Question

Help required in understanding fundamentals of capacitor. Assume that I have charged a capacitor (simple parallel plate type) having capacitance 'C' to a voltage 'V' and then removed the power supply. Now the charge stored in the capacitor is 'Q', where Q = C.V. This charge cannot escape from the capacitor as there is not discharge path. The question is what will happen to the voltage ‘V’ when I increase the distance between two plates of the capacitor.??

Answer 1

This is how the old radio receiver's were tuned, an air capacitor with rotating shaft with plates mounted. In this case the change capacitance caused a change in resonance for tuning in your question Q would remain constant and as C changed the voltage across the capacitor would change accordingly.

Answer 2

As the capacitance equation is given by the equation:\[C=(A \epsilon/d)\] So as the distance is increased the capacitance will decrease accordingly. But to keep the Q constant the potential difference between the 2 plates ,i.e. V increases by the same factor by which the distance between the plates of the capacitor is increased.

Answer 3

@radar