Answer to Question #312975 in Electric Circuits for BxJ

Question

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?

Answer 1

We know that

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

Force inverse square proportional to distance

$\frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}$

$\frac{F}{6F}=\frac{r_2^2}{0.06^2}$

$\frac{36\times10^{-4}}{6}=r_2^2$

$r_2=\sqrt{6\times10^{-4}}=0.0244m\\r_2=2.44cm$