A <math> <mn>60</mn> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">F</mi> </math> capacitor is fully charged by a <math> <mn>20</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math> supply. It is then disconnected from the supply and is connected to another uncharged <math> <mn>60</mn> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">F</mi> </math> capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in <math> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">J</mi> </math> )

6Sol. Common potential after connection. V common = C 1 V 1 + C 2 V 2 C 1 + C 2 = 60 × 20 + 0 120 = 10 V o l t  Loss of energy = 1 2 C V 2 - 1 2 ( 2 C ) × V Common 2  = 1 2 × 60 × 10 - 12 × ( 20 ) 2 - 60 × 10 - 12 × ( 10 ) 2 = 60 × 10 - 12 ( 200 - 100 ) = 6000 × 10 - 12 = 6 n J

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Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

A $60 p F$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $n J$ )

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alok kumar singh | Contributor-Level 10

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Sol. Common potential after connection.

$V_{common} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{60 \times 20 + 0}{120} = 10 V o l t$

Loss of energy $= \frac{1}{2} {C V}^{2} - \frac{1}{2} (2 C) \times V_{Common}^{2}$

$\begin{array}{r} = \frac{1}{2} \times 60 \times 10^{- 12} \times (20)^{2} - 60 \times 10^{- 12} \times (10)^{2} \\ = 60 \times 10^{- 12} (200 - 100) \\ = 6000 \times 10^{- 12} \\ = 6 n J \end{array}$

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A $60 p F$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $n J$ )

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A 60 p F capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 p F capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in n J )

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A $60 p F$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 p F$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $n J$ )