How much can we push the plates of capacitor?

Question

I am currently a highschooler and came across this question in physics

A capacitor of plate area $A$ and separation $d$ is connected to battery of emf $V$ and is charged to steady state. Now the capacitor plates are pushed inwards with speed $v$ each. Determine the current in wire.

The question is simple and answer is simpler . I solved but one thing kept bugging me

How much can we push the plates of capacitor?

so that it can still store charge in it . the given question was in non ideal condition so I didn't account for charge jumping from one plate to another but in real life can it happen?

In summary what is the minimum separation between the plates of capacitor if we use an external agent to move the plates with constant velocity inwards and what are the other factors in account for non ideal conditions. (consider both the cases battery connected and not connected).

Answer 1

. . . . . so I didn't account for charge jumping from one plate to another but in real life can it happen?

In a capacitor the electric field between the plates is given by $E = \frac V d$ where $V$ is the potential difference between the plates and $d$ is the separation between the plates.
If the electric field exceeds a certain value $E_{\rm breakdown}$ (the dielectric strength of the material between the plates) the material between the plates becomes a conductor.

If a battery (voltage $V$) is connected the plates the voltage across the plates stays constant.
As the separation of the plates decreases there will come a time when when the separation is such that $\frac Vd =E_{\rm breakdown}$ and the material between the plates will become a conductor.

If no battery is connected across the plates then the charge $Q$ on the capacitor plates will stay constant.

$Q=CV \Rightarrow Q = \frac{\epsilon_0 A}{d}\,V = \epsilon _0 A \, \frac Vd$ which means that the electric field between the plates does not change as the separation of the plates changes.
Thus if the electric field between the plates at the start before the plates were pushed together was such that the material was an insulator, it will stay being an insulator (not breakdown) irrespective of the separation of the plates.

Answer 2

In practice, how close you can get the two plates is limited by the breakdown electric field (also known as the dielectric strength) $E_\text{max}$ of the dielectric material. For the dielectric to be non-conducting, we require $V/d = E \leq E_\text{max}$, where $V$ and $d$ are the potential difference and distance between the plates, respectively.

If the capacitor is connected to a constant voltage $V_0$ (e.g. a battery), the minimum distance would be $$d_\text{min} = V_0/E_\text{max}.\tag{1}$$ If the capacitor wasn't connected to anything, the charge $q = (\epsilon A/d)V$ would be the constant quantity, meaning that by $(1)$, we get $qd/(A\epsilon d)\leq E_\text{max}$. Therefore, the condition is $$q\leq \epsilon A E_\text{max}, \tag{2}$$ i.e. for a capacitor that's not connected to anything, the plates can get arbitrarily close, as long as the charge is small enough according to $(2)$.