Đáp án:

 \(\dfrac{{31250}}{3}\left( {V/m} \right)\)

Giải thích các bước giải:

a) Lực tương tác giữa 2 điện tích là: 

\(F = \dfrac{{k\left| {{q_1}{q_2}} \right|}}{{{r^2}}} = \dfrac{{{{9.10}^9}.{{\left( {{{2.10}^{ - 8}}} \right)}^2}}}{{0,{1^2}}} = 3,{6.10^{ - 4}}N\)

b) Xét tại M, ta có:

\(\begin{array}{l}
{E_1} = \dfrac{{k\left| {{q_1}} \right|}}{{r_1^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{05}^2}}} = 72000\left( {V/m} \right)\\
{E_2} = \dfrac{{k\left| {{q_2}} \right|}}{{r_2^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{05}^2}}} = 72000\left( {V/m} \right)\\
{E_M} = {E_1} + {E_2} = 144000\left( {V/m} \right)
\end{array}\)

Xét tại N, ta có: 

\(\begin{array}{l}
{E_1} = \dfrac{{k\left| {{q_1}} \right|}}{{r_1^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{3^2}}} = 2000\left( {V/m} \right)\\
{E_2} = \dfrac{{k\left| {{q_2}} \right|}}{{r_2^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{2^2}}} = 4500\left( {V/m} \right)\\
{E_M} =  - {E_1} + {E_2} = 2500\left( {V/m} \right)
\end{array}\)

Xét tại P, ta có: 

\(\begin{array}{l}
{E_1} = \dfrac{{k\left| {{q_1}} \right|}}{{r_1^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{08}^2}}} = 28125\left( {V/m} \right)\\
{E_2} = \dfrac{{k\left| {{q_2}} \right|}}{{r_2^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{06}^2}}} = 50000\left( {V/m} \right)\\
{E_M} = \sqrt {E_1^2 + E_2^2}  = 57367,4\left( {V/m} \right)
\end{array}\)

Xét tại Q, ta có: 

\(\begin{array}{l}
{E_1} = \dfrac{{k\left| {{q_1}} \right|}}{{r_1^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{12}^2}}} = 12500\left( {V/m} \right)\\
{E_2} = \dfrac{{k\left| {{q_2}} \right|}}{{r_2^2}} = \dfrac{{{{9.10}^9}{{.2.10}^{ - 8}}}}{{0,{{12}^2}}} = 12500\left( {V/m} \right)\\
\cos \dfrac{\alpha }{2} = \sin \dfrac{Q}{2} = \dfrac{5}{{12}}\\
{E_Q} = 2{E_1}\cos \dfrac{\alpha }{2} = 2.12500.\dfrac{5}{{12}} = \dfrac{{31250}}{3}\left( {V/m} \right)
\end{array}\)