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4.5 micro joules

M.Sc. Biomedical Genetics, B.Sc. Biotechnology

The charge on the capacitor is Q = CV = 900 1012 F 100 V = 9 108 C The energy stored in the capacitor = (1/2) CV2 = (1/2) QV = (1/2) 9 108 C 100 V = 4.5 10% J

Working as student Bachelor of Technology with mechanical engineer .And i will completed my 10+2 from Kendriya Vidyalaya.

as we know electrostatic energy = 1/2 CV2 where c is capacitance v is potential so keeping the values in the formula we get (1/2) X 900 X 10-12 X (100)2 = 9 X 10-6 J this answers first part f the problem now after disconnecting it is joined to another 900pF capacitor in series therefore now new capacitance... read more
as we know electrostatic energy = 1/2 CV2 where c is capacitance v is potential so keeping the values in the formula we get (1/2) X 900 X 10-12 X (100)2 = 9 X 10-6 J this answers first part f the problem now after disconnecting it is joined to another 900pF capacitor in series therefore now new capacitance is 1/C' = 1/900 + 1/900 hence C'= 450 pF so now new electrostatic energy = 1/2 C'V2 = 4.5 X 10-6 J hence lost energy is (9 - 4.5) X 10-6 J . read less

School & Engineering tuition, Hindi tution

Its 4.5 micro joules as the stored energy is given by E=(1/2)*C*V^2

u=1/2 cV2 U=9x10pow -6 joule

The Wizard Of Math-Science

Energy stored is C.V.V/2 Where C is 900pf which is 900*(10)^(-12) N V is 100 v Thus answer is 4.5 micro joules.

B.SC MATHEMATICS(PURSUING)

.5*c*v^2

ans is 2.25*10^(-6)

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