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  • Confused! kindly explain, - Electrostatics - JEE Main-14

A 6 \mu F capacitor is charged from 10 volts  to 20 volts  . Increase in energy will be

  • Option 1)

    18\times 10^{-4}J

  • Option 2)

    9\times 10^{-4}J

  • Option 3)

    4.5\times 10^{-4}J

  • Option 4)

    9\times 10^{-6}J

 

Answers (1)

best_answer

As we have learned

Loss of Energy -

\Delta U=U_{i}-U_{f}=\frac{C_{1}C_{2}}{2\left ( C_{1} +C_{2}\right )}\left ( V_{1}-V_{2} \right )^{2}

-

 

 \Delta E= E_{final}- E_{initial}= 1/2C(V_{final}^{2}-V_{initial}^{2})

= 1/2*6*(20^2-10^2)*10^{-6}

= 3*(400-100)*10^{-6}= 3*300*10^{-6}= 9*10^{-4}

 


Option 1)

18\times 10^{-4}J

Option 2)

9\times 10^{-4}J

Option 3)

4.5\times 10^{-4}J

Option 4)

9\times 10^{-6}J

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Plabita

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