Question

Two point charges $Q$ and $q$ are placed at distance $x$ and $\frac{x}{2}$ respectively from a positive charge $4q$ all along the same line. If the net force on $q$ is zero then, find $Q$ in terms of $q$.

Answer 1

The distance between $Q$ and $4q$ is $x$ and distance between $4q$ and $q$ is $\frac{x}{2}$.

Distance between $q$ and $Q$ is $x+\frac{x}{2}=\frac{3x}{2}$

The force applied by $Q$ on $q$ be $F_1$ and by 4q on q be $F_2$

So,

$\begin{array}{c}{F}_1=\frac{kQq}{{\left(\frac{3}{2}x\right)}^2}=\frac{4kQq}{9x^2}\\ F_2=\frac{k\left(4q\right)q}{{\left(\frac{x}{2}\right)}^2}=\frac{4k{q}^2}{x^2}\\ F_1+F_2=0\\ F_1=-F_2\\ \frac{4kQq}{9x^2}=-\frac{4k{q}^2}{x^2}\\ Q=-9q\end{array}$