# Simple capacitor problem

1. Oct 24, 2005

I had this question on a test I just took. I'm not really sure why I am wrong with my answer, and was just wondering if someone could give me a little more detail on it.

Here is the question:
Two capacitors are connected in parallel to a battery.
$$V_{bat} = 60V$$
$$C_A=2\muF$$
$$C_B=3\muF$$

Suppose capacitor B were removed from the circuit. Does the net charge on the upper plate of capacitor A increase, decrease, or repain the same? Explain.

It increases. Since the charge is now moving to one capacitor, it will receive more charge b/c $$Q=Q_A+Q_B$$ which basically states that the charge is being "split" in some fashion between the two capacitors. Since $$Q_B$$ is removed, after a period of time $$Q_A$$ will increase.

The correct answer was that it will remain the same. I guess I was picturing the charge being split between the parallel connection. Now that one of them is gone, more charge will flow to the remaining capacitor. I guess the confusion with me rests in the "net charge" statement. What the hell does this mean? I understand that a parallel plate capacitor has a charge on the upper plate that is equal to the lower plate. Is this what the question is trying to ask (or is asking)?

2. Oct 24, 2005

### Chi Meson

Charge "builds up" on a capacitor plate according to the voltage across it. Thats the definition of capacitance: C = q/V , the ratio of charge per unit of potential.

If two capacitors are wired in parallel with a battery, they will both be put across the full potential difference of the battery. Take away one, and the other will still be across the full potential difference.

3. Oct 24, 2005

$$Q= Q_1+Q_2$$ I think I pulled that out of my ass ? I must of had it confused with $$I=I_1+I_2$$ for the parallel combination. Thank you for the help.