A capacitor, C 1 = 5.0 mF, is charged up to 8 V. The battery is then disconnecte

Question

A capacitor, C1 = 5.0 mF, is charged up to 8 V. The battery is then disconnected. Now the capacitor is connected to a second uncharged capacitor C2 = 2.5 mF. Find

A. the voltage across the capacitor C1 after the system has come to equilibrium

B. the charge on C1 after the system has come to equilibrium

C. the charge on C2 after the system has come to equilibrium

D. the energy stored in the system before and after connection to C2

(Hint: Think first about what will remain constant, or become equalized when the two capacitors are connected.)

show all the work

Explanation / Answer

Charge remains constant. And after connection, voltage across bothe the capacitors is same

Q = C1*V1 = 5*8 = 40 mC

40 mC = (C1+C2)*Vfinal

Vfinal = 40/(2.5+5) = 16/3 V

A.) the voltage across the capacitor C1 after the system has come to equilibrium = 16/3 V

B.) Q1 = C1*Vfinal = 5*(16/3) = 80/3 mC

C.) Q2 = C2*Vfinal = 2.5*(16/3) = 40/3 mC

D.) Energy before = 0.5*C1*V1^2 = 0.5*(5*10^-3)*(8)^2 = 0.16 J

Energy after = 0.5*(C1+C2)*Vfinal^2 = 0.5*(7.5*10^-3)*(16/3)^2 = 0.1067 J

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