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Question

The capacitance of a parallel plate capacitor is C0 when the plates have air between them. This region is now filled with a dielectric slab of dielectric constant K and capacitor is connected with a battery of emf E and zero internal resistance. Now slab is taken out, then during removal of slab:

1.) Charge EC0(K−1) flows through the cell.

2.) Energy E2C0(K−1) is absorbed by cell.

3.) The energy stored in the capacitor is reduced by E2C0(K−1).

4.) The external agent has to do E2C0(K−1) amount of work to take out the slab.

Select the correct phenomenon happening during the process:

A
1 & 3
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B
2 & 3
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C
2,3 & 4
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D
1 & 2
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Solution

The correct option is D 1 & 2
Initial charge on capacitor when slab inside the plates,

Qi=KC0E

Charge after removing dielectric slab,

Qf=C0E

Amount of charge that flows through the cell is,

ΔQ=KC0EC0E=C0E(K1)

After removing dielectric, the charge on capacitor is decreased. Thus more charge will be supplied by the battery

Now, energy absorbed by cell (workdone on battery),

ΔE=ΔQ×E

ΔE=C0E2(K1)

Initial energy stored in capacitor,

Ui=12KC0E2

Final energy stored in capacitor,

Uf=12C0E2

loss of energy will be

ΔUlost=12KC0E212C0E2

ΔUlost=12C0E2(K1)

Workdone by external agent is equal to the energy dissipated in circuit.

Wext=12C0E2(K1)

Correct phenomenon (1) & (2) only.

Hence, option (d) is correct.
Why this question?
Caution: If a positive charge flows through a cell such that it leaves the +ve terminal of battery and enters through its -ve terminal, then work is done by battery i.e. Wbattery>0 and for the reverse case Wbattery<0.

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