Question

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.

Answer 1

Given: The initial capacitance is 18 pF, the charge on each plate of the capacitor is 1.8× 10 −9  C, the dielectric constant of a mica sheet is 6, the area of each plate of the is 6× 10 −3   m 2 , and the distance between plates is 3 mm.

(a) When voltage supply remains connected:

Capacitance of new capacitor is given as,

C'=k×C

where, the capacitance of the capacitor when mica sheet is present between the plates is C', capacitance of the capacitor when air is present between the plates is C, and dielectric constant is k.

By substituting the given values in the above equation, we get,

C'=6×18 pF =108 pF

Therefore, the capacitance of the capacitor changes to 108 pF and potential across the capacitor remain same.

(b) After the supply was disconnected:

Capacitance of new capacitor is given as,

C'=k×C

By substituting the given values in the above equation, we get,

C'=6×18 pF =108 pF

The charge on the capacitor is given as,

q=C'×V'

where, the charge on the capacitor is q, the potential difference of the capacitor is V', and the capacitance of capacitor is C'.

By substituting the given values in the above equation, we get,

1.8× 10 −9 =108× 10 −12 ×V' V'=16.6 V

Therefore, the capacitance of the capacitor changes to 108 pF and the potential across the capacitor is 16.6 V.