InterviewSolution
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InterviewSolution
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Two particles having charges Q_(1) and Q_(2) , when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles. |
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Answer» Solution :`F = (1)/(4 PI epsilon_(0)) (Q_(1)Q_(2))/(R^(2))` If the distance is EDUCED byhalf and two particles of charges are doubled. F. = `(1)/(4pi epsilon_(0)) (2Q_(1)2Q_(2))/((r//2)^(2)) = (1)/(4pi epsilon_(0)) (4Q_(1)Q_(2))/((r^(2)//4)^(2))` `F. = (1)/(4pi epsilon_(0)) (16(Q_(1)Q_(2)))/(r^(2)) = 16 [ (1)/(4pi epsilon_(0)) (Q_(1)Q_(2))/(r^(2))]` F. = 16 F |
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