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The force of repulsion between two charges +1 and 6C is 12N. A third charge -4C add between them, now the force between them will be …

Answer»

4 N repulsive
4 N attractive
8 N repulsive
8 N attractive

Solution :Here, `F = (kq_(1)q_(2))/R^(2)`
`F_(1) = (K(2)(6))/gamma^(2)` and `F_(2) = (k(-2)(2))/r^(2)`
`therefore F_(1) = (12k)/gamma^(2)`…….(i), `therefore F_(2) = (-4K)/r^(2)`……….(ii)
`therefore F_(2)/F_(1) = -4/12 = -1/3 F_(2)` is a FORCE between t after addint `-4C` charge.
`therefore q_(1)^(.) = -2C` and `q_(2)^(.) = 2C`
`therefore F_(2) = -1/3 xx F_(1) =-1/3 xx 12 = -4 N`
Negative sign indicates attractive force.


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