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Q.

In a capacitor of capacitance 20 μ F, the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is

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a

2 μ F

b

15.5 μ F

c

26.6 μ F

d

32 μ F

answer is C.

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Detailed Solution

C=ε0Ad and C′=ε0Ad−t+tK⇒CC′=d−t+tKd⇒20C′=2×10−3−1×10−3+1×10−322×10−3⇒C′=26.6μF
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