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
