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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 mmLet the distance of separation be d, increase be x and dielectric length the be t.When the dielectric is inserted:1Cnet=(d−t+x)(Aϵ0)+t(κAϵ0)And before the dielectric is inserted:1Cinitial=d(Aϵ0)Now, as per given condition:1Cnet=1Cinitial⇒d−t+x+tκ=d⇒x=t(1−1κ)Therefore, increase in distance to restore original capacitance =t(1−1κ)=4×0.8 mm=3.2 mm

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