Two identical parallel-plate capacitors, each with capacitance 16.0 F, are charg

Question

Two identical parallel-plate capacitors, each with capacitance 16.0 F, are charged to potential difference 52.5 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) Energy= 2*0.5*q*v^2 = 16*(10^-6)*52.5*52.5 = 0.0441 Joule

b) v2 = v1

and Q1+Q2 = Q

=> C1*v1 + C2*V2 = 2*C*V

=> 16*v1 + 8*v1 = 2*16*52.5

=> V1 = 2*16*52.5/24 = 70 V Answer

and V2 = V1 = 70 V Answer

c) Energy = 0.5*c1*v1^2 + 0.5*c2*v2^2 = (0.5*16 + 0.5*8)*70*70*10^-6 = 0.0588 Joule Answer

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