Why Do Capacitor Plates Have Equal and Opposite Charges?

[vanhees71]

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.

 

[Delta2]

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.

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]

Of course, we use Maxwell equations all the time when dealing with electrodynamics. What else should we use?

 

[rude man]

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.

 

[feynman1]

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.

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]

Many thanks for agreeing on the argument of Gauss' law. How did you calculate inside faces charge = (q1 + q2)/2 and -(q1 + q2)/2?

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 $\sigma 's$.

 

[Delta2]

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 $\sigma 's$.

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.

 

[feynman1]

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 $\sigma 's$.

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.

 

[Delta2]

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.

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]

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)

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

 

[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.