Two conductors of capacitance 1muF and 2muF are charged to +10V and -20 V 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.
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The the common potential is 20 V and the redistribution of charges is 20 μC.
Explanation:
Given data:
Q = 10 μC
Q' = - 40 μC
Net charge "q" = ( 40 -10 ) = 30 μC
Potential is common to V.
Let charge distribution
q' = (q - q')2
q' = 2q -2q'
3q' = 2q
q' = 60 / 3 = 20 μC
q - q' = 10 μC
V = q' / c
V = 20μC / 1 μF = 20 V
Hence the the common potential is 20 V and the redistribution of charges is 20 μC.
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