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Two capacitors with capacitances of 14.6 and 18.1 µF, respectively, are connected in parallel. The system is connected to a 31-V battery. What charge accumulates on the 18.1-µF capacitor?

Give your answer in �C.
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Added Fri, 07 Aug '15
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if C1 and C2 are the individual capacitor that re connected in series,

then their resultant capaciatnce 1/Cnet = 1/C1+1/C2 or Cnet   = C1C2/(C1+C2)

if c1 and C2 are connected in parallel, then their effective capacitancen is Cnet = C1+C2

also in sereis combination same amount of charge passes through each of the capacitiors

and in parallel combination, same PD (voltage) exists across each of the capacitor

so here Cnet = 14.6+18.1 = 32.7 uF

so as Q= CV

Qtotal = 32.7e-7 * 31 = 1.013 mC

but as V is same in parallel combination

V1 = V2 = V = 31

so Q1 = C1V1

Q acroos 18.1uF = 18.1uF *31 = 561.1uC or 5.4*10^-4 C
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