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Two particles experience an attractive electric force of 5 N. If one of the charges triples and the distance between them doubles, wha
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Answers ( )
3.75 N
Explanation:
F = k(q)(q)/r^2 = 5 N
F’ = k(q)(3q)/(2r)^2
= k×3(q)(q)/4r^2
= (3/4)[k(q)(q)/r^2]
= (3/4)F
= (3/4)(5 N)
= 3.75 N