Đáp án:

 (q1,q2)=(${8.10^{ - 8}}$,$ - {2.10^{ - 8}}$);($-{8.10^{ - 8}}$,$  {2.10^{ - 8}}$));(${2.10^{ - 8}}$,$ - {8.10^{ - 8}}$);($-{2.10^{ - 8}}$,$  {8.10^{ - 8}}$)

đơn vị: C

Giải thích các bước giải:

$\begin{array}{l}
F = \frac{{k\left| {{q_1}{q_2}} \right|}}{{{r^2}}}\\
 \Rightarrow 3,{6.10^{ - 4}} = \frac{{{{9.10}^9}.\left| {{q_1}{q_2}} \right|}}{{0,{2^2}}}\\
{q_1}{q_2} =  - 1,{6.10^{ - 15}}\left( 1 \right)
\end{array}$

sau khi tiếp xúc điện tích 2 quả cầu

$\begin{array}{l}
q = \frac{{{q_1} + {q_2}}}{2}\\
F' = \frac{{k{{\left( q \right)}^2}}}{{{r^2}}}\\
 \Rightarrow 2,{025.10^{ - 4}} = {\frac{{{{9.10}^9}.\left( {\frac{{{q_1} + {q_2}}}{2}} \right)}}{{0,{2^2}}}^2}\\
 \Rightarrow {\left( {\frac{{{q_1} + {q_2}}}{2}} \right)^2} = {9.10^{ - 16}}\\
 \Rightarrow {q_1} + {q_2} =  \pm {6.10^{ - 8}}\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{q_1}{q_2} =  - 1,{6.10^{ - 15}}\\
{q_1} + {q_2} = {6.10^{ - 8}}
\end{array} \right.\\
\left\{ \begin{array}{l}
{q_1}{q_2} =  - 1,{6.10^{ - 15}}\\
{q_1} + {q_2} =  - {6.10^{ - 8}}
\end{array} \right.
\end{array} \right.
\end{array}$

q1,q2 là nghiệm của các pt:

$\left[ \begin{array}{l}
{q^2} - {6.10^{ - 8}}q - 1,{6.10^{ - 15}} = 0\\
{q^2} + {6.10^{ - 8}}q - 1,{6.10^{ - 15}} = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
q = {8.10^{ - 8}};q =  - {2.10^{ - 8}}\\
q =  - {8.10^{ - 8}};q = {2.10^{ - 8}}
\end{array} \right.$

vậy

(q1,q2)=(${8.10^{ - 8}}$,$ - {2.10^{ - 8}}$);($-{8.10^{ - 8}}$,$  {2.10^{ - 8}}$));(${2.10^{ - 8}}$,$ - {8.10^{ - 8}}$);($-{2.10^{ - 8}}$,$  {8.10^{ - 8}}$)

đơn vị: C