No, I never diverted.Delta2 said:I don't know its just a forum's thread, conversation can divert (a little or a lot) from the original topic.
I think you did, you start with one capacitor and then you speak about capacitors in series.feynman1 said:No, I never diverted.
I think that you are also using the assumption that the electric field outside the capacitor is zero (together with Gauss's law). But this is not always a valid assumption, especially in the time dependent case with high frequency -low wavelength current. It can be valid if you assume ideal capacitor and ideal connecting wires as you say, but this is not the case in real world situations.feynman1 said:where the so called +2Q and -2Q could be +3Q and -3Q etc, if there were no Gauss' law.
i changed the recent post with a pic to restrict our discussion to DC+ideal wires+ideal capacitors (no electric field leaking out).Delta2 said:I think that you are also using the assumption that the electric field outside the capacitor is zero (together with Gauss's law). But this is not always a valid assumption, especially in the time dependent case with high frequency -low wavelength current. It can be valid if you assume ideal capacitor and ideal connecting wires as you say, but this is not the case in real world situations.
I think under DC and under ideal wires +ideal capacitor (No E-field escapes outside the capacitor) you are right that conservation of charge alone doesn't prevent the configuration that you show in post #53. It is gauss's law that prevents it (or alternatively the circuit theory assumption that the current along the same branch is everywhere the same, which is valid for the DC-case).feynman1 said:i changed the recent post with a pic to restrict our discussion to DC+ideal wires.
Then you cannot treat the problem in quasistationary approximation and it's far from the DC case discussed in this thread since for this you need the full Maxwell equations and their retarded solutions. This would be an entirely different topic and thus should be discussed in a separate new thread.Delta2 said:Yes this is the issue except that I had not in mind a DC battery but rather the AC case with very high frequency such that the wavelength of current in the circuit is in comparison with the dimension of the branch containing the capacitors in series.
With only "ideal wires" cicruit theory becomes singular either. You should have some finite resistance to not overcomplicate things.feynman1 said:i changed the recent post with a pic to restrict our discussion to DC+ideal wires+ideal capacitors (no electric field leaking out).
Ok let's assume finite resistances, and charge the capacitors with a DC voltage source, how can conservation of charge only prevent the configuration of charges shown in post #53?vanhees71 said:With only "ideal wires" cicruit theory becomes singular either. You should have some finite resistance to not overcomplicate things.
I agree, but we are using gauss's law here don't we? how else can we infer there would an e-field outside the capacitors (because there would be net charge enclosed).vanhees71 said:As discussed several times, then you'd have an electric field outside the capacitors and a current would flow in the wire connecting the capacitors, which by assumption (electrostatic situation) should not be the case.
Many thanks for agreeing on the argument of Gauss' law. How did you calculate inside faces charge = (q1 + q2)/2 and -(q1 + q2)/2?rude man said:I don't have the temperament to read all the posts, but:
it is perfectly possible for a capacitor to have differing amounts of charge on its plates. You just put separate charge magnitudes on each plate the way you would normally charge a plate - by a charged object held against each plate.
Take two identical plates, charge each of them separately with different charge q1 and - q2, ## q2 \neq q1 ##. Then bring them close to each other.
The sides of plates facing each other will have the same magnitude of charge, just like in a capacitor carged by current. This as you point out can be shown by Gauss's theorem, or by the fact that ## \nabla \cdot \bf E = ~0 ##.
The outside faces will have differing charge densities:
The total picture is:
outside faces charge = (q1 - q2)/2
inside faces charge = (q1 + q2)/2 and -(q1 + q2)/2
If the capacitor is charged the normal way i.e. by current, q2 = q1 and the inside faces will have all the accumulated charge while the outside faces will have zero charge.
Glad you asked.feynman1 said:Many thanks for agreeing on the argument of Gauss' law. How did you calculate inside faces charge = (q1 + q2)/2 and -(q1 + q2)/2?
Just to add one little thing: The net E-field inside the plate is zero, so total force is zero, so because the force from 2 and 3 cancel, the forces from 1 and 4 have to cancel too, otherwise the total force and the e-field wouldn't be zero.rude man said:The last equation is the interesting one. Imagine a unit test charge inside plate 1. This charge will see a force from each of the four face charges:
Charges 2 and 3 forces cancel.
Charge 1 and charge 4 forces also cancel (charge1 pushes the test charge to the right, charge 4 pushes the test charge to the left). So ## \sigma1 - \sigma4 = 0 ##.
That gives you 4 equations in the four unknown ## \sigma 's ##.
Many thanks. I agree with your 4 equations. The last 1 simply ensures both plates be conductors.rude man said:Glad you asked.
OK, we have +q1 on plate 1 and -q2 on plate 2. There are 4 sides: the two outside ones and the two inside ones. Say the plates are of unit area and lined up left to right.
Let
## \sigma1 = ## left plate outside face charge density
## \sigma2 = ## left plate inside face charge density
## \sigma3 = ## right plate inside face charge density
## \sigma4 = ## right plate outside face charge density
By Gauss, ##\sigma3 = - \sigma2 ##. I think you got that already.
Then,
##\sigma1 + \sigma2 = q1 ##
## \sigma 3 + \sigma4 = -q2 ##
The last equation is the interesting one. Imagine a unit test charge inside plate 1. This charge will see a force from each of the four face charges:
Charges 2 and 3 forces cancel.
Charge 1 and charge 4 forces also cancel (charge1 pushes the test charge to the right, charge 4 pushes the test charge to the left). So ## \sigma1 - \sigma4 = 0 ##.
That gives you 4 equations in the four unknown ## \sigma 's ##.
I think they vanish only in the static case and in the quasi-static approximation(DC and low frequency-high wavelength AC). In the full dynamic case I believe they don't necessarily vanish (neither that the inside charges are opposite and equal)feynman1 said: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.
By dynamic, do you mean AC circuits or LRC ones? Even in these, they still vanish.Delta2 said:I think they vanish only in the static case and in the quasi-static approximation(DC and low frequency-high wavelength AC). In the full dynamic case I believe they don't necessarily vanish (neither that the inside charges are opposite and equal)
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.feynman1 said:By dynamic, do you mean AC circuits or LRC ones? Even in these, they still vanish.
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.feynman1 said: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.
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 said: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
No particular motivation. Jus FYI. As I said, no, if in a circuit the current will always balance capacitor charge + and - equally.feynman1 said: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?
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 said: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?
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 said:If I make a "circuit" consisting of a capacitor and a wire, then it will manifestly not be true.
.Delta2 said:The inner charges will be approximately opposite and equal even in this case and regardless how you charge the capacitor.
You might be right here, fringe fields will make rudeman post not exactly correct but approximately correct.hutchphd said:There will be fringe fields and they will all point in the same directions (out if positive charges) . That is the point.
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.hutchphd said: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.
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.rude man said: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.
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 solutionDelta2 said: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'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.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).
OK it's late here. I will look at post #81 later today.Delta2 said:@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.Delta2 said:@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.
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.rude man said:I looked at post 81 and unlike you I did not see any reasoning that made any sense to me.
I think the first equation ##\sigma_2+\sigma_3=0## also becomes an approximation due to fringe field flux.rude man said: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 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.Delta2 said:I think the first equation ##\sigma_2+\sigma_3=0## also becomes an approximation due to fringe field flux.
rude man said: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.)
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 said: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.
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 said: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 ##.
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!Delta2 said: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.
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 said: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!
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 said: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 ##
That is my reasoning.Delta2 said: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.