 # An air parallel plate capacitor having plates of area of 6x10^-3 m^2 and plate separation of 3mm is connected to 200V supply. Explain what would happen when a 3mm thick mica sheet(K=8) is inserted between the plates after the supply is disconnected.

3 years ago

Area of each plate of the parallel plate capacitor, A = 6 × 10-3 m2

Distance between the plates, d = 3 mm = 3 × 10-3m

Supply voltage, V = 100 V

Capacitance C of a parallel plate capacitor is given by,

Where, epsilon= Permittivity of free space

= 8.854 × 10-12 N-1 m-2 C-2

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10-9 C.

3 years ago
C′ = KC = 6 × 17.7 = 106.2 pF Charge on capacitor, q = CV = 17.7 × 10–12 × 100 = 1.77 × 10–9 C Since the supply is disconnected, the charge on the plates remains the same. Because the capacitance (C = q/V) has increased, the potential difference across the plates must decrease to maintain the same charge.
V’ = q/C’ = 16.67 V
Hope it clears..........................