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RE: How many volts?

On Thu, 3 Apr 1997, Lemmerhirt, Fred wrote:

Just in case anyone is still interested in the "How many volts?" problem,
here is one more approach that I understand may be just a bit different from
those already submitted:

Assume that the plates are horizontal, with plate 1 on top; that e represents
epsilon-nought; and that the charges on the plates are distributed uniformly
in the horizontal directions. Then the field between the plates and also
just above the top plate due to the charge Q1 alone is (Q1/A)/(2e). (It
seems to me that this doesn't depend on how the charge Q1 is distributed
between the upper and lower surfaces of the top plate.) Similarly, the field
between the plates and also just below the bottom plate due to the charge on
the bottom plate alone is (Q2/A)/(2e). (I mean these are the magnitudes of
the fields, taking Q1 & Q2 as absolute values.) Since between the
plates these fields point in the same direction, the resulting value there is
(Q1+Q2)/(2eA), which is then multiplied by the plate separation to get the
potential difference. Using the specified values gives:

(10nC+100nC)(.01m)/(1m^2)/(.00885nC^2/Nm^2)/2 = 62V


Fred Lemmerhirt
FredL@wccb.wcc.cc.il.us


I believe that this is basically correct, but it is an approximation.
The approximation is probably well-founded in this case, but we
have to show that the field is perpendicular to the plates everwhere--
even at the edges, and that the additional 90nC of net charge does not
change the potential difference of the plates, but only acts as an additive
potential term for the whole device.


W. Barlow Newbolt 540-463-8881 (telephone)
108 Parmly Hall 540-463-8884 (fax)
Washington and Lee University newbolt.w@fs.science.wlu.edu
Lexington, Virginia 24450 wnewbolt@liberty.uc.wlu.edu

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