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

• I
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
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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.

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

Delta2
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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.

vanhees71
There will be fringe fields and they will all point in the same directions (out if positive charges) . That is the point.

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

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hutchphd
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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.

vanhees71 and feynman1
Delta2
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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 doesnt 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).

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rude man
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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 doesnt 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
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@rude man check post #81 I cant 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
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@rude man check post #81 I cant 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.

Delta2
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@rude man check post #81 I cant 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.

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
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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
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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
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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
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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
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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 cant 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.

hutchphd
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I cant 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
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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
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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
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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
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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.

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rude man
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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?