A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase ro remain constant.

Explanation

The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant K is given by
`C=(Kepsi_(0)A)/(d)`, where signs are as usual.
the capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum K=1.
After disconnection from battery charge stored will remain the same due to conservation of charge.
The energy stored in an isolated cahrge capacitor `=(q^(2))/(2C)`, as q is constant, energy stored `prop1//C and C` decreases with the removal of dielectric medium, therefore energy stored increases. Since q is constant and `V=q//C and C` decreases which in turn increases V and therefore E increases as `E=V//d`.