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- Hướng dẫn giải

Trước khi tiếp xúc

$$\begin{gathered} F=k\frac{\left|q_1{q}_2\right|}{\epsilon r^2}\stackrel{}{\to }\left|q_1{q}_2\right|=\frac{\epsilon r^{2}F}{k} \\ \to \left|q_1{q}_2\right|=\frac{2,{7.10}^{-4}.0,{02}^2}{{9.10}^9}=1,{2.10}^{-17} \end{gathered}$$

Do 2 điện tích đẩy nhau nên $q_1{q}_2>0$ thì $q_1{q}_2=1,{2.10}^{-17}$

Sau khi tiếp xúc: $q_1^{/}=q_2^{/}=\frac{q_1+q_2}{2}$

$$\begin{gathered} F^{/}=k\frac{\left|q_1^{/}{q}_2^{/}\right|}{\epsilon r^2}\stackrel{}{\to }\left|q_1^{/}{q}_2^{/}\right|=\frac{\epsilon r^2{F}^{/}}{k} \\ \to {\left(\frac{q_1+q_2}{2}\right)}^2=\frac{3,{6.10}^{-4}.0,{02}^2}{{9.10}^9} \end{gathered}$$

$\stackrel{}{\to }{q}_1+q_2=\pm {8.10}^{-9}$

Khi $q_1+q_2={8.10}^{-9}$  ta có hệ

$$\begin{gathered} 〈\begin{array}{l}{q}_1{q}_2=1,{2.10}^{-17}\\ q_1+q_2={8.10}^{-9}\end{array}\to q^2-{8.10}^{-9}q+1,{2.10}^{-17}=0 \\ \to 〈\begin{array}{l}{q}_1={6.10}^{-9}\\ q_2={2.10}^{-9}\end{array}hoac\to 〈\begin{array}{l}{q}_1={2.10}^{-9}\\ q_2={6.10}^{-9}\end{array} \end{gathered}$$

Khi $q_1+q_2=-{8.10}^{-9}$ ta có hệ

$$\begin{gathered} 〈\begin{array}{l}{q}_1{q}_2=1,{2.10}^{-17}\\ q_1+q_2=-{8.10}^{-9}\end{array}\to q^2+{8.10}^{-9}q+1,{2.10}^{-17}=0 \\ \to 〈\begin{array}{l}{q}_1=-{6.10}^{-9}\\ q_2=-{2.10}^{-9}\end{array}hoac\to 〈\begin{array}{l}{q}_1=-{2.10}^{-9}\\ q_2=-{6.10}^{-9}\end{array} \end{gathered}$$