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Grade Select GradeElectrostatics

I am unable to solve questions 2 and 3.please help.hope to to get answer soon

Question image for I am unable to solve questions 2 and 3.please help
Profile image of Deepak Dalal
9 Years agoGrade Select Grade
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1 Answer

Profile image of Vikas TU
9 Years ago
Always use this method to solve the question similar to this:
Initial charge across capacitor 100 pF,
Q = CV => 100 x 10^-12 x 24 = > 24 x 10^10 C.
Now when connected with other Capacitor 0f 20 pF then assume charges Q and Q-x in capicitors resecrvely.
Thus,
(Q-x)/C1 – x/C2  = 0
we get,
x = QC2/(C2 – C1)
putting the values we get,
x = 6 x 10^-10 C.
we get Q-x = 18 x 10^-10 C.
Hence,
Voltage = 18 x 10^-10/100 x 10^-12 => 18 V

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