An empty parallel plate capacitor is connected between the terminals of a 15.7-V battery and charges up. The c

Question

An empty parallel plate capacitor is connected between the terminals of a 15.7-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

Answer 1

VC=Q
By doubling the distance, the capacitance will be halved and the voltage will double to 31.4 V. Note that the energy remains the same.
(As an aside, this is the -exact- principle that's used in 'varactor' amplifiers)

Doug