[tex]\displaystyle \sf \underline{\text{For{\c c}a el{\'e}trica}} : \\\\ F_e = \frac{k\cdot Q_1\cdot Q_2 }{d^2} \\\\\ temos : \\\\ Q_1 = 5\cdot 10^{-6}\ C\\\\ Q_2= 3\cdot 10^{-6}\ C \\\\ d= 250 \ cm = 250\cdot 10^{-2} \ m \\\\ \text{Colocando uma carga q a uma dist{\^a}ncia x de }\ Q_2 . Temos :\\\\ \frac{k\cdot Q_1\cdot q}{x^2}=\frac{k\cdot Q_2\cdot q}{(d-x)^2} \\\\\\ \frac{Q_1}{x^2}=\frac{Q_2}{(d-x)^2}\\ \\\\ Q_1\cdot (d-x)^2=Q_2\cdot x^2 \\\\[/tex]
[tex]\displaystyle \sf 5\cdot 10^{-6}\cdot (250\cdot 10^{-2}-x)^2 = 3\cdot 10^{-6}\cdot x^2 \\\\ \frac{3}{5}\cdot x^2 = (25\cdot 10^{-1}-x)^2 \\\\\\ \sqrt{\frac{3}{5}\cdot x^2} = \sqrt{(25\cdot 10^{-1}-x)^2} \\\\\\ \frac{\sqrt{3}}{\sqrt{5}}\cdot x= \pm\ 2,5 -x \\\\\ \frac{\sqrt{15}}{5}\cdot x=\pm \ 2,5-x \\\\\\ 1^{\circ} caso : \\\\ \frac{\sqrt{15}}{5}\cdot x = 2,5-x \\\\ x\cdot \left(\frac{\sqrt{15}}{5}+1\right) = 2,5 \\\\\\[/tex]
[tex]\displaystyle \sf x \cdot \frac{(\sqrt{15}+5)}{5} = 2,5 \\\\\\ x = \frac{5\cdot 25\cdot 10^{-1}}{\sqrt{15}+5} \to x = \frac{5\cdot 2,5 \cdot (5-\sqrt{15}) }{(5+\sqrt{15})\cdot (5-\sqrt{15})} \\\\\\ x = \frac{5\cdot 2,5\cdot (5-\sqrt{15})}{25-15} \\\\\\ x = \frac{5\cdot 2,5\cdot (5-\sqrt{15}) }{10}\cdot \\\\\\ x = 1,25\cdot (5-\sqrt{15})\ m[/tex]
[tex]\text{Dist{\^a}ncia at{\'e} Q = 3}\mu \ C : 250\cdot 10^{-2} - \sf \left[1,25\cdot (5-\sqrt{15})\ m \right] \\\\\ 2,5-6,25 +1,25\cdot \sqrt{15} \\\\\ 1,25\cdot \sqrt{15} - 3,75\ m \\\\ Portanto: \\\\ \boxed{\begin{array}{I} \displaystyle \sf \text{A uma dist{\^a}ncia de }\left[1,25\cdot (5-\sqrt{15}) \ \right]\ m\ \text{da carga 5}\mu C \\\\ \displaystyle \sf \text{A uma dist{\^a}ncia de }\left[1,25\cdot \sqrt{15}-3,75 \ \right]\ m\ \text{da carga 3}\mu C \end{array} }\checkmark[/tex]
vou deixar o 2º caso para você, que seria :
[tex]\displaystyle \sf \frac{\sqrt{15}}{5}\cdot x= -(2,5 - x)[/tex]
