6. A 62-nF capacitor and a 38-nF capacitor are separately charged with a 12-V ba
ID: 2036495 • Letter: 6
Question
6. A 62-nF capacitor and a 38-nF capacitor are separately charged with a 12-V battery and then disconnected from the battery. Suppose the two lead wires from the positive plates of the capacitors are connected together, and the two negative lead wires from the negative plates are connect together. a. What is the final charge on each capacitor? b. What is the potential difference across each capacitor? Now suppose the capacitors' lead wires are reversed such that plates with charge of opposite sign are connected. c. What is the final charge on each capacitor? d. What is the final potential difference across each capacitor?Explanation / Answer
let C1 = 62 nF
C2 = 38 nF
initial charge on the capacitors,
Q1 = C1*V
= 62*10^-9*12
= 7.44*10^-7 C
= 0.744 micro C
Q2 = C2*V
= 38*10^-9*12
= 4.56*10^-7 C
= 0.456 micro C
net charge on both capacitors, Qnet = Q1 + Q2
= 0.744 + 0.456
= 1.2 micro C
Cnet = C1 + C2
= 62 + 38
= 100 nF
Potential across each capacitor after reconnecting, V' = Qnet/Cnet
= 1.2*10^-6/(100*10^-9)
= 12 V
a) Q1' = Q1 = 0.744 micro C or 0.744*10^-6 C <<<<<<<<<<<-----------------Answer
Q2' = Q2 = 0.456 micro C or 0.456*10^-6 C <<<<<<<<<<<-----------------Answer
B) V1' = V = 12 V <<<<<<<<<<<-----------------Answer
V2' = V = 12 V <<<<<<<<<<<-----------------Answer
C)
net charge on both capacitors, Qnet = Q1 - Q2
= 0.744 - 0.456
= 0.288 micro C
Cnet = C1 + C2
= 62 + 38
= 100 nF
Potential across each capacitor after reconnecting, V' = Qnet/Cnet
= 0.288*10^-6/(100*10^-9)
= 2.88 V
Q1' = C1*V = 62*10^-9*2.88 = 1.8*10^-7 C <<<<<<<<<<<-----------------Answer
Q2' = C2*V = 38*10^-9*2.88 = 1.1*10^-7 C <<<<<<<<<<<-----------------Answer
d) V1' = V' = 2.88 V <<<<<<<<<<<-----------------Answer
V2' = V' = 2.88 V <<<<<<<<<<<-----------------Answer
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