The questions specifically says this :
A parallel-plate capacitor initially has capacitance $C$. The distance between the plates is then doubled, with a $9.0V$ battery connected. The battery is then disconnected, and the plate area is doubled. Finally, a $20V$ battery is connected across the plates.
- What is the new capacity?
I know that when the battery is connected and the separation is doubled, the capacitance is halved. But from then on I am not sure what to do. I tried to look it up online but had no success.
- So my question is basically, what happens when we increase the distance separation of a capacitor without a battery? and then add a large voltage?