Lời giải:
a) Ta thấy AB = AC + BC
\[\overrightarrow {{F_3}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} \]
\[{F_3} = {F_1} + {F_2} = k.\frac{{\left| {{q_1}.{q_3}} \right|}}{{A{C^2}}} + k.\frac{{\left| {{q_2}.{q_3}} \right|}}{{B{C^2}}}\]
\[{F_3} = {9.10^9}.\left( {\frac{{\left| {{{8.10}^{ - 8}}{{.8.10}^{ - 8}}} \right|}}{{{{0,04}^2}}} + \frac{{\left| {{{8.10}^{ - 8}}.\left( { - {{8.10}^{ - 8}}} \right)} \right|}}{{{{0,02}^2}}}} \right) = 0,18\,\,(N)\]
b, \[CB = AB + AC = 6 + 4\]
\[{F_3} = \left| {{F_1} - {F_2}} \right|\]
\[{F_3} = \left| {k.\frac{{{q_1}.{q_3}}}{{A{C^2}}} - k.\frac{{{q_2}.{q_3}}}{{B{C^2}}}} \right|\]
\[{F_3} = \left| {{{9.10}^9}.{{({{8.10}^{ - 8}})}^2}\left( {\frac{1}{{{{0,04}^2}}} - \frac{1}{{{{0,1}^2}}}} \right)} \right| = 0,03024(N)\]
c. \[CA = CB = 5\,\,(cm)\]
\[\cos \alpha = \frac{{{5^2} + {5^2} - {6^2}}}{{2.5.5}} = \frac{7}{{25}}\]
\[F_3^2 = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \beta \]
\[F_3^2 = 2F_1^2(1 - \cos \alpha )\]
\[{F_3} = {F_1}\sqrt {2(1 - \cos \alpha )} = k.\frac{{{q_1}.{q_3}}}{{{r^2}}}\sqrt {2(1 - \cos \alpha )} \]
\[{F_3} = {9.10^9}.\frac{{{{\left( {{{8.10}^{ - 8}}} \right)}^2}}}{{{{0,05}^2}}}\sqrt {2\left( {1 - \frac{7}{{25}}} \right)} = 0,028(N)\]
