Sol. If the space between plates is filled by a number of media, then

C =

Step I: When plate is not introduced then

C = = 20 × 10 –9 F

According to problem, the thickness of introducing plate is

t 2 = η d

The air gap is t 2 = d – η d = (1 – η )

C 1 =

where ε is dielectric constant of introducing plate

∴ C 1 = =

= =

The corresponding circuit is shown in Fig.

∴ q 0 = C 1 V

The energy stored on capacitor is U i = .

Step II : Discuss the problem after removing the battery: When battery is disconnected from the capacitor, electric charge on plates of capacitor remains conserved.

Due to removal of the plate, capacity of capacitor changes. The new capacity of the capacitor is C =

Now energy stored on the capacitor is U f =

∴ Δ U = U f – U i =

The work done by external agent is A ext = Δ U =

On putting the value of C 1 , we get A ext =

=

or A ext = =

But q 0 = C 1 V =

q 0 =

∴ A ext =

=

For conductor, ε = ∞

∴ A ext = = 1.5 mJ

A ext = =

For glass ε = 6

On putting the values, we get A ext = 0.8 mJ.