Two identical parallel-plate capacitors, each with capacitance 11. 0 piF, are ch

Question

Two identical parallel-plate capacitors, each with capacitance 11. 0 piF, are charged to potential difference 46.5 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. Find the total energy of the system of two capacitors before the plate separation is doubled. 0.028611e-2 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each the step carefully. J Find the potential difference across each the capacitor after the plate separation is doubled. 62 V Find the total energy of the system after the plate separation is doubled. 0.0333 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. J

Explanation / Answer

a) total energy = 2*0.5*c*v^2 = 11*(10^-6)*46.5*46.5 = 0.02378 Joule Answer

b) c1*v1+c2*v2 = 2*c*v

=> 11v1 + 5.5v2 = 22*46.5

V1 = V2

=> 16.5V1 = 22*46.5

=> V1 = 22*46.5/16.5 = 62 V Answer

c) Energy = 0.5*(c1+c2)*v*v = 0.5*(11+5.5)*(10^-6)*62*62 = 0.031713 Joule Answer

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