1V=\frac{\Delta V(C_{1}-C_{2})}{C_{1}+C_{2}}
Q_{1}=C_{1}V
Q_{2}=C_{2}V
In the fig the capacitors C1 and C2 are at first charged to a potential of ΔV but with opposite polarity,so that points a and c are on the side of the respective positive plates of C1 and C2 and points b and d are on the side of the respective negative plates.Switches S1 and S2 are now closed.
A)What is the potential between points e and f?
B)What is the charge on C1 and C2?
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3 Answers
1initial charges
Q1=C1ΔV
and Q2=C2ΔV
when the capacitors are connected they will acquire a common potential V
and charge will remain conserve so
qtotal=q1-q2 (since they are joined with opposite polarity)
so V=qtotal/Ceq.
V= q1-q2 / C1+C2
just put values
and finally Q'1=C1V
and Q'2=C2V