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  • How to solve this problem- A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinse

A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential  V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

  • Option 1)

    ZERO

  • Option 2)

    \frac{1}{2}\left ( K-1 \right )CV^{2}

  • Option 3)

    \frac{CV^{2}\left ( K-1 \right )}{K}

  • Option 4)

    \left ( K-1 \right )CV^{2}

 

Answers (2)

best_answer

As we learnt in

Energy Stored -

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

-

 

 The potential energy of a charged capacitor

U_{i}= \frac{q^{2}}{2C}

where U_{i} is the initial potential energy. If a dielectric slab is slowly introduced, the energy

= \frac{q^{2}}{2KC}

Once is taken out, again the energy increases to the old value. Therefore after it is taken out, the potential energy come back to the old value.

Total work done =ZERO


Option 1)

ZERO

Correct

Option 2)

\frac{1}{2}\left ( K-1 \right )CV^{2}

Incorrect

Option 3)

\frac{CV^{2}\left ( K-1 \right )}{K}

Incorrect

Option 4)

\left ( K-1 \right )CV^{2}

Incorrect

Posted by

divya.saini

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