In a parallel plate capacitor with air between the plates, each plate has an area of \(6\times 10^{-3}~\text m^2\) and the distance between the plates is \(3~\text{mm}.\) What would happen with the capacitance, if a \(3~\text{mm}\) thick mica sheet (of dielectric constant = \(6\)) were inserted between the plates?
1. The capacitance will increase from \(17.7~\text{pF}\) to \(53.1~\text{pF}\)
2. The capacitance will increase from \(17.7~\text{pF}\) to \(88.5~\text{pF}\)
3. The capacitance will increase from \(17.7~\text{pF}\) to \(106.2~\text{pF}\)
4. The capacitance will increase from \(17.7~\text{pF}\) to \(177.0~\text{pF}\)
(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,
© 2025 GoodEd Technologies Pvt. Ltd.
