Đáp án:

\(\begin{array}{l}
7.A\\
\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{k\dfrac{{|q.q'|}}{{{\varepsilon _1}r_1^2}}}}{{k\dfrac{{|q.q'|}}{{{\varepsilon _2}r_2^2}}}} = \dfrac{{{\varepsilon _2}r_2^2}}{{{\varepsilon _1}r_1^2}} = \dfrac{{2.{{(\frac{{{r_1}}}{2})}^2}}}{{1r_1^2}} = \dfrac{2}{9} = \dfrac{1}{{4,5}}\\
 \Rightarrow {F_2} = 4,5{F_1}\\
8.A.F\\
{F_1} = {F_2} = k\dfrac{{|{q_1}{q_2}|}}{{{r^2}}}\\
9.C.16F\\
\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{k\dfrac{{|q.q'|}}{{r_1^2}}}}{{k\dfrac{{|q.q'|}}{{r_2^2}}}} = \dfrac{{r_2^2}}{{r_1^2}} = \dfrac{{{1^2}}}{{{4^2}}} = \dfrac{1}{{16}}\\
 \Rightarrow {F_2} = 16{F_1}\\
10.B.8,1N\\
{q_1}' = {q_2}' = \dfrac{{{q_1} + {q_2}}}{2} = \dfrac{{{{8.10}^{ - 6}} + ( - {{2.10}^{ - 6}})}}{2} = {3.10^{ - 6}}C\\
F = k\dfrac{{|{q_1}'.{q_2}'|}}{{{r^2}}} = {9.10^9}.\dfrac{{|{{3.10}^{ - 6}}{{.3.10}^{ - 6}}|}}{{0,{1^2}}} = 8,1N\\
11.B
\end{array}\)