a much larger single sphere (to a very rough approximation).

Not very "much" larger. When two spheres of radius R are in contact,
they fit within a single imaginary of size 2R. The capacitance of
the imaginary sphere is only twice the capacitance of each individual
sphere. The imaginary sphere carries twice the charge, sphere, so
its voltage would be exactly 5V.

Clearly the dumbbell-shaped combination of two small spheres is smaller
than the imaginary large sphere, so its capacitance is less, so the
voltage is greater. To say the same thing another way, if you start
with the large sphere and squeeze it into a dumbbell shape, you do
work against it, pushing the charges into a less-favorable configuration,
increasing the voltage.

=============================

There are lots of "static electricity" generator machines that work
on this principle, moving things around while holding constant charge
(not constant voltage), doing work against the electric field so as
to increase the energy per charge (i.e. voltage).
http://www.av8n.com/physics/contact-electrification.htm

Also I revised my article on capacitance, adding a new section on how
to calculate the energy of a multi-terminal capacitor, in terms of
the elastance matrix:
http://www.av8n.com/physics/capacitance.htm#sec-energy

====================

Tangential suggestion: I recommend talking about _the voltage_ rather
than «the electric potential». That's because not every voltage is a
potential. Not every electric field is the gradient of some potential.

In this thread, we are talking about electrostatics, so the voltage is
in fact a potential ... but we don't want students to develop bad habits.
Treating voltage as synonymous with potential is a bad habit, and will
cause confusion later on. There is something to gain and nothing to lose
by formulating this problem in terms of voltage.
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