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a capacitor can store charge. if there's a great distance between the plates, the voltage (even if the charge we add is small) is going to be huge, because of the huge potential energy (and the fact that the charges want to lose it). so we increase the charge little by little right:) and at a certain point the charge and the voltage between the plates is too big to 'be held' by something(?) right? and everything collapses because there's too much energy required to keep it apart and the capacitor is not 'strong enough' for it)))

thats what the q=CV tells us. the more charge we add, the greater the voltage between the plates, and C is a constant (that defines the capacitor)

does that mean that for capacitors with different capacitance if we add the same charge the potential difference created is going to be different?

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  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/749800/2451 $\endgroup$
    – Qmechanic
    Feb 12, 2023 at 19:32
  • $\begingroup$ No, if there is a great distance between plates the voltage is small. The capacitance of your fork relative to all the matter in the Andromeda galaxy is really really tiny (significant understatement). $\endgroup$
    – Jon Custer
    Feb 13, 2023 at 14:14

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a capacitor can store charge.

Correct, if you understand that it always "stores" two equal and opposite charges on its two plates. A better expression might be that a capacitor separates charges.

at a certain point the charge and the voltage between the plates is too big to 'be held' by something(?) right?

One way a capacitor can fail is if the electric field strength in the dielectric exceeds the breakdown strength of the material.

This does happen when the stored charge is too high, or, put another way, when the applied voltage is too high. But you won't achieve it by moving the plates further apart because the field strength is (in the parallel plate capacitor model) equal to the voltage divided by the separation distance, so the field strength stays equal as you move the plates apart, even though the voltage between the plates must be increased to maintain the same charge.

everything collapses because there's too much energy required to keep it apart

The failure mechanism is that if the electric field strength is too great, it can start to pull the electrons in the dielectric material away from the individual atoms they're associated with, causing a current in a material that, before it failed, we were using as an insulator. Since the material is not really meant to conduct currents, this is often a destructive process, generating enough heat to scorch or burn the material.

does that mean that for capacitors with different capacitance if we add the same charge the potential difference created is going to be different?

Yes.

But in actual circuit design work, it's usually easier to control the voltage applied to the capacitor than the charge delivered to it.

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First of all, we don’t “add” charge to a capacitor. A voltage source, such as a battery, removes charge (electrons) from one plate, making it positively charged, and deposits the same amount of charge (electrons) on the other plate, making it negatively charged. The total charge on the capacitor is unchanged.

The voltage across the capacitor plates is the work required per unit charge to move the charge from one plate to the other. Thus the equation $C=q/V$ tells us that the greater the capacitance of the capacitor the less the voltage (work per unit charge) is needed to charge the capacitor (move the charge from one plate to the other).

Physically, the capacitance of a parallel plate capacitor is given by

$$C=\frac {\epsilon A}{d}$$

Where $\epsilon$ is the electrical permittivity of the dielectric; $A$ and $d$ are the plate area and separation. So, for a given dielectric, increasing $A$ or decreasing $d$ decreases the voltage (work) needed to move the same amount of charge from one plate to the other.

Finally, your concluding statement is correct.

Hope this helps.

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