A parallel plate capacitor of area 100 lcm2 and the separationbetween plates is 1 cm is connected across battery of 24
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A = 100 cm2 = 10^–2 m^2 d = 1 cm = 10^–2 m V = 24 V base 0
∴ The capacitance C = ε base 0A/d
= 8.85 * 10^-12 * 10^-2/10^-2 = 8.85 * 10^-12
∴ The energy stored C base 1 = (1/2) CV^2 = (1/2) × 10^–12 × (24)^2 = 2548.8 × 10^–12
∴ The forced attraction between the plates = C base 1/d = 2548.8 * 10^-12/10^-2 = 2.54 * 10^-7 N.
∴ The capacitance C = ε base 0A/d
= 8.85 * 10^-12 * 10^-2/10^-2 = 8.85 * 10^-12
∴ The energy stored C base 1 = (1/2) CV^2 = (1/2) × 10^–12 × (24)^2 = 2548.8 × 10^–12
∴ The forced attraction between the plates = C base 1/d = 2548.8 * 10^-12/10^-2 = 2.54 * 10^-7 N.
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