# A parallel-plate 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   Rate this question :

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