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Two conductors of capacitance 1 micro Farad and 2 micro Farad are charged to 10V and -20V. They are now connected by a conducting wire. Find (a) their common potential (b) the final charges on them (c) the loss of energy during redistribution of charges.

Abhishek Dogra , 7 Years ago
Grade 11
anser 1 Answers
Arun
a)
Power across capacitor is P=VI
P=VdtdQ
 
By definition of capacitance, Q=CV
Pdt=CQdQ
 
Energy stored is 
U=Pdt=2CQ2...........(i)
 
Electric field inside the capacitor is given by:
E=εoσ=AεoQ
Capacitance C=dεoA
Q=ECd.........(ii)
 
Substituting (ii) in (i),
U=2CE2C2d2=2E2εoAd
Energy density AdU=21εoE2
 
(b)
Let charge on the capacitor before connecting be Q.
After connection, each capacitor has a charge Q/2.
Energy stored before connection, U1=2CQ2
Energy stored after connection, U2=22C(Q/2)2=4CQ2
Hence, energy stored in the combination is less than that of the single capacitor.

 
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