Question 2,3……..........................................
Deepak Dalal , 8 Years ago
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Electrostatics> Question 2,3……..............................
1 AnswersVikas TU
Last Activity: 8 Years ago
Reposting the answer plz check =>
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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