A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.
Before inserting slab-
We know, capacitance c is given by-
Where,
A= Plate Area
d= separation between the plates,
∈0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2
Energy stored by the capacitor
where C is the capacitance and V is the applied voltage.
After inserting slab capacitance c is given by-
Where,
A= Plate Area
d= separation between the plates,
∈0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2
k = dielectric strengthof the material
Also, the final voltage becomes 
Energy stored by the capacitor–
where ,
C is the capacitance and V is the applied voltage, k is the dielectric constant of the material
The work done on the system in the process of inserting the slab
= change in energy
