In Fig. 21-23, particles 1 and 2 are fixed in place, but particle 3 is free to move. If the net electrostatic force on particle 3 due to particles 1 and 2 is zero and ${\mathbf{L}}_{\mathbf{23}}\mathbf{=}\mathbf{2}\mathbf{.00}{\mathbf{L}}_{\mathbf{12}}$, what is the ratio${\mathbf{q}}_{\mathbf{1}}\mathbf{/}{\mathbf{q}}_{\mathbf{2}}$?

1. Separation values,${\mathrm{L}}_{23}=2.00{\mathrm{L}}_{12}$
2. The net force on particle 3 is zero.

Using the concept of Coulomb's law, the net electrostatic force on particle 3 due to particles 1 and 2 can be found. Hence, for the net force as zero, we can get the required ratio of the charges of particle 1 and particle 2.

Formula:

The magnitude of the electrostatic force between any two particles, $\mathrm{F}=\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (i)

Regarding the forces on ${\mathrm{q}}_3$exerted by ${\mathrm{q}}_1$and ${\mathrm{q}}_2$, one must “push” and the other must “pull” in order that the net force is zero; hence, ${\mathrm{q}}_1$and ${\mathrm{q}}_2$have opposite signs. For individual forces to cancel, their magnitudes must be equal, thus, the forces value using equation (i) is given as:

$\begin{array}{c}\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_3}{{({\mathrm{L}}_{12}+{\mathrm{L}}_{23})}^2}=\mathrm{k}\frac{{\mathrm{q}}_2{\mathrm{q}}_3}{{\left({\mathrm{L}}_{23}\right)}^2}\\ \frac{{\mathrm{q}}_1}{{({\mathrm{L}}_{12}+2{\mathrm{L}}_{12})}^2}=\frac{{\mathrm{q}}_2}{{\left(2{\mathrm{L}}_{12}\right)}^2}\\ \frac{{\mathrm{q}}_1}{9}=\frac{{\mathrm{q}}_2}{4}\\ {\mathrm{q}}_1=-\frac{9\left({\mathrm{q}}_2\right)}{4}\\ \frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=-2.25\end{array}$

Hence, the value of the required ratio is$-2.25$