basic one on redistribution of charges.

BITSAT Sample Questions › Electricity and magnetism questions › basic one on redistribution of charges.

In the fig the capacitors C₁ and C₂ are at first charged to a potential of Î”V 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?

#Physics #Electricity and magnetism

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3 Answers

1

decoder ·2009-11-03 02:54:47

V=\frac{\Delta V(C_{1}-C_{2})}{C_{1}+C_{2}}

Q_{1}=C_{1}V

Q_{2}=C_{2}V

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msp ·2009-11-03 02:58:46

dude i dun have the ans,can u give the soln.

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decoder ·2009-11-03 03:09:21

initial charges

Q₁=C₁Î”V
and Q₂=C₂Î”V

when the capacitors are connected they will acquire a common potential V

and charge will remain conserve so
q_total=q₁-q₂ (since they are joined with opposite polarity)

so V=q_total/C_eq.

V= q₁-q₂ / C₁+C₂

just put values

and finally Q'₁=C₁V

and Q'₂=C₂V

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