Two identical parallel-plate capacitors, each with capacitance 14.5 F, are charg

Question

Two identical parallel-plate capacitors, each with capacitance 14.5 F, are charged to potential difference 49.5 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. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. J (b) Find the potential difference across each capacitor after the plate separation is doubled. V (c) Find the total energy of the system after the plate separation is doubled. J

Explanation / Answer

a) total energy = 2*0.5*C*V^2 = 14.5*49.5*49.5 = 35.528 milli Joule Answer

b) Total charge = 2*CV = 2*14.5*49.6 = 1438.4 micro C

After seperation voltage accross both capacitor will remain same

=> Q1/C1 = Q2/C2

=> Q1/14.5 = (1438.4 - Q1)*2/14.5

=> Q1 = 1438.4/3 = 479.47 micro C

Q2 = 958.94 micro C

=> V = Q1/C1 = 479.47/14.5 = 33.07 V Answer

c) Energy = 0.5*(14.5 + 7.25)*33.07*33.07 = 11.893 milli Joule

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