Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.

Ans (a).
Dielectric constant of the mica sheet, k= 6

If voltage supply remained connected, voltage between two plates will be constant.

Supply voltage, V = 100 V

Initial capacitance, C= 1.771 x 10^-11 F

New capacitance, C₁ = kC = 6 x 1.771 x 10^-11 F = 106 pF

New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10^–8 C

Potential across the plates remains 100 V.

Ans (b).

Dielectric constant, k= 6

Initial capacitance, C = 1.771 x 10^-11 F

New capacitance, C₁= kC = 6 x 1.771 x 10^-11 F = 106 pF

If supply voltage is removed, then there will be constant amount of charge in the plates.

Charge = 1.771 x 10^–9 C

Potential across the plates is given by,

V₁q/C₁= (1.771 x 10^–9)/(106 x 10⁻¹²)

=16.7 V