Answer: [tex]18.22\ cm[/tex] from [tex]3.5\ nC[/tex] charge.
Explanation:
Given
The magnitude of the first charge is [tex]Q_1=3.5\ nC[/tex]
The magnitude of the second charge is [tex]Q_2=5\ nC[/tex]
[tex]-6\ nC[/tex] charge must be placed in between the two charges to establish equilibrium
The electrostatic force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Equilibrium will be established when force by both the charges balance out each other. Suppose [tex]-6\ nC[/tex] is placed at a distance of [tex]x[/tex] cm from [tex]3.5\ nC[/tex] . So, we can write
[tex]\Rightarrow \dfrac{k(3.5)(-6)}{x^2}=\dfrac{k(5)(-6)}{(40-x)^2}[/tex]
Canceling similar terms
[tex]\Rightarrow \left [ \dfrac{40-x}{x}\right ]^2=\dfrac{10}{7}\\\\\Rightarrow \dfrac{40-x}{x}=\sqrt{\dfrac{10}{7}}=1.195\\\\\Rightarrow 40-x=1.195x\\\Rightarrow 40=2.195x\\\Rightarrow x=18.22\ cm[/tex]
Thus, the equilibrium position is [tex]18.22\ cm[/tex] from [tex]3.5\ nC[/tex] charge.
