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

• I
ZapperZ
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Sorry for being unfair I blame @feynman1  he has the tendency of diverging from the thread's original topic in the threads he makes.
You're slowly catching up, grasshopper. :)

Zz.

• Delta2
I don't know its just a forum's thread, conversation can divert (a little or a lot) from the original topic.
No, I never diverted. where the so called +2Q and -2Q could be +3Q and -3Q etc, if there were no Gauss' law (under the umbrella of DC+ideal wires+ideal capacitors (no electric field leaking out) premise).

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Delta2
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No, I never diverted.
I think you did, you start with one capacitor and then you speak about capacitors in series.
Also you did divert in that other thread about current not being a vector but current density being a vector.

Its ok, it happens alot in these forums you are not the only one.

Delta2
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where the so called +2Q and -2Q could be +3Q and -3Q etc, if there were no Gauss' law.
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 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 changed the recent post with a pic to restrict our discussion to DC+ideal wires+ideal capacitors (no electric field leaking out).

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• Delta2
Delta2
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i changed the recent post with a pic to restrict our discussion to DC+ideal wires.
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
vanhees71
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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.
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.

• feynman1
vanhees71
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i changed the recent post with a pic to restrict our discussion to DC+ideal wires+ideal capacitors (no electric field leaking out).
With only "ideal wires" cicruit theory becomes singular either. You should have some finite resistance to not overcomplicate things.

Delta2
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With only "ideal wires" cicruit theory becomes singular either. You should have some finite resistance to not overcomplicate things.
Ok lets 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
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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
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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
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Of course, we use Maxwell equations all the time when dealing with electrodynamics. What else should we use?

rude man
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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.

• etotheipi, feynman1 and Delta2
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
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Many thanks for agreeing on the argument of Gauss' law. How did you calculate inside faces charge = (q1 + q2)/2 and -(q1 + q2)/2?
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 ##.

• feynman1 and Delta2
Delta2
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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
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
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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)

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
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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.

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rude man
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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

• feynman1 and Delta2
Delta2
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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.

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
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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.