Answer to Question #280519 in Electricity and Magnetism for Anonymous005

Question

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Ans: 90 N to the right

Answer 1

2.

$F=F_2-F_1=\frac{kqq_2}{r_2^2}-\frac{kqq_1}{r_1^2}=90~N.$

3.

$F_2=F_1,$

$\frac{kqq_2}{(r_1+r_2-x)^2}=\frac{kqq_1}{x^2},$

$x=3.2~cm.$