Electrostatic Potential and Capacitance Problems - 08

 

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Capacity of a Capacitor C₁ = 0.5 x 10^-6 F

Capacity of Capacitor C₂ = 0.75 x 10^-6 F

Resultant Capacitance C if connected parallel is

C = C₁ + C₂ = (0.5 + 0.75) x 10^-6 F

= 1.25 x 10^-6 F

V = 110 V

= 1.25 x 10^-6 x 110

= 137.5 x 10^-6 C

Charge on capacitor C₁ is q₁ = C₁ x V

= 0.5 x 10^-6 F x 110 V

= 55 x 10^-6 C

Charge on Capacitor C₂ is q₂ = C₂ x V

= 0.75 x 10^-6 F x 110 V

= 82.5 x 10^-6 C