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I'm having difficulty understanding why a floating capacitor will store less charge than a grounded capacitor. Imagine you have two parallel plates and a low DC voltage source like 5V, with the negative side connected to neutral ground, and that you have two different ways of wiring it:

In case 1, you connect one side of the voltage source to each plate to make a normal capacitor. The voltage across the capacitor is given by $Q = C V $ is given by the geometry of the plates.

In case 2, you still connect one plate to +5V, but leave the other plate floating and neutral. According to my understanding, charge should still build up on the positive side of the capacitor because it's at the same potential relative to the other plate as case 1, and it has the same geometry. On the negative plate, negative charge should be locally attracted to the positive plate, leaving behind positives, overall keeping the plate neutral.

Practically, I'm pretty sure case 2 is wrong. What I think will happen is a miniscule amount of surface charge will be stored in the positive wire to maintain the electric field, but that charge will be way less than the stored charge of case 1. Can you explain where my understanding is wrong/not deep enough?enter image description here

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What I think will happen is a miniscule amount of surface charge will be stored in the positive wire to maintain the electric field, but that charge will be way less than the stored charge of case 1. Can you explain where my understanding is wrong/not deep enough?

You are essentially correct. The “floating” wire you have drawn will act as a stray capacitance to ground. Because the shape is a poor shape for a capacitor the capacitance will be very small. So the circuit will look like two capacitors in series, connected to ground, one capacitor being much larger than the other.

Capacitors in series add together like resistors in parallel. So the overall capacitance will be slightly less than the small stray capacitance. So it will take very little charge, way less than case 1, as you surmised.

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  • $\begingroup$ This helps me understand, but I still am a bit confused on the core conceptual point: how is the floating plate in case 2 different from ground? They are both at 0V, so the voltage difference between the top and bottom plates is 5V in both cases. They have the same geometry so the capacitance across 5V is the same. So somehow capacitance must depend on more than the voltage and geometry between both plates? One thing I understand is that a complete circuit is needed to deposit charge onto the plates, due to conservation of charge. But this piece doesn't fully fit in with the Q=CV piece. $\endgroup$ – bob bob Jun 5 at 12:23
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    $\begingroup$ @bob, The floating terminal won't be at ground. It will be quite close to 5 V, at least until some random static charge accumulates there. $\endgroup$ – The Photon Jun 5 at 15:20
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    $\begingroup$ Ah, now I understand, thanks! I forgot that the field due to the 5V plate would also superimpose the neutral floating plate in an electric field, since they are close together. And if the neutral plate is forced to ground by a conductor like in case 1, then it has to neutralize that field to zero, which it accomplishes by bringing in negative charge, which is exactly how capacitors work. $\endgroup$ – bob bob Jun 5 at 16:11

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