Đáp án:
Ta có \(F = k\dfrac{{\left| {{q_1}{q_2}} \right|}}{{{r^2}}} \Rightarrow F \sim \dfrac{1}{{{r^2}}} \Rightarrow \dfrac{{{F_2}}}{{{F_1}}} = \dfrac{{r_1^2}}{{r_2^2}} \Leftrightarrow 4 = \dfrac{{0,{{03}^2}}}{{r_2^2}} \Rightarrow {r_2} = 0,015m = 1,5cm\)