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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answer is C.

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Detailed Solution

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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