Why Do Capacitor Plates Have Equal and Opposite Charges?

[rude man]

I think the first equation $\sigma_2+\sigma_3=0$ also becomes an approximation due to fringe field flux.

I don't think so. That would violate $\nabla \cdot \bf E = 0$ since the E field would have to be different near one plate than the other.

(As I said, I assume equal plate geometries.)

 

[Delta2]

I don't think so. That would violate $\nabla \cdot \bf E = 0$ since the E field would have to be different near one plate than the other.

(As I said, I assume equal plate geometries.)

We can have $\nabla\cdot \mathbf{E}=0$ even in non-uniform E-fields e.g the field of a point charge located at the origin, it is $\nabla\cdot \mathbf{E}=0$ everywhere except the origin.

 

[rude man]

We can have $\nabla\cdot \mathbf{E}=0$ even in non-uniform E-fields e.g the field of a point charge located at the origin, it is $\nabla\cdot \mathbf{E}=0$ everywhere except the origin.

Right. But in this case a correct gaussian surface would be a conical section so as to make the cone's side flux = 0 , then the net cross-sectional flux would also be zero and $\nabla \cdot \bf E = 0$ still.. But a properly shaped gaussian surface for the parallel plates might e.g. be a right circular cylinder running from inside one plate to inside the other. The net charge inside the cylinder would be net non-zero if $\sigma3 \neq -\sigma2$ which would violate $\nabla \cdot \bf E =0$.

 

[Delta2]

Right. But in this case a correct gaussian surface would be a conical section so as to make the cone's side flux = 0 , then the net cross-sectional flux would also be zero and $\nabla \cdot \bf E = 0$ still.. But a properly shaped gaussian surface for the parallel plates might e.g. be a right circular cylinder running from inside one plate to inside the other. The net charge inside the cylinder would be net non-zero if $\sigma3 \neq -\sigma2$ which would violate $\nabla \cdot \bf E =0$.

I can't be sure on how exactly you take that cylinder unless a figure is provided but I think that $\nabla\cdot\mathbf{E}$ not necessarily zero everywhere inside that cylinder. AND i think you are working with a "silent" assumption in your mind (if i can read your mind hehe) that the E-field in the region between the capacitor's plates will be perfectly perpendicular to the plates. I think this assumption though seems very logical, especially if you take the plates dimensions to be much larger than the plate separation distance, it generally does not hold, not in the case of post #81.

 

[rude man]

I can't be sure on how exactly you take that cylinder unless a figure is provided but I think that $\nabla\cdot\mathbf{E}$ not necessarily zero everywhere inside that cylinder. AND i think you are working with a "silent" assumption in your mind (if i can read your mind hehe) that the E-field in the region between the capacitor's plates will be perfectly perpendicular to the plates. I think this assumption though seems very logical, especially if you take the plates dimensions to be much larger than the plate separation distance, it generally does not hold, not in the case of post #81.

There is no way $\nabla \cdot \bf E \neq 0$ in any electric field - anywhere, at any point, anytime, - even in an electrodynamic field - unless charge is present there! Just ask Maxwell!

Even if the plates are small, $\sigma3 = -\sigma2.$ In that case $\sigma = \sigma(x,y)$ if the plates are parallel to the xy plane. Then $\sigma3(x,y) = - \sigma2(x,y).$ since there is perfect charge symmetry including the fringing fields as long as the plates have the same geometry. So no, I don't need to assume perfectly perpendicular E fields to the plates.

 

[Delta2]

There is no way ∇⋅E≠0∇⋅E≠0 \nabla \cdot \bf E \neq 0 in any electric field - anywhere, at any point, anytime, - even in an electrodynamic field - unless charge is present there! Just ask Maxwell!

Doesn't your gaussian cylinder enclose at least partially portion of $\sigma_2$ and $\sigma_3$ that's why $\nabla\cdot\mathbf{E}\neq 0$ there, exactly because there is surface charge there as you say.

 

[rude man]

Put a right circular cylder of small area running just outside each plate to just outside the other plate. So contained charge = 0 and $\iiint_V ( \nabla \cdot E) ~dv = 0$

Assume plate 2 is more negative than plate 3 is positive. Then there would be more flux leaving the cylinder near the negative plate than entering near the positive plate, violating $\iiint_V ( \nabla \cdot E) ~dv = 0$

This is true even in the fringe field. There would be more net flux leaving and entering the end and side of the surface near the negaive plate than entering and leaving the surface near the positive plate:

 

[rude man]

Or maybe you like this fringe field picture better: I can shape my gaussian closed surface any way I want. So I wll shape it such that the only net flux entering or leaving the volume is at the flat ends which are located just outside each plate, where the two fields would be different if the charge densities are different (in magnitude) since $\sigma = \bf D \cdot \bf n$.
This would violate $\iint_S \bf D \cdot \bf dS = 0$

 

[Delta2]

Or maybe you like this fringe field picture better: I can shape my gaussian closed surface any way I want. So I wll shape it such that the only net flux entering or leaving the volume is at the flat ends which are located just outside each plate, where the two fields would be different if the charge densities are different (in magnitude) since $\sigma = \bf D \cdot \bf n$.
This would violate $\iint_S \bf D \cdot \bf dS = 0$

Your reasoning seems to me like that any two charge densities should be made equal otherwise they would violate $\oint_s\mathbf{D}\cdot\mathbf{dS}=0$. I doubt that there is such a gaussian surface as you describe. In my opinion what happens when the two charge densities are unequal is that the fringe field flux from the sides balances out the field flux from the top and bottom.

 

[rude man]

Your reasoning seems to me like that any two charge densities should be made equal otherwise they would violate $\iint_s\mathbf{D}\cdot\mathbf{dS}=0$. I doubt you can choose such a gaussian surface as you describe. In my opinion what happens when the two charge densities are unequal is that the fringe field flux from the sides balances out the field flux from the top and bottom.

That is my reasoning.
All you have to do to construct such a surface is to define it to conform to the E field at every point on the surface, whatever its direction happens to be, except at the ends by the plates. So at every point there is no flux in or out except right at the plates where the surface is abruptly made parallel to each plate.

This in fact is what you do to come up with a useful gaussian surface to demonstrate that the total integrated flux from a single charge (cf. your post 92 ) is zero. The side of the surface forms a section of a cone, thus conforming to the E field.

It makes no difference the shape of the E field; a closed surface can always be hypothesized such that net side flux is zero. Why not?

 

[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!