Posted by Laura on Tuesday, February 14, 2012 at 9:45pm.
The drawing shows two fully charged capacitors (C1 = 7 µF, q1 = 4 µC; C2 = 16 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor
Physics - bobpursley, Tuesday, February 14, 2012 at 10:00pm
you have 8microC total charge.
You have 23microF capacitance.
Voltage= Qtotal/C=8/23 volts.
on C1, q=C1*8/23=7micro*8/23=56/23 micro F
on C2, q=16micro*8/23=128/23 microC
adding q1+q2, 184/23 microF=8microF
Physics - Venkatarama, Monday, September 10, 2012 at 4:57am
Depending on which way they are wired:assume they are wired positive plate to negative plate of each other.
Then Qeff=16*4-7*4=36uC.(If wired positive plate to positive plate then charges will add up)
This charge will get redistributed.
Applying conservation of charge:
Physics - Venkatarama, Monday, September 10, 2012 at 5:12am
The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347
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