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  • Help me answer: A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

  • Option 1)

     Reduction of charge on the plates and increase of potential difference across the plates

  • Option 2)

    Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

  • Option 3)

    Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

  • Option 4)

    None of the above

 

Answers (1)

best_answer

As we have learned

Energy Stored -

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

-

 

 Battery in disconnected so Q will be constant asC\propto K . So with introduction of dielectric slab capacitance will increase using Q = CV, V will decrease and using U= \frac{Q^2}{2C} , energy will decrease.

 


Option 1)

 Reduction of charge on the plates and increase of potential difference across the plates

Option 2)

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

Option 3)

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

Option 4)

None of the above

Posted by

Plabita

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