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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

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

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