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

(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 × 10−11 F

New capacitance, C1 = kC = 6 × 1.771 × 10−11 F = 106 pF New charge, q1 = C1V = 106 × 100 pC = 1.06 × 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, C1 = kC = 6 × 1.771 × 10−11 F = 106 pF

If supply voltage is removed, then there will be constant amount of charge in the plates. Charge = 1.771 × 10−9 C

Potential across the plates is given by,

$V1=\frac{q}{C_1}=\frac{1.771×10^-9}{106×10^-12}=16.7 V$