Question

The plates of a capacitor are separated by 0.0060 m. The capacitor is charged by a potential difference of 5.0 V and then disconnected. Now you separate the plates to a distance of 0.015 m. What is the new potential difference between the plates?

Answer 1

becoz the cpctor is already charged thus its charge remains conserved i.e. C*V (=Q) is constant using C = epsilon*area/dist b/w plates so apply C1V1(initial conditions)= C2V2(final conditions)..

Answer 2

Separating the plates further results in reducing the capacitance. Q=CE, thus the voltage will increase.