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  • Need clarity, kindly explain! A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system:

  • Option 1)

    increases by a factor of 4

  • Option 2)

    decreases by a factor of 2

  • Option 3)

    remains the same

  • Option 4)

    increases by a factor of 2

 

Answers (1)

As we discussed in

Energy Stored -

U=frac{1}{2}CV^{2}=frac{1}{2}QV=frac{Q^{2}}{2C}

-

 

 when the capacitor is charged by a battery of potential V, then energy stored in the capaciter

V_{i}=\frac{1}{2}Cv^{2}

When the battery is removed and another identical unchanged capacitor is connected in parallel

common potential  

V^{1}=\frac{Cv}{C+c}=\frac{V}{2}

V_{f}=\frac{1}{2}(2c)(\frac{v}{2})^{2}

\frac{1}{4}CV^{2}

V_{f}=\frac{V_{1}}{2}

 


Option 1)

increases by a factor of 4

Option is incorrect

Option 2)

decreases by a factor of 2

Option is correct

Option 3)

remains the same

Option is incorrect

Option 4)

increases by a factor of 2

Option is incorrect

Posted by

Vakul

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