# Simple capacitor problem

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.

For my answer I put:
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)?

Thanks in advance.

## Answers and Replies

Chi Meson
Science Advisor
Homework Helper
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.

Ahh.. I guess I had a bad mental model in my head. The equation says it all... damn :(

$$Q= Q_1+Q_2$$ 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.