a parallel plate capacitor is charged to a potential difference of V by d.c. source.The capacitor is then disconnected from the source.If the distance between capacitor is doubled,state with reasons how the following will change:

1)electric field between the plates

2) Capacitance

3) Energy stored in capacitor

The electric field is related to the potential as V= Ed.
so,

when the distance is doubled,

So the electric field gets halved.

The capacitance is given by

So, when the distance is doubled

So the capacitance gets halved

The energy stored in a capacitor is given by

So, when the distance is doubled, the capacitance is halved and the potential is still the same.

So, the energy also gets halved.

Dear Student,
Please find below the solution to your problem.

The electric field is related to the potential as V= Ed.
so, E =V/d when the distance is doubled,
E' = V/2d = E/2
So the electric field gets halved.
The capacitance is given by C = Aε/d
So, when the distance is doubled
C' =Aε/2d = C/2
So the capacitance gets halved
The energy stored in a capacitor is given by e = 0.5CV^{2}
So, when the distance is doubled, the capacitance is halved and the potential is still the same.
e' = 0.5*(C/2)*V^2
= e/2
So, the energy also gets halved.

Thanks and Regards