Re: How many volts ?

["Donald E. Simanek" <dsimanek@eagle.lhup.edu>]

On Tue, 25 Mar 1997, W. Barlow Newbolt wrote:

> I'm as confused as a Physics major taking the graduate record exam.  I 
> think that John has the solution right for the difference in potential 
> for two plates which may be considered infinite.  That is, those charges 
> were charge densities!
>
> There are two things here that this solution doesn't tell us much about: 
>  
> 		1.  What happens if the plates are not infinite. I think 
>                     this is at least part of Ludwik's objection.

The problem had several quirks. One was the distractor, the comment about
the capacitance, calculated from the parallel-plate capacitor formula, a
result not necessary to solving the problem. But it does suggest that the
student is to *make the usual assumptions* which went into the dervation
of the parallel plate capacitor: That when equal and opposite charges are
placed on the plates, the field is mostly between the plates and
negligible elsewhere.

For such a pair of plates, if equal charges (same) sign, are placed on the
plates, their field will be mostly *outside* the plates and negligible
inside. The plates are essentially acting as a shielding enclosure, with
very little field leakage near the edges.

I think John's solution, and answer are correct, and insightful, using
Gauss' superposition principle.

> 		2.  What happens to the potential difference between the 
>                     plates as they are moved together.  After all, they
>                     were infinitely far apart originally.  I presume that 
>                     the total potential would go down since they attract 
>                     each other.  
> 		    other and work will be done by the field.

That's another distractor, in my understanding of this problem, for the
potential of any charge configuration is the (work done)/charge in
assembling it from charges initially infinitely separated. it doesn't
matter how the assembly is carried out. You might as well start with the
plates in place and put the given charges on them. I think the problem did
say or imply "in a vacuum" so once the charges are on the plates they
won't go anywhere else as you move the plates. The potential will change
as the plates are moved together, of course, but that fact is, I think,
irrelevant to this problem, since it merely spoke of charges placed on the
plates, not on their changes in potential. Only the *final* potential was
wanted, and this will be independent of how the charges got there to that
final configuration. It's really a very neat problem which forces one to
keep one's wits intact, understand concepts, avoid irrelevant
disctractions, and pick up on subtle clues (such as the assumption of a
vacuum being mentioned only in an otherwise irrelevant piece of
information about the capacitance). I give high marks to whoever invented
and worded this clever problem. It could have been a straightforward
problem worded: 

Two parallel metal plates of small separation, much smaller than their
dimension, are in a vacuum. Charge Q1 is placed on one plate. And charge
of opposite sign and different magnitude, -Q2, is placed on the other
plate. What is the approximate potential between the plates.

But that would have been too easy!

And I'm glad I don't have to take graduate oral exams ever again.

	-- Donald

......................................................................
Dr. Donald E. Simanek                            Office: 717-893-2079
Prof. of Physics                    Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745          CIS: 73147,2166 
Home page: http://www.lhup.edu/~dsimanek            FAX: 717-893-2047
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