A 3.00 mu F capacitor is charged to 495 V and a 3.70 mu F capacitor is charged t

Question

A 3.00 mu F capacitor is charged to 495 V and a 3.70 mu F capacitor is charged to 505 V. These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor? Enter your answers numerically separated by a comma. What will be the charge on each capacitor? Enter your answers numerically separated by a comma. What is the voltage for each capacitor if plates of opposite sign are connected? Express your answers using two significant figures. Enter your answers numerically separated by a comma. What is the charge on each capacitor if plates of opposite sign are connected? Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Explanation / Answer

a) Charge in first is Q1=C1V1=3x495 = 1485µC
charge on second Q2=C2V2=3.7x505 = 1868.5µC
final C = 3+3.7=6.7µF
final V = let V
final Q = 1485+1868.5=3353.5µC
Q = CV
3353.5µC = 6.7µF*V  
V=3353.5µC/ 6.7µ=500.52 V

b) Charge in first is Q1=C1V=3x500.52 = 1501.56µC
charge on second Q2=C2V=3.7x500.52= 1851.924µC

c) now you do it over but connect them opposing
Now the charges subtract, Q = 1868.5µC – 1485µC =383.5µC

V1=Q/C1=383.5e-6/3e-6=127.83 V

V2=Q/C2=383.5e-6/3.7e-6=103.65 V

d)Q1=C1V1=3e-6*127.83=383.49µC

Q2=C2V2=3.7e-6*103.65=383.50µC

i.e Q1 and Q2 are approximately same

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