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Question

The plates of a parallel plate capacitor are charged to 200 V and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant 5 and thickness 4 mm is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:


A
1.6 mm
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B
3.2 mm
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C
0.8 mm
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D
4.8 mm
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Solution

The correct option is B 3.2 mm
Let the distance of separation be d, increase be x and dielectric length the be t.

When the dielectric is inserted:
1Cnet=(dt+x)(Aϵ0)+t(κAϵ0)

And before the dielectric is inserted:
1Cinitial=d(Aϵ0)

Now, as per given condition:
1Cnet=1Cinitialdt+x+tκ=dx=t(11κ)

Therefore, increase in distance to restore original capacitance =t(11κ)=4×0.8 mm=3.2 mm

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