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Two similar point charges q1 and q2 are placed at a distance r apart in air.  Assume that a slab of thickness one third the separation between the charges is placed between the charges and it is observed that the ratio of Coulomb’s repulsive force before and after placement of dielectric is 25:9. Then the dielectric constant K of such a slab is

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By Expert Faculty of Sri Chaitanya
a
1
b
4
c
9
d
25

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detailed solution

Correct option is C

F1=q1q2d214πεoIn the second case effective distance2d3+kd3F2=14πEoq1q22d3+kd32F1F2=2.59=q1q2/d2q1q2/2d3+kd322.59=2d3+kd32d2=53=23+k133=kK=9


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In the circuit shown, the base current is 30 µA. The value of R1 is (Neglect VBE)

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