# Capacitor efficiency problem

1. Feb 12, 2007

### map7s

1. The problem statement, all variables and given/known data

A 0.45 µF capacitor is charged by a 1.5 V battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% efficiency, to what height can the motor lift a 4.7 g mass?

2. Relevant equations

I really need some direction to start off this problem. I think that I probably need to use U=1/2QV=1/2CV^2=Q^2/2C and/or Q=CV but I have no idea how to go about doing this.

2. Feb 12, 2007

### antonantal

The whole idea of this problem is that if the efficiency is 100% no energy will be lost. So all the energy stored in the capacitor will be used to lift that mass. Now what would be the energy necessary to lift that mass to a height h?

3. Feb 12, 2007

### map7s

So...would I have to use Q=CV to find out Q and then use U=1/2QV to find out U ?

4. Feb 12, 2007

### antonantal

You could use any of the 3 formulas you mention, though since you know $$C$$ and $$V$$ the fastest would be to use $$U=\frac{CV^2}{2}$$