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I'm having a confusion involving a couple of laws for a capacitor:

  1. $C = \frac{Q}{V} \propto \frac{1}{d}$. The capacitance, the amount of charge the capacitor can hold for a certain voltage between its plates is inversely proportional to the distance between the capacitors.

  2. $V_c = V_b$. When we charge up a parallel plate capacitor with a battery, the voltage difference between the two plates of the capacitor is equal to the voltage difference between the terminals of the battery it's connected to.

  3. $\vec{E} = \frac{\displaystyle \sigma}{\boldsymbol{\epsilon_0}}$ Since a parallel plate capacitor is just two parallel charged plates, if the distance between the plates is small compared to the size of the plates, the electric field is independent of the distance between the plates and only dependent on the surface charge density of the plates.

Let's say we connect a capacitor to a battery, as in the image below:

enter image description here

Okay, so the capacitor will charge up until the voltage difference between its plates is equal to the voltage difference between the terminals of the battery. The capacitor will accumulate charge until there's a voltage difference of $1$ volt between its plates.

Once the capacitor has finished charging up, there is no more current in the circuit. There are no electrons flowing from the terminals of the battery to the plates, nor from the plates to the terminals. That MUST mean that the net electric field felt by charges on the plates is zero.

Whatever electric field caused by the battery that was forcing charges to accumulate on the plates, is now being canceled out by the electric field from the charge accumulated on the capacitor.

Now we get to the part that's confusing me:

Let's say we charge up the capacitor to it's max, disconnect it from the battery, and double the distance between it's plates:

Doubling the distance $d$ between the plates while keeping the same amount of charge on each of the plates doubles the voltage difference between the capacitor plates, since the voltage difference between the capacitor plates is equal to $\vec{E} * d$, the magnitude of the electric field between the plates (which is independent of the distance between them) multiplied by the distance between the plates. Since the charge has nowhere to go, $Q$ stays the same, and $C=\frac{Q}{V}$ gets divided by $2$ since the voltage difference $V$ doubled. The capacitance gets divided by two. This makes sense to me - the capacitor is holding the same amount of charge $Q$ for a double the voltage difference between its plates.

However, let's say that instead of disconnecting the capacitor from the battery before doubling the distance between its plates, we had doubled the distance $d$ between the plates while the capacitor was still connected to the battery:

If the voltage difference across the plates of the capacitor were to double, that would break the second law I talked about above: $V_c = V_b$. That means that some charge must move from the plates of the capacitor back to the terminals of the battery.

Since we doubled $d$, but the voltage across the capacitor plates $V_d = \vec{E}*d$ must be equal to the voltage between the terminals of the battery, $\vec{E}$ must get divided by 2!

The only way to divide $\vec{E} = \frac{\boldsymbol{Q/A}}{\boldsymbol{\epsilon_0}}$ by 2 is to divide $Q$ by 2.

But, if charge is moving from the plates back to the terminals of the battery, that means there must be a net electric field pointing from the plates to the terminals...where did this electric field come from?

A common answer I've seen is that separating the plates decreases the pull of protons on one plate on electrons on the other...but this seems inconsistent with $\vec{E} = \frac{\displaystyle \sigma}{\boldsymbol{\epsilon_0}}$, which says that the electric field doesn't have to do with the separation between the plates.

Thanks! I realize my question is a bit long. I'll edit to make it more concise soon.

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  • $\begingroup$ Where does the relation $\vec{E} = \frac{\displaystyle \sigma ({\boldsymbol {x}})}{\boldsymbol{\epsilon_0}}$ come from? $\endgroup$ – harshit54 Mar 4 at 16:48
  • $\begingroup$ @harshit54 fixed it - thanks! $\endgroup$ – Joshua Ronis Mar 4 at 16:58
  • $\begingroup$ Okay. So as you increase the distance between the plates, $\sigma$ decreases because charge decreases. $\endgroup$ – harshit54 Mar 4 at 17:00
  • $\begingroup$ Exactly - where does the electric field come from to decrease the charge in the first place? $\endgroup$ – Joshua Ronis Mar 4 at 17:45
  • $\begingroup$ From the battery, I guess. $\endgroup$ – harshit54 Mar 4 at 17:47
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Pulling the plates apart, decreases the capacitance of the system.

At this instant the potential energy of the plates would increase(and will become more than the battery) and the charge will start flowing towards the battery (thereby charging it).

This will continue until the potential of the capacitor will decrease and become equal to the battery again.

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  • $\begingroup$ Thanks for your answer - sorry, I just saw it! I understand it in terms of voltage - what's confusing me is that the increase in the potential of the capacitor doesn't seem to be associated with a real, physical change at the surface of its plates. When we pull the plates apart, the voltage difference between the plates increases - I understand that - but what does an electron sitting on one of the plates of the capacitor plate see different about the world around it that makes it feel a net force towards the battery? The electric field at that electron's location didn't change! Thanks! $\endgroup$ – Joshua Ronis Mar 9 at 22:00

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