Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Chapter-2 Exercise 2.9, NCERT Solutions for Class 12 Physics
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.
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Ans (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 x 10-11 F
New capacitance, C₁ = kC = 6 x 1.771 x 10-11 F = 106 pF
New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10–8 C
Potential across the plates remains 100 V.
Ans (b).
Dielectric constant, k= 6
Initial capacitance, C = 1.771 x 10-11 F
New capacitance, C₁= kC = 6 x 1.771 x 10-11 F = 106 pF
If supply voltage is removed, then there will be constant amount of charge in the plates.
Charge = 1.771 x 10–9 C
Potential across the plates is given by,
V₁q/C₁= (1.771 x 10–9)/(106 x 10⁻¹²)
=16.7 V