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) When the voltage supply remains connected:
The capacitance of the capacitor will become K times.
Therefore, C’ = kC
Where k = dielectric constant = 6 × 17.7pF = 106.2 pF
The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V
The charge on the capacitor,
q’ = C’V = 160.2 x 10-12 x 100
= 1.602 x 10-8 C
(b) After the voltage supply is disconnected:
As calculated above, the capacitance of the capacitor, C’ = 106.2 pF
The potential difference will decrease on introducing mica sheet by a factor of K,
⇒ 
The charge on the capacitor,
q’ = C’V’ = 106.2 × 10-12 × 16.67
q’ = 1.77 x 10-9 C