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A paraller plate capacitor having capacity 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a poecelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

 

  • Option 1)

    560 pJ

  • Option 2)

    600\: \: pJ

  • Option 3)

    508\: \: pJ

  • Option 4)

    692\: \: pJ

Answers (1)

best_answer

 

Energy Stored -

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

 

 

 

If K filled between the plates -

{C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

U_{i} = \frac{1}{2}\frac{V^{2}}{c} = 600J   (initial energy)

 

Since battery is disconnected charge remains same.

V_{f} = \frac{1}{2}\frac{V^{2}}{C}   (final energy)

        = \frac{1}{2}\times \frac{120^{2}}{12\times 6.5} = 92

W = U_{i} - U_{f} = 508J

 


Option 1)

560 pJ

Option 2)

600\: \: pJ

Option 3)

508\: \: pJ

Option 4)

692\: \: pJ

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