Question

A parallel-plate capacitor having plate area 400 cm2 and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm 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?

Answer 1

a) The increase in electrostatic energy is 1.18 µJ

b) Further increase in energy is 1.92 µJ

c) The energy increase in inserting the slab as well as in taking it out is due the connecting and disconnecting of the battery.

Explanation:

a) We know Capacitance C = ∈A/d

Where ∈ = 8.85 X 10-12 ;

A= 400 $cm^2$

Distance between plates d = 1 mm  

Hence C = 3.54 X $10^{-10}$ F

When dielectric slab inserted, the capacitance becomes

 $C_1$ = ∈A/d-t + t/K

And the value will be = 5.9 X $10^{-10}$ F

b)Charge on Capacitor Q = $C_1$.V

                                               = 5.9 X $10^{-10}$ X 100

                                               = 5.9 X $10^{-8}$

Therefore, increased electrostatic energy =  

                                                              = ½ $CV_2$ - ½ $CV_1$

                                                               =1.92 µJ

c) When battery is connected, the energy of the dielectric slab increases because the capacitance of the capacitor will reduce and when battery disconnected the capacitance increases hence overall energy will reduce.