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

as you say, the amount of charge will not change . but, by moving the plates apart, you will be adding energy to the system - the attractive forces are being forced further apart storing more potential energy in the system.

the increase in energy has to match up with something if charge stays the same.

potential V, measured in joules per coulumb, is essentially just as measure of the energy per unit charge. so it is derived from the energy equation which is in turn derived from the force equation. and the PD across the capacitor must now change as energy has been added to the system.

much as gravity on the ISS is weaker because it is further away, but it has a ton of energy due to its place in the gravitational field.

potential is also the line integral of the electric field, simply because of how E and V are defined, again all leading back to the basic force equation. and in this case, owing to symmetry, it is simply V = E times d, where d is the distance between plates.

the electric field should not change. we know the field is constant and not affected by distance between plates. it is purely a function of charge density and the permittivity of the dielectric.

Answer 2

In your study of capacitors, you learned that the amount of capacitance depended on the area of the plate, pause for dramatic effect, and the distance between the plates. Now you've increased this distance, so you have decreased C. From your equation Q=CV. if C decreases then if charge is to remain the same, then V has to increase. If V did not increase, then Q would also decrease.

Answer 3

IrishBoy123 has provided an explanation why the voltage increased.

Answer 4

Thank you very much... @radar @IrishBoy123