Q. 594.8( 6 Votes )

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.

Answer :


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





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