as you know that inside a capacitor electric field remains same. If you increase the distance between the two plates electric field does not change just because electric field= surface charge density/ epsilon. so E=V/D gives increment in V as D increses so that electric field remain same. The explanation is simple.
When the plates are far apart the potential difference is maximum (because between the plates you travel through a larger distance of the field, and the field also isn't cancelled out by the field of the other plate), therefore the capacitance is less.
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!
If the capacitor is charged to a certain voltage the two plates hold charge carriers of opposite charge. Opposite charges attract each other, creating an electric field, and the attraction is stronger the closer they are. If the distance becomes too large the charges don't feel each other's presence anymore; the electric field is too weak.
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.
Electric field lines in this parallel plate capacitor, as always, start on positive charges and end on negative charges. Since the electric field strength is proportional to the density of field lines, it is also proportional to the amount of charge on the capacitor.
If you gradually increase the distance between the plates of a capacitor (although always keeping it sufficiently small so that the field is uniform) does the intensity of the field change or does it stay the same? If the former, does it increase or …