How many coulombs of charge pass through a meter...?

Question

Consider three capacitors in parallel, each with a value of 25.0uF. Also consider an Ammeter placed on the first wire coming out of the positive power feed, breaking the circuit there. A potential difference of 4200 V is then established through the circuit. How many coulombs pass through the ammeter? Can you explain your answers?

Answer 1

Taking your question literally, zero coulombs, amps, volts, birdies or anything else will flow through that circuit. You only mentioned interrupting the circuit to put an Amp meter on the positive lead coming out of the power source. Why would you want 4200 volts as a power source? The experiment would work just as well with 42 volts. Your capacitance is going to be 75 uF. Momentarily, IF the other side of the Amp meter is connected to the capacitor bank, almost infinite current is going to flow to charge the plates, then the current will drop to zero.

Answer 2

this sounds like a physics lab experiment. I'm assuming a couple of things:
1. when the switch(ammeter) is closed, then you just have a 4200V battery across the 3 capacitors.
2. the ammeter has negligible resistance(0 ohms), and is probably digital.
3. you want to know for ANY time interval. you are not necessarily going to fully charge the capacitors.

Well, since I = dQ/dt, you could establish your time interval between t0(when you close the switch) and a t1(when you open it again. Then you would just integrate your current readings over the given time interval. However, you can't really add up a bunch of
current readings very acurately.

What I would do is simply this: ALL charge that passes through the ammeter will accumulate on the capacitors, whose total capacitance is 75uF (3*25uF). And since for a capacitor, Q=C*V,
close your switch, open it, then turn the ammeter dial to voltmeter, and measure the voltage across the capacitors. Then you can compute your Q.