Answer:
0.64 m from the first charge
Explanation:
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}[/tex]
[tex]F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}[/tex]
These forces are equal
[tex]\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m[/tex]
The distance that charge should be placed is 0.64 m from the first charge
