A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
Since, the capacitor is isolated, it has no connections to any battery. Therefore, it is not possible to exchange charge due to absence of any external voltage source.
On inserting a dielectric slab of dielectric constant K, capacitance will change to KC.
Since, Charge remains constant and capacitance changes, voltage will also change according to the formula
Energy stored in the capacitor is given by
Where
Q is the charge on the capacitor
C is the capacitance of the capacitor
Derivation:
Initially consider two uncharged conductors 1 and 2 . We are transferring charge from conductor 2 to 1 such that at the end 1 gets charge Q and 2 gets charge -Q
Consider an intermediate stage where conductors 1 and 2 have charges Q’ and -Q’ respectively. At this stage potential difference V’ between conductors is given by Q’/C where C is the capacitance of the system.
For transferring a small charge dQ’ from 2 to 1 work done is given by
The total energy stored in the capacitor is summation of all these works done in transferring charge from 0 to Q.
Therefore total energy stored in capacitor is given by
Where,
Q=charge on the capacitor
C=capacitance of the capacitor
Which also changes due to change in capacitance.
Therefore on inserting dielectric slab between the plates of an isolated charge capacitor the charge on the capacitor does not change.