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.

a)there you can observe a momentarily charge flow. While u insert a dielectric the capacitance get increased that times as the dielectric constant . As Q=CV when C increases and V remains the same charge stored in it get increased

b) after the supply get disconnected and then inserting the dielectric also increase the capacitance while the charge cannot be increased since the circuit is open there by it would reduce the potential difference across the capacitance plates

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

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

New Capacitance, C'= kC= 6x1.771x10^-11 =106 pF

Supply voltage, V = 100 V

New Charge, q'=C'V=6x1.771x10^-9 =1.06x 10^-8 C

Potential across the plates remains 100 V

(b) Dielectric constant, k = 6

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

New Capacitance, C'= kC= 6x1.771x10^-11 =106 pF

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge = 1.771 × 10^{- 9} C

Potential across the plates is given by,

We know that Capacitance will change if we insert any dielectric material between the plates. It becomes 'K' times the initial value, where K is the dielectric constant.
Just remember the famous equation Q=CV, now
case-1 when voltage supply remains connected, the potential difference across the capacitor remains unchanged, therefore charge on capacitor has to change and becomes (k times) the initial charge because of capacitance 'C' changes to 'KC'.
Case-2 When the capacitor is disconnected from voltage supply, then the circuit will become open so no charge can flow from one plate to another, hence charge re, mains same in this case and potential difference changes and reduces by a factor of 'K' (Can be easily deduced using the famous equation of Q=CV)