# Capacitor efficiency problem

## Homework Statement

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?

## Homework 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.

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