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Question

A parallel-plate capacitor is connected to a battery, which maintains a potential difference V0 between its plates. A slab of dielectric constant K is slowly inserted between the plates, completely filling the space between them. If Q0 is the charge on the plates before the dielectric is inserted, how much work is done by mechanical forces on the slab in the whole process ?

A
KQ0V0
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B
12Q0V0(K1)
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C
KQ0V02
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D
12Q0V0(K1)
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Solution

The correct option is D 12Q0V0(K1)

Suppose the capacitance before the dielectric is inserted is C0

The initial Charge, Q0=C0V0

Now when the dielectric is inserted, charge
Q=KC0V0

As K>1Q>Q0

As we know that work done by battery will be

W=QV

Wbattery=(QQ0)V0=(K1)C0V20

Wbattery=(K1)Q0V0

Using conservation of energy, we have

Wmech+Wbatt=ΔU

No kinetic energy present because slab inserted slowly.

Wmech+(K1)Q0V0=12KC0V2012C0V20

Wmech=12C0V20(K1)(K1)Q0V0

Wmech=12Q0V0(K1)Q0V0(K1)

Wmech=12Q0V0(K1)

Observe that Electric force does positive work and in turn the applied mechanical forces do negative work on the dielectric slab.

Hence, option (d) is correct.
Key concept- Work done by external forces on the dielectric when it is inserted in a parallel plate capacitor with battery connected.

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