Charges \(3 \mathrm{Q}, \mathrm{q}\) and Q are placed along x -axis at positions \(\mathrm{x}=0, \mathrm{x}=\frac{1}{3}\) and \(\mathrm{x}=1\) respectively. When the force on charge Q is zero, the value of \(q\) is

Force \(F_1\) due to charge \(3 Q\) on \(Q\) will be,
\(\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{3 \mathrm{Q}^2}{l^2}\)
Force \(F_2\) due to charge \(q\) on charge \(Q\) will be,
\(\begin{aligned}
& \mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\left(\frac{2 l}{3}\right)^2} \\
\therefore \quad & \mathrm{~F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{9 \mathrm{Qq}}{4 l^2}
\end{aligned}\)
For the force on charge Q to be zero,
\(\begin{array}{ll}
& \mathrm{F}_1+\mathrm{F}_2=0 \text { or } \mathrm{F}_1=-\mathrm{F}_2 \\
\therefore & \cdot \frac{3 \mathrm{Q}^2}{l^2}=-\frac{9 \mathrm{Qq}}{4 l^2} \\
\therefore & \mathrm{Q}=-\frac{3 \mathrm{q}}{4} \\
\therefore \quad & \mathrm{q}=-\frac{4}{3} \mathrm{Q}
\end{array}\)