Why Do Capacitor Plates Have Equal and Opposite Charges?

[Delta2]

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 can't 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.

 

[rude man]

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!

 

[rude man]

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.

 

[rude man]

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!