Question #312975

The attractive force between two equal but opposite charges is F newtons when the charges are 6 cm apart. What should be their distance apart to increase the attractive force to 6F newtons?

1
Expert's answer
2022-03-21T12:47:45-0400

We know that

F=kq1q2r2F=\frac{kq_1q_2}{r^2}

Force inverse square proportional to distance

F1F2=r22r12\frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}

F6F=r220.062\frac{F}{6F}=\frac{r_2^2}{0.06^2}

36×1046=r22\frac{36\times10^{-4}}{6}=r_2^2

r2=6×104=0.0244mr2=2.44cmr_2=\sqrt{6\times10^{-4}}=0.0244m\\r_2=2.44cm


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