Người ta đặt 3 điện tích q1 = 8.10^-9 C, q2 = q3 = -8.10^-9 C tại 3 đỉnh của một tam giác đều cạnh 6 cm trong không khí. Xác định lực tác dụng lên điện tích q0 = 6.10^-9 C đặt ở tâm O của tam giác.

2 câu trả lời

Đáp án:

${F_0} = 3,{28.10^{ - 4}}N$

Giải thích các bước giải:

 

Ta có:

+ Lực do \({q_1}\) tác dụng lên \({q_0}\)  

+ Lực do \({q_2}\) tác dụng lên ${q_0}$ ${F_{20}} = k\dfrac{{\left| {{q_2}{q_0}} \right|}}{{B{O^2}}} = 1,{64.10^{ - 4}}N$

+ Lực do \({q_3}\) tác dụng lên \({q_0}\) ${F_{30}} = k\dfrac{{\left| {{q_3}{q_0}} \right|}}{{C{O^2}}} = 1,{64.10^{ - 4}}N$

$\overrightarrow {{F_0}}  = \overrightarrow {{F_{10}}}  + \overrightarrow {{F_{20}}}  + \overrightarrow {{F_{30}}} $

Ta có: $\overrightarrow {{F_{23}}}  = \overrightarrow {{F_{20}}}  + \overrightarrow {{F_{30}}} $

Mặt khác: \(\left\{ \begin{array}{l}{F_{20}} = {F_{30}}\\\left( {\overrightarrow {{F_{20}}} ,\overrightarrow {{F_{30}}} } \right) = {120^0}\end{array} \right. \Rightarrow {F_{23}} = {F_{20}} = {F_{30}} = 1,{6.10^{ - 4}}N\)

Lại có: \(\overrightarrow {{F_{23}}}  \uparrow  \uparrow \overrightarrow {{F_{10}}} \)

\( \Rightarrow {F_0} = {F_{10}} + {F_{23}} = 3,{28.10^{ - 4}}N\)

Đáp án:

Lời giải:

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