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Two point charges placed at a distance of 0.20m in air repel each other with a certain force. When a dielectric slab of thickness 0.08m is introduced in between the charges, the force of interaction is half of its previous value. Find the dielectric constant. |
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Answer» Solution :r= 0.2m, x= 0.08m `F= (1)/(4PI in_(0)) (q_(1)q_(2))/((0.2)^(2))` …….in air `F^(1)= (1)/(4pi in_(0))(q_(1)q_(2))/([(0.2-0.08) + 0.08 sqrtK]^(2)) " Here " F^(1)= (1)/(2)F` SUBSTITUTING we get `K ~= 4` |
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