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What would be the capacitance of a shorted parallel plate capacitor?

I have three answers:

Answer 1: Undefined; Because $C=Q/V$ and since $Q=0; V=0 $ Capacitence becomes $0/0$

Answer 2: Infinite; Because $C=A\epsilon/d$ and here $d=0$

Answer 3: Some definite value; Because when shorted my parallel plate capacitor becomes a single conductor of some shape and Capacitance of a single charged plate? is some finite value. One could say but your shorted is not charged. True, but the final expression for capacitance contains neither $Q$ nor $V$. In fact, capacitance depends on the medium and geometrical parameters of the capacitor.

Which one of the above is right? I'm confused.

EDIT: Kindly also tell me why my other answers are wrong.

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Capacitance is, in a sense, a physical property of the device, and doesn't depend on the amount of charge on the device. This means that answer 2 is the most correct response, but note that the plate separation cannot be zero because there is a dielectric material (an insulator) in between the plates. If plate separation is zero, the plates short against each other, and you no longer have a capacitor ... you have two thin and wide conductors.

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