The Sketch of the figure will bear the parameters of the charge that lies at (4,0,0)
Generally, the equation for is mathematically given as
1)
Since both of the other points, "q1"<"q2," are on the x-axis, the third one must also be on the x-axis, and it must be close to the charge of lesser magnitude. off q1 by x cm
Now, forces on q3 due to q1
[tex]\vec{F}_{1}=\frac{k q_{1} q_{3}}{x^{2}} \hat{x}[/tex]
Force on q3 due to q2
[tex]\vec{F}_{2}=\frac{k q_{2} q_{3}}{(6+x)^{2}}(-\hat{x})[/tex]
[tex]\quad \vec{F}_{1}+\vec{G}_{2}=0$[/tex]
[tex]&\frac{k q_{1} q_{3}}{x^{2}}-\frac{k q_{2} q_{3}}{(x+6)^{2}}=0 \\&\frac{2}{x^{2}}=\frac{8}{(x+6)^{2}} \Rightarrow \frac{x+6}{x}=\sqrt{\frac{8}{2}}=2 \\&x+6=2 x \Rightarrow x=6 \mathrm{~cm}[/tex]
In conclusion, the charge lies at (4,0,0)
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