I have a question/comment to Bob Sciamanda and a question/comment to John
Denker.
To Bob...
Thanks for your message in which you detailed Q1, Q2, q1, q2, q3, q4 etc.
This was very helpful for me to see what you are doing. But I don't get
part of your reasoning.
I agree that we can show q2=-q3, and I assume we are choosing Q1 = -Q2, and
it then follows that q1 = -q4.
But I don't see how you get q1 -q2 -q3 -q4 = 0 and q1 +q2 +q3 -q4 =0.
Something has to be wrong with these. q1 + q2 = Q1 = -Q2 = -(q3+q4) = -q3 -
q4. Therefore q1 + q2 + q3 +q4 =0.
If you are somehow getting these from E = sigma/2epsilon then I think that
is incorrect because that relation does not apply to this situation.
Outside a conductor with surface charge density sigma the field is
sigma/1epsilon.
Are you trying to apply the superposition principle at various points and
trying to use all four surfaces? If so, see next comment.
To John...
I think Bernard's comment about adding fields, to which you responded "alway
always always" add fields was really a mistated problem. Yes, the
superposition principle always holds for the electric field, so we can
always add all the fields from the contributing charges to find the net
field. The key word is "contributing." What Bernard was really saying was
both sides of charge on a finite-thickness insulating sheet with charge on
both sides will contribute to the electric field on one side of the sheet
(i.e. we use all charges when we apply the superposition principle in this
case) whereas only the near-side charges of the finite-thickness conductive
plate contribute to the field on one side of the plate when we apply the
superposition principle.
This strikes me both true and false. In the true sense... Literally we only
consider the charge on the near side because the electric field from the far
side does not penetrate the conductor. So there is no field to "always
always always" add from the far side of the conductor. We only add the
fields from the charges on the near side. In the false sense... in reality
the charges on the far side are important in the case of the conductor,
because if they weren't there, the charges on the near side would spread out
and the charge density on the near side will drop by a factor of two. So
the presence of the charge on the far side of the conductor is what keeps
the charge density on the near side as high as it is.
The presence or absence of charge on the far side of a finite-thickness
insulating sheet has no bearing on the amount of charge on the near side
because those near-side charges are immobile. Take the far side charges
away from the insulating sheet and the near side density remains at the
sigma it was. Take the far side charges away from a conductive sheet and
the near side charge density drops to half its original value.
So, when we apply the superposition principle to the conductive sheet, are
we summing the fields from the charges on both sides? Literally no. But in
a way yes.
Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817