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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)

• Option 3)

• Option 4)

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N

As we learnt in

Energy Stored -

$\dpi{100} U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}$

-

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)

Incorrect

Option 3)

Incorrect

Option 4)

Incorrect

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