The real reason for a capacitor having the same amounts of + and - charges on the two plates

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  • #101
Delta2
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No sorry, I don't think that the construction of such a closed surface is possible. You got me here though since topology is my weak area and I cant find a good argument to back it up. All I can say is that if such a surface was possible then we would argue that any two charge densities are equal. Take for example a surface charge density ##\sigma_1## that is defined on the xy-plane (z=0) and another ##\sigma_2## on the plane z=5. Using your argument we can conclude that they are always equal.
 
  • #102
rude man
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Take for example a surface charge density σ1 that is defined on the xy-plane (z=0) and another σ2 on the plane z=5. Using your argument we can conclude that they are always equal.
I think that if (1) plates have same geometry and (2) there were no other charges (EDIT:) "nearby" then that seemingly absurd assumption would still be true. Any excess charge ## (\sigma2 vs. sigma3) ## would have to be on the other side of whatever plate had the greater charge magnitude.

But I know I'm going far afield so I'm not at all sure about that. Thanks for presenting me with a reality check!
 
  • #103
rude man
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I'm going to look at this from a different viewpoint: forget about gaussian surfaces and focus on the charge situation between two identical plates removed a finite distance apart. I will look at the fact that the E field inside both plates is zero and what that implies regarding charge distributions if one plate has greater charge magnitude than the other. Maybe I'll get a better insight into the stuation from that.

In any case I agree that ## \sigma3 = -\sigma2 ## is an approximaton subject to ## d << a ## for two identical square plates of sides ##a## and separation ##d##. I should have stuck to that all along. :confused:
 
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  • #104
rude man
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OK, I recomputed the charge densities for finite separation distance of plates. The basic assumption is that at distance ##d## there is an attenuation factor ##a## for the force on a unit test charge in either plate. ##a## is a function of ##d## i.e. ## a=a(d) ## but the exact relationship is difficult to determine. So this is more like a qualitative analysis but should be accurate for ## a=1 ## (my old approximations) and ## a=0 ## (plates separated by large distance).

In this view we have, for unit area plates,
## \sigma1 + \sigma2 = Q1 ##
## \sigma3 + \sigma4 = Q2 ##
## \sigma1 - \sigma2 - a\sigma3 - a\sigma4 = 0##
## a\sigma1 + a\sigma2 + \sigma3 - \sigma4 = 0 ##

This solves to
## \sigma1 = \frac {Q1+aQ2} {2} ##
## \sigma2 = \frac {Q1-aQ2} {2} ##
## \sigma3 = \frac {Q2-aQ1} {2} ##
## \sigma4 = \frac {aQ1+Q2} {2} ##

For ## a=1 ## (close-in plates) we get my previous values, wth ##\sigma3 = -\sigma2## etc.

For ## a=0 ## (widely separated plates) we get
## \sigma1 =\frac { Q1} {2} ##
## \sigma2 = \frac {Q1} {2} ##
## \sigma3 = \frac {Q2} {2} ##
## \sigma4 = \frac {Q2} {2} ##
as expected.

Too bad ## a(d) ## is so hard to determine,at least for introductory physics! But it does I think give at least a feel for how charges are rearranged as plate distance varies.

Hope you like this post better than my previous few!
 
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