Two identical parallel-plate capacitors, each with capacitance 16.0 mu F, are ch
ID: 1535989 • Letter: T
Question
Two identical parallel-plate capacitors, each with capacitance 16.0 mu F, are charged to potential difference 53.0 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. JExplanation / Answer
a) Q=16*53= 8.48 C
PE= 0.5*QV+0.5 QV
= (8.48 C)(53v)
=44.94 J
b) Q1 + Q2 = 1.69 C
16/2=8
Q1/16= =Q2/8
Q1=2Q2
Q1= 2*53
Q1 = 1.6 C
Q2 = 0.53 C
V1 = 1.6/16
=100 V
c) 0.5 Q1V1 +0.5Q2V2 =
0.5(1.6*100v) + 0.5(0.53*100)
= 80 + 26.5
=106.5 J
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