Why Do Capacitor Plates Have Equal and Opposite Charges?

[Delta2]

By dynamic, do you mean AC circuits or LRC ones? Even in these, they still vanish.

I don't think they vanish if we have a RLC circuit in which we apply a very high frequency voltage source such that the wavelength $\lambda=\frac{c}{f}$ ($f$ the frequency of the source) is comparable to the distance between the capacitor's C plates. The reason (I think) is that the current and the electric field inside the circuit's wires(and inside the capacitor) become function of position and not only time(e.g the electric field inside the capacitor i believe it would be given by an equation like $$E(x,t)=E_0\sin(2\pi ft+2\pi\frac{x}{\lambda})$$ in the high frequency case) and not just as $$E(t)=E_0\sin(2\pi ft)$$ which is the equation we use in the quasi static approximation.
So because the electric field will be different inside the two plates, the whole reasoning with Gauss's law that $$\oint_S\vec{E}\cdot d\vec{S}=0$$ falls apart.

 

[rude man]

Many thanks. I agree with your 4 equations. The last 1 simply ensures both plates be conductors.

So in what circumstances will outside faces charge not vanish? I think when the wires connecting the capacitor are ideal, they should vanish whatsoever, unsteady charging/discharging included.

As long as the capacitor or capacitors are charged/discharged in a circuit, $q2 \neq q1$ is impossible. The only way I can think of doing $q2 \neq q1$ is by separately charging the plates as I described. So yes the outside charges vanish.

If you're interested in what happens at ever higher frequencies I suggest reading Vol. II chapt. 23 of the Feynman Lectures on Physics. It gets VERY complicated and eventually you wind up with a cavity resonator. Not "introductory physics" IMO.

https://www.feynmanlectures.caltech.edu/II_23.html

 

[Delta2]

Thanks @rude man for sharing with us this wonderful link. So I tried to read this as fast as possible, the electric field between capacitor's plate is no longer uniform (as we assume is the case in static and quasi static) but becomes a Bessel function. Still he claims that it is symmetrical around the center of the capacitor ( a rather natural conclusion now that I think of it), so I think I was wrong at post #71 in my final conclusion.

 

[feynman1]

As long as the capacitor or capacitors are charged/discharged in a circuit, $q2 \neq q1$ is impossible. The only way I can think of doing $q2 \neq q1$ is by separately charging the plates as I described. So yes the outside charges vanish.

If you're interested in what happens at ever higher frequencies I suggest reading Vol. II chapt. 23 of the Feynman Lectures on Physics. It gets VERY complicated and eventually you wind up with a cavity resonator. Not "introductory physics" IMO.

https://www.feynmanlectures.caltech.edu/II_23.html

Thanks for sharing the high frequency scenario. What's the motivation of charging both plates of a capacitor separately? If they are in 1 circuit, is it even possible?

 

[rude man]

Thanks for sharing the high frequency scenario. What's the motivation of charging both plates of a capacitor separately? If they are in 1 circuit, is it even possible?

No particular motivation. Jus FYI. As I said, no, if in a circuit the current will always balance capacitor charge + and - equally.

 

[hutchphd]

As was pointed out earlier even in a circuit charge can be added to the entire circuit making the argument using Gauss patently incorrect. What is the point?

 

[Delta2]

As was pointed out earlier even in a circuit charge can be added to the entire circuit making the argument using Gauss patently incorrect. What is the point?

Yes that is correct, but check @rude man post, even in this case the charges in the inner surfaces of the capacitors plates will be opposite and equal.

 

[hutchphd]

If I make a "circuit" consisting of a capacitor and a wire, then it will manifestly not be true.

 

[Delta2]

If I make a "circuit" consisting of a capacitor and a wire, then it will manifestly not be true.

The inner charges will be approximately opposite and equal even in this case and regardless how you charge the capacitor. Gauss's law will yield $\oint_S\vec{E}\cdot d\vec{S}=0=\sum Q_{inner}/\epsilon_0$ for a gaussian surface $S$ that passes through the interior of the plates (assuming the plates have a small finite thickness, in the interior of the plates the E-field is zero at least in the static and quasi static case) and encloses the capacitor and assuming that no field fringes out from the capacitor's edges.

 

[hutchphd]

There will be fringe fields and they will all point in the same directions (out if positive charges) . That is the point.

 

[hutchphd]

The inner charges will be approximately opposite and equal even in this case and regardless how you charge the capacitor.

.
To reiterate: the capacitor leads are connected. One quatloo of positive charge applied to them
The "inner" charges will be approximately zero because between the plates you are very nearly inside a conductor! This can be considered symmetric, so which "plate" would be negative and why?
The charge will be all of one sign (positive) and reside mostly on the very outside edges of the plates. There is no spontaneous magic separation of charge.

 

[Delta2]

There will be fringe fields and they will all point in the same directions (out if positive charges) . That is the point.

You might be right here, fringe fields will make rudeman post not exactly correct but approximately correct.

To reiterate: the capacitor leads are connected. One quatloo of positive charge applied to them
The "inner" charges will be approximately zero because between the plates you are very nearly inside a conductor! This can be considered symmetric, so which "plate" would be negative and why?
The charge will be all of one sign (positive) and reside mostly on the very outside edges of the plates. There is no spontaneous magic separation of charge.

Do you mean that we charge the capacitor so it has charge +Q (or -Q) to both of its plates? Then indeed all the charge resides on the outside faces.

But the inner charges are still equally and opposite charged because 0 is the opposite of itself

Yes ok i see now in this case most of the charge will reside in the outside faces while the inner faces will have a small fraction of charge of the same sign. The reasoning of posts #64,66,79 fails because of the existence of non negligible fringe fields.

 

[rude man]

Fringe fields have nothing to do with it. There will be an exact q and -q on the two plates. That's because charge is the time integral of current and the same currnet flows from the - plate to the + plate and nowhere else.

All the charge will be on the INNER faces. Zero on the outer faces. cf. my post #64.

 

[Delta2]

Fringe fields have nothing to do with it. There will be an exact q and -q on the two plates. That's because charge is the time integral of current and the same currnet flows from the - plate to the + plate and nowhere else.

All the charge will be on the INNER faces. Zero on the outer faces. cf. my post #64.

In his example the capacitor is not charged by a current but by some other process. I am afraid its all about fringe fields in this case of "peculiar" charging of capacitors.

Your post #64 is correct if we ignore fringe fields, if fringe fields are not negligible the reasoning doesn't work. Especially if the fringe fields are coming from a capacitor that has the same charge (same sign) in both plates then there would be fringe field flux which does not cancel out (like it might cancel out if the fringe field is due to equal and opposite charge).

 

[rude man]

In his example the capacitor is not charged by a current but by some other process. I am afraid its all about fringe fields in this case of "peculiar" charging of capacitors.

I doubt it, bt I'd have to know exactly how the plates are charged. I gave an example of separately charged plates with the attendant solution

Your post #64 is correct if we ignore fringe fields, if fringe fields are not negligible the reasoning doesn't work. Especially if the fringe fields are coming from a capacitor that has the same charge (same sign) in both plates then there would be fringe field flux which does not cancel out (like it might cancel out if the fringe field is due to equal and opposite charge).

I'm sorry, I don't understand your reasoning. Charge is charge, and once you define the amount of charge fringing plays no role I can see. Maybe a concrete example might help.

 

[Delta2]

@rude man check post #81 I can't find any flaw behind his reasoning, fringe fields is the only explanation I can come up with. he doesn't tell in this post how exactly he charges the capacitors. Now that I reread your post #64 it mentions q1 and -q2 charge, in post #81 we have the same sign charge in both plates.

 

[rude man]

@rude man check post #81 I can't find any flaw behind his reasoning, fringe fields is the only explanation I can come up with. he doesn't tell in this post how exactly he charges the capacitors. Now that I reread your post #64 it mentions q1 and -q2 charge, in post #81 we have the same sign charge in both plates.

OK it's late here. I will look at post #81 later today.
I may have assumed large plates where fringing is by definition not a factor. Let me look at all this some more.

 

[rude man]

@rude man check post #81 I can't find any flaw behind his reasoning, fringe fields is the only explanation I can come up with. he doesn't tell in this post how exactly he charges the capacitors. Now that I reread your post #64 it mentions q1 and -q2 charge, in post #81 we have the same sign charge in both plates.

I looked at post 81 and unlike you I did not see any reasoning that made any sense to me.

However, its conclusion as to the distribution of charge densities is correct.

If you go back to my post 64, substitute -q2 = q1 so that both plates have the same charge, you get that result. My post 64 is completely universal; you can pick any charge magnitude or polarity you want on either plate, and you get the correct result.

The only assumptions I made are that the plates have the same geometry and that the spacing between plates is << plate size (length and width).

If the latter is not the case then my 4th equation becomes an approximation.

 

[hutchphd]

I looked at post 81 and unlike you I did not see any reasoning that made any sense to me.

To put charge on the shorted capacitor, I shuffle across the acrylic carpet, pick it up and put it back down. In winter it will now contain excess charge. As I explained previously.

I'm sorry that my explanation was insufficient. It is certainly correct and I cannot be any more clear.

 

[Delta2]

The only assumptions I made are that the plates have the same geometry and that the spacing between plates is << plate size (length and width).

If the latter is not the case then my 4th equation becomes an approximation.

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

 

[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!