A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect ?

  • Option 1)

    The change in energy stored is 

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

  • Option 2)

    The charge on the capacitor is not conserved.

  • Option 3)

    The potential difference between the plates decreases K times.

  • Option 4)

    The energy stored in the capacitor decreases K times.

 

Answers (1)
V Vakul

As we learnt in 

If K filled between the plates -

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

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

 

 AND

 

Energy Stored -

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

-

 

 c=\frac{q}{v}

After inserting dielectric

c'=kc

New potential difference

v'=\frac{v}{k}

U_{i}=\frac{1}{2}cv^{2}=\frac{q^{2}}{2c}

U_{f}=\frac{q^{2}}{2c'}=\frac{q^{2}}{2Kc}=\frac{c^{2}v^{2}}{2Kc}=\frac{u_{i}}{K}

\Delta U=U_{f}-U_{i}=\frac{1}{2}cv^{2} \left (\frac{1}{K}-1 \right )

New potential difference v'=\frac{q}{cK}=\frac{v}{K}


Option 1)

The change in energy stored is 

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

Incorrect

Option 2)

The charge on the capacitor is not conserved.

Correct

Option 3)

The potential difference between the plates decreases K times.

Incorrect

Option 4)

The energy stored in the capacitor decreases K times.

Incorrect

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