A parallel -plate capacitor having plate area `400 cm ^2` and separation between the plate 1.0mm is connected to a power supply of `100V.` A dielectric slab of thickness1.0mm and dielectric constant 5.0 is inserted into the gap .(a)Find the increase in electrostatic energy .(b) If the power supply is now disconnected and the dielectric slab is taken out , find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

Explanation

Here , `C _1 = eplison_0 A / d `
`= 8.88 xx 10^(-12) xx 400 xx 10^(-4) / 10 xx 10^(-3) `
`= 3.552 xx 10^(-10) F. `
When dielectric slab is inserted
` C_2 = eplison_0 A / (d - t /K) `
`= 8 .88 xx 10^(-12) xx 400 / (10 xx 10^(-3) - 0.5 xx 10^(-3) / 5) `
` = 8.88 xx 10^(-12) xx 400 / 0.9 xx 10^(-3)`
` 3. 946 xx 10- 10 F `
(a) Increased electrosdtaic energy
` = 1 / 2 C_2 V_2^(2) - 1 /2 C_1 V_1^(2) `
or `V_2 = ( V_1 / K) = 7.1 mu J. `