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  1. Why can the parallel plate capacitor not be charged with unlimited charges?

    Since the zero resistance wire do not have the electric field inside, and the two parallel metal plates with an equal magnitude and opposite charges have the field only between the plates, perhaps it is not the capacitor's field that resists the charges charging. Then what force resists the charges charge on the capacitor when the voltage equals the battery's voltage, except following Kirchhoff circuit laws (KVL)?

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  2. What if the two plates of the capacitor have a different area? How can I calculate its capacitance?

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  • $\begingroup$ Isn't there a canonical question about the capacitor (in its purest form, a very large plate capacitor in a "pure" vacuum (huge void (intergalactic space))? $\endgroup$ – Peter Mortensen Jun 16 at 11:56
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Since the zero resistance wire do not have the electric field inside

Real wire either has resistance or it is a superconductor. Both may have electric fields inside them. It is an approximation to the truth that there is no electric field inside a wire, and in this case that approximation does not help to clarify what is happening in the wire.

and two parallel metal plate with equal magnitude and opposite charges have the field only between the plates

This is an another approximation to the truth that does not help you understand what is going on in the wire.

Why the parallel plate capacitor cannot be charged the unlimited charges?

What happens is this. When the battery is connected and there is no charge in the capacitor plates, there is an electric field in the wire between the battery and the capacitor. This electric field causes electrons to flow down the wire and into the capacitor on one side, and out of the capacitor toward the battery on the other side. As the charges build up on the capacitor plates, the create a counter electric field in the wire. As long as this counter electric field is less than the electric field created by the battery, there will be a net electric field in the wires. This net electric field will cause current to continue flowing. The charges in the capacitor plates will continue to grow until the field generated in the wires by the plates exactly counters the field in the wires generated by the battery. Then, and only then, will there be no net electric field in the wire. When there is no electric field in the wire, then the current will stop (assuming the wire is not a superconductor).

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  • $\begingroup$ Even the normal wire, the function of the electric field is to allow electrons to overcome the resistance inside the wire to pass through the wire, so the resistance has canceled out the effect of the electric field. I still think it's weird. $\endgroup$ – steve Jun 16 at 15:38
  • $\begingroup$ I'm not sure I would say that an electric field allows electrons to "overcome resistance". The dissipation of power that is associated with resistance only happens because electrons are flowing. The flowing electrons interact with (collide with) the conductor lattice and become randomized (thermalized). If there were no current, there are no electrons to become thermalized. They all already are. $\endgroup$ – Math Keeps Me Busy Jun 16 at 15:47
  • $\begingroup$ Nor would I say "resistance cancels out the effect of an electric field." Resistance does not "cancel" electric fields. The only way an electric field is "canceled" is with an equal but opposite electric field. Resistance per se is just a propensity of a conductor to thermalize flowing electrons. $\endgroup$ – Math Keeps Me Busy Jun 16 at 15:53
  • $\begingroup$ Then how to explain if the wire is a superconductor, there is sure no electric field inside the superconductor, I wonder to find a unified interpretation. $\endgroup$ – steve Jun 16 at 16:03
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    $\begingroup$ In a normal conductor, if there is current, there must be an electric field. In a superconductor, there is an electric field if and only if the current is changing. $\endgroup$ – Math Keeps Me Busy Jun 16 at 17:01
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A capacitor has capacity. A capacitor is such a device that can store only so much charge per unit voltage. In theory, the amount of charge the capacitor can store is not restricted by its plate area but rather by the voltage applied. Remember $Q=CV$.

In real the real-world, even the most ideal capacitors have an upper limit to how much voltage that can be charged or the amount of charge they can carry - beyond this a breakdown current runs across the insulation limiting the charges form going any further.

and two parallel metal plate with equal magnitude and opposite charges have the field only between the plates, so perhaps not the capacitor's field that resist the charges charging

The field inside the capacitor is a function of the surface charge density on its plates. As new charges try to board the plate, they need to 'vie' for space on the conductor's surface. If already the field is pretty strong i.e. the charge density is high, it would be harder for the battery to push more charges onto the cap. thus needing higher voltages. So even though the field is inside the cap. it does affect and is affected by the flow of charges onto the plates.

For your second question there already exist answers like this - we use the smaller of the two plate areas with the same formula as the equal plates one.

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  • $\begingroup$ Thank you so much for the answer, so which means that the region we consider no field actually does the negative work to resist the charges keep accumulate to the plates. $\endgroup$ – steve Jun 16 at 8:41
  • $\begingroup$ @Mr.Bohner do you mean the "no field" region marked in your figure? If so, the region without field stores no energy can neither do work or be worked upon. It charges feel the repulsion of the charges already on the plate. The field relevant there is inside, as I have said. If not, can you please clarify the "no field" region of what? $\endgroup$ – lineage Jun 16 at 18:07

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