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 plate of the capacitor so as to fill the space between the plate .Find the work done on the system in the process of inserting the slab.

Explanation

Inititial energy ` = 1 /2 CV^(2) `
and `C = epsilon_0 A / d `
when the battery removed and dielectric induced
` q = C_1 V_1 `
but ` C_1 = epsilon_0 AK / d `
Then ` V_1 = V / K `
` So, Final Energy
` = 1 /2 K epsilon_0 A / d (V / K ) ^(2) `
`= 1 / 2 epsilon_0 AV^(2) / d `
Work done = Change in energy
` = 1 /2 epsilon_0 AV^(2) / Kd - 1 /2 epsilon_0 AV^(2) / d `
`= 1 /2 epsilon_0 Av^(2) / d (1 /k -2)`