Q 2.9: 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) 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,
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