Answer:
Explanation:
Given
Charge  on particles
[tex]q_1=+9\ nC[/tex]
[tex]q_2=-3\ nC[/tex]
[tex]x_1=-2\ cm[/tex]
[tex]x_2=13\ cm[/tex]
Third charge [tex]q_3[/tex] must be placed right side of [tex]q_2[/tex] as [tex]q_2[/tex] will attract and [tex]q_1 will repel the [tex]q_3[/tex] so net force will be zero
Electrostatic force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
suppose [tex]q_3[/tex] is at a distance of x cm from [tex]q_2[/tex]
[tex]F_{13}=\frac{kq_1q_3}{(13+2+x)^2}[/tex]
[tex]F_{23}=\frac{kq_2q_3}{(x)^2}[/tex]
[tex]F_{13}+F_{23}=0[/tex]
[tex]\frac{k(9)(q_3)}{(15+x)^2}=\frac{k3q_3}{(x)^2}[/tex]
[tex]x(\sqrt{3}-1)=15[/tex]
[tex]x=\frac{15}{\sqrt{3}-1}[/tex]
[tex]x=20.49\ cm[/tex]
