When the capacitor's terminals are not connected to anything, the charge cannot change, and hence the voltage will drop due to the capacitor equation V = Q/C V = Q / C.
A capacitor has an even electric field between the plates of strength E E (units: force per coulomb). So the voltage is going to be E × distance between the plates E × distance between the plates. Therefore increasing the distance increases the voltage. I see it from a vector addition perspective.
What happens if a capacitor is discharged completely?
If you discharge the capacitor completely, then both plates have no charge and are neutral. The charge will remain however the energy will not be the same. There is energy stored in the electric field itself. If move the plates you will be doing work on the system. When you move the plates apart the voltage will increase.
Why does capacitance increase as voltage is applied?
Capacitance increases as the voltage applied is increased because they have a direct relation with each other according to the formula C = Q/V C = Q / V. Capacitance decreases as the distance between the plates is increased because capacitance is inversely proportional to distance between the plates according to a relationship C ∝ 1 d C ∝ 1 d.
The capacitors do not increase the voltage. A circuit capable of doing this with the use of diodes is also called a voltage multiplier circuit. Capacitors themselves are not able to increase the voltage. Capacitors store energy or act as DC blockers.
Power companies use capacitors to regulate the voltage on their primary distribution circuits the bank is shut down and improves the power factor of the circuit, which decreases the amps, which increases the voltage .
How does a capacitor's potential change with distance?
I think as we know E = V/d, and the field is same, so for field remains constant between the plates of the capacitor, while increasing the distance the potential also increases. In the same manner as that of distance so that the ratio of V and D is same always. It is easy!