O módulo (ou intensidade) da força elétrica entre duas cargas é dada por:
[tex]\boxed{\sf F~=~k\cdot \dfrac{q_1\cdot q_2}{d^2}}\\\\\\\sf Onde:~~\left\{\begin{array}{ccl}\sf k&:&\sf Constante~Eletrostatica~(=~9\cdot 10^9~N\cdot m^2/C^2~no~vacuo)\\\sf q_1&:&\sf Carga~eletrica~1\\\sf q_2&:&\sf Carga~eletrica~2\\\sf d&:&\sf Distancia\end{array}\right.[/tex]
Vamos substituir os valores fornecidos no texto e achar o que é solicitado em cada um dos três questionamentos feitos.
1)
Lembrando que, como estamos interessados na intensidade da força, não consideramos o sinal das cargas elétricas.
[tex]\sf F~=~9\cdot 10^9\cdot \dfrac{5\cdot 10^{-6}\cdot 2\cdot 10^{-6}}{0,6^2}\\\\\\F~=~ \dfrac{9\cdot 10^9\cdot5\cdot 10^{-6}\cdot 2\cdot 10^{-6}}{0,36}\\\\\\F~=~ \dfrac{9\cdot5\cdot 2\cdot 10^{9+(-6)+(-6)}}{36\cdot 10^{-2}}\\\\\\F~=~\dfrac{90\cdot 10^{-3-(-2)}}{36}\\\\\\F~=~\dfrac{5\cdot 10^{-1}}{2}\\\\\\\boxed{\sf F~=~2,5\cdot 10^{-1}~N} ~~ou~~\boxed{\sf F~=~0,25~N}[/tex]
2)
[tex]\sf 0,4~=~9\cdot 10^9\cdot \dfrac{5\cdot 10^{-6}\cdot 5\cdot 10^{-6}}{d^2}\\\\\\4\cdot 10^{-1}~=~ \dfrac{9\cdot 10^9\cdot5\cdot 10^{-6}\cdot 5\cdot 10^{-6}}{d^2}\\\\\\d^2~=~ \dfrac{9\cdot 5\cdot 5\cdot 10^{9+(-6)+(-6)}}{4\cdot 10^{-1}}\\\\\\d^2~=~ \dfrac{225\cdot 10^{-3-(-1)}}{4}\\\\\\d~=~ \sqrt{\dfrac{225\cdot 10^{-2}}{4}}\\\\\\d~=~ \dfrac{\sqrt{225}\cdot \sqrt{10^{-2}}}{\sqrt{4}}}\\\\\\d~=~ \dfrac{15\cdot 10^{-1}}{2}\\\\\\\boxed{\sf d~=~ 7,5\cdot 10^{-1}~m}~~ou~~ \boxed{\sf 0,75~m}[/tex]
3)
Lembrando de converter a distancia para o S.I, teremos 0,2 m entre as duas cargas.
[tex]\sf F~=~9\cdot 10^9\cdot \dfrac{5\cdot 10^{-5}\cdot 2\cdot 10^{-6}}{0,2^2}\\\\\\F~=~ \dfrac{9\cdot 10^9\cdot5\cdot 10^{-5}\cdot 2\cdot 10^{-6}}{0,04}\\\\\\F~=~ \dfrac{9\cdot5\cdot 2\cdot 10^{9+(-5)+(-6)}}{4\cdot 10^{-2}}\\\\\\F~=~\dfrac{90\cdot 10^{-2-(-2)}}{4}\\\\\\F~=~22,5\cdot 10^{0}\\\\\\\boxed{\sf F~=~22,5~N}[/tex]
[tex]\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio[/tex]
