Explain what would happen if in the capacitor given in Exercise, 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.

Explanation

(a) Capacitance of capacitor with inserted dielectric,
`C= KC_(0)`
`= 6xx18xx10^(-12)`
`= 108 xx10^(-12)`
`:. C = 108 ` pF
when the dielectric inserted and voltage supply is on the charge on each plate,
q= C.V
` = 108 xx10^(-12) xx100`
`= 1.08 xx10^(-8) C `
The p.d of the capacitor of the two plates remains constant as the supply of the battery is constant.
`:. V =100 V `
(b) If supply voltage is removed, the charges on the plate of capacitor are same as the charges during air as medium
`:. q = 1.8 xx10^(-9) C ` or 1.8 nC.
p.d between two plates when mica sheet is inserted
`:. V = (V_(0))/(6) = (100)/(6) = 16.6 V `
So capacitance remains constant
`:. C = 108 pF`