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?
Best Answer
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.