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Q.

The plates of a parallel plate capacitor are charged upto 200 Volts. A dielectric slab of thickness 4 mm is inserted between its plates. Then to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of the dielectric slab is

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a

1

b

6

c

4

d

5

answer is C.

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Detailed Solution

∆d=t1−1K;⇒ 3.2×10−3=4×10−3×1−1K⇒1K=0.2; ⇒ K=5

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