posted by Laura on .
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
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
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:
The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347