Đáp án:
\(\begin{array}{l}a){F_M} = \dfrac{{32}}{3}{.10^{ - 3}}N\\b){F_N} = 0,01125N\end{array}\)
Hướng dẫn giải chi tiết:
a) \(AM = MB = \dfrac{{AB}}{2} = 4,5cm\)
Lực điện tác dụng lên \({q_3}\) (như hình vẽ)
\(\overrightarrow {{F_3}} = \overrightarrow {{F_{13}}} + \overrightarrow {{F_{23}}} \)
Ta có: \(\left\{ \begin{array}{l}{F_{13}} = k\dfrac{{\left| {{q_1}{q_3}} \right|}}{{A{M^2}}} = {9.10^9}.\dfrac{{\left| {{{4.10}^{ - 8}}{{.3.10}^{ - 8}}} \right|}}{{0,{{045}^2}}} = \dfrac{{16}}{3}{.10^{ - 3}}N\\{F_{23}} = k\dfrac{{\left| {{q_2}{q_3}} \right|}}{{B{M^2}}} = {9.10^9}.\dfrac{{\left| { - {{4.10}^{ - 8}}{{.3.10}^{ - 8}}} \right|}}{{0,{{045}^2}}} = \dfrac{{16}}{3}{.10^{ - 3}}N\end{array} \right.\)
Lại có: \(\overrightarrow {{F_{23}}} \uparrow \uparrow \overrightarrow {{F_{13}}} \Rightarrow {F_3} = {F_{23}} + {F_{13}} = \dfrac{{32}}{3}{.10^{ - 3}}N\)
b) \(\overrightarrow {{F_N}} = \overrightarrow {{F_{13}}} + \overrightarrow {{F_{23}}} \)
Ta có: \(\left\{ \begin{array}{l}{F_{13}} = k\dfrac{{\left| {{q_1}{q_3}} \right|}}{{A{N^2}}} = {9.10^9}.\dfrac{{\left| {{{4.10}^{ - 8}}{{.3.10}^{ - 8}}} \right|}}{{0,{{03}^2}}} = 0,012N\\{F_{23}} = k\dfrac{{\left| {{q_2}{q_3}} \right|}}{{B{N^2}}} = {9.10^9}.\dfrac{{\left| { - {{4.10}^{ - 8}}{{.3.10}^{ - 8}}} \right|}}{{0,{{012}^2}}} = 7,{5.10^{ - 4}}N\end{array} \right.\)
Lại có: \(\overrightarrow {{F_{13}}} \uparrow \downarrow \overrightarrow {{F_{23}}} \)
\( \Rightarrow {F_N} = \left| {{F_{13}} - {F_{23}}} \right| = 0,01125N\)
