A 900 pF battery is charged by 100 V battery. How much electrostatic energy is stored by the capacitor ?

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Rajesh N v

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4.5 micro joules
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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...
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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 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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School & Engineering tuition, Hindi tution

Its 4.5 micro joules as the stored energy is given by E=(1/2)*C*V^2
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M.Sc. Biomedical Genetics, B.Sc. Biotechnology

The charge on the capacitor is Q = CV = 900 × 1012 F × 100 V = 9 × 108 C The energy stored in the capacitor = (1/2) CV2 = (1/2) QV = (1/2) × 9 × 108 C × 100 V = 4.5 × 10% J
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Physics Tutor

u=1/2 cV2 U=9x10pow -6 joule
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Excellent Coaching for CBSE, ICSE, IGCSE and IAL Physics, Chemistry +

U= 1/2 CV^2 =4.5x10^-6 joules
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Problem Cracker

Electrostatic energy E=0.5*C*V^2 so your answer is E=0.5*900*10000=4.5*10^6 pJ
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B.SC MATHEMATICS(PURSUING)

.5*c*v^2
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ans is 2.25*10^(-6)
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pico farad
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