Đáp án:
a) \(8000\left( {V/m} \right)\)
b) \(10125\left( {V/m} \right)\)
Giải thích các bước giải:
a) Ta có:
\(\begin{array}{l}
{E_A} = k.\dfrac{Q}{{O{A^2}}}\\
{E_B} = k.\dfrac{Q}{{O{B^2}}}\\
\Rightarrow \dfrac{{{E_A}}}{{{E_B}}} = {\left( {\dfrac{{OB}}{{OA}}} \right)^2} = \dfrac{1}{4} \Rightarrow \dfrac{{OB}}{{OA}} = \dfrac{1}{2}\\
\Rightarrow OA = 2.OB\\
OM = \dfrac{{OA + OB}}{2} = \dfrac{{3OB}}{2} = 1,5.OB\\
{E_M} = k.\dfrac{Q}{{O{M^2}}}\\
\Rightarrow \dfrac{{{E_M}}}{{{E_A}}} = {\left( {\dfrac{{OA}}{{OM}}} \right)^2} = {\left( {\dfrac{2}{{1,5}}} \right)^2} = \dfrac{{16}}{9}\\
\Rightarrow {E_M} = 8000\left( {V/m} \right)
\end{array}\)
b) Ta có:
\(\begin{array}{l}
MA = 2MB\\
\Rightarrow MO - OA = 2\left( {OB - OM} \right)\\
\Rightarrow 3MO = 2OB + OA\\
\Rightarrow OM = \dfrac{{2OB + OA}}{3} = \dfrac{4}{3}.OB\\
{E_M} = k.\dfrac{Q}{{O{M^2}}}\\
\Rightarrow \dfrac{{{E_M}}}{{{E_A}}} = {\left( {\dfrac{{OA}}{{OM}}} \right)^2} = {\left( {\dfrac{2}{{\dfrac{4}{3}}}} \right)^2} = \dfrac{9}{4}\\
\Rightarrow {E_M} = 10125\left( {V/m} \right)
\end{array}\)
