Câu hỏi 21

Khi vật treo chưa tích điện: \({T_0} = 2\pi \sqrt {\dfrac{l}{g}}  = 2s\)

Khi vật treo tích điện q₁: \({T_1} = 2,4s > {T_0} \Rightarrow {T_1} = 2\pi \sqrt {\dfrac{l}{{g - \dfrac{{\left| {{q_1}} \right|E}}{m}}}}  = 2,4s \Rightarrow {q_1} < 0\)

Khi vật treo tích điện q₂: \({T_2} = 1,6s < {T_0} \Rightarrow {T_2} = 2\pi \sqrt {\dfrac{l}{{g + \dfrac{{\left| {{q_2}} \right|E}}{m}}}}  = 1,6s \Rightarrow {q_2} > 0\)

Có: \(\left\{ \begin{array}{l}{\left( {\dfrac{{{T_0}}}{{{T_1}}}} \right)^2} = \dfrac{{g - \dfrac{{\left| {{q_1}} \right|E}}{m}}}{g} = \dfrac{{25}}{{36}} \Rightarrow \dfrac{{\left| {{q_1}} \right|E}}{m} = \dfrac{{11}}{{36}}.g\,\,\,\,\left( 1 \right)\\{\left( {\dfrac{{{T_0}}}{{{T_2}}}} \right)^2} = \dfrac{{g + \dfrac{{\left| {{q_2}} \right|E}}{m}}}{g} = \dfrac{{25}}{{16}} \Rightarrow \dfrac{{\left| {{q_2}} \right|E}}{m} = \dfrac{9}{{16}}.g\,\,\,\,\left( 2 \right)\end{array} \right.\)

Lấy \(\dfrac{{\left( 1 \right)}}{{\left( 2 \right)}} \Leftrightarrow \left| {\dfrac{{{q_1}}}{{{q_2}}}} \right| = \dfrac{{44}}{{81}} \Rightarrow \dfrac{{{q_1}}}{{{q_2}}} =  - \dfrac{{44}}{{81}}\)

Chọn A.