Đáp án:
a.$385,2N$
b.$832,5N$
c.$1217,7N$
Giải thích các bước giải:
a$\begin{array}{l}
a.{E_2} = \frac{{k{q_2}}}{{d_1^2}} = \frac{{{{9.10}^9}.1,{{5.10}^{ - 6}}}}{{0,{{03}^2}}}\\
{E_3} = \frac{{k{q_3}}}{{{{\left( {{d_1} + {d_2}} \right)}^2}}} = \frac{{{{9.10}^9}{{.22.10}^{ - 6}}}}{{0,{{05}^2}}}\\
\overrightarrow {{E_1}} = \overrightarrow {{E_2}} + \overrightarrow {{E_3}} \\
\Rightarrow {E_1} = {E_3} - {E_2}\\
{F_1} = {E_1}{q_1} = 385,2N\\
b.{E_1} = \frac{{k{q_1}}}{{d_1^2}} = \frac{{{{9.10}^9}{{.6.10}^{ - 6}}}}{{0,{{03}^2}}}\\
{E_3} = \frac{{k{q_3}}}{{d_2^2}} = \frac{{{{9.10}^9}{{.22.10}^{ - 6}}}}{{0,{{02}^2}}}\\
\overrightarrow {{E_2}} = \overrightarrow {{E_1}} + \overrightarrow {{E_3}} \\
\Rightarrow {E_2} = {E_1} + {E_3}\\
{F_2} = {E_2}{q_2} = 832,5N\\
c.{E_2} = \frac{{k{q_2}}}{{d_2^2}} = \frac{{{{9.10}^9}.1,{{5.10}^{ - 6}}}}{{0,{{02}^2}}}\\
{E_1} = \frac{{k{q_1}}}{{{{\left( {{d_1} + {d_2}} \right)}^2}}} = \frac{{{{9.10}^9}{{.6.10}^{ - 6}}}}{{0,{{05}^2}}}\\
\overrightarrow {{E_1}} = \overrightarrow {{E_2}} + \overrightarrow {{E_3}} \\
\Rightarrow {E_3} = {E_1} + {E_2}\\
{F_3} = {E_3}{q_3} = 1217,7N
\end{array}$
