In Fig. 21-23, three charged particles lie on an x-axis. Particles 1 and 2 are fixed in place. Particle 3 is free to move, but the net electrostatic force on it from particles 1 and 2 happens to be zero. If ${\mathbf{L}}_{\mathbf{23}}\mathbf{=}\mathbf{}{\mathbf{L}}_{\mathbf{12}}$, what is the ratio ${\mathbf{q}}_{\mathbf{1}}\mathbf{/}{\mathbf{q}}_{\mathbf{2}}$?

The net force on particle 3 is zero.

The separation between particles 2 and 3 is equal to the separation between particles 1 and 2 with all three particles lying on the x-axis,${\mathrm{L}}_{23}={\mathrm{L}}_{12}$

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them.

Formula:

The magnitude of the electrostatic force between any two particles,

${\mathrm{F}}_1=\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (1)

With rightward as positive magnitude for force, the net force acting on charge $q_3$is given using equation (1) as:

$\begin{array}{c}{\mathrm{F}}_3={\mathrm{F}}_{13}+{\mathrm{F}}_{23}\\ =\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_3\right|}{{({\mathrm{L}}_{12}+{\mathrm{L}}_{23})}^2}+\mathrm{k}\frac{\left|{\mathrm{q}}_2\right|\left|{\mathrm{q}}_3\right|}{{\left({\mathrm{L}}_{23}\right)}^2}\end{array}$

$\text{(sincealllieinx-axis,theseparationbetween1and3is}{\mathrm{L}}_{12}+{\mathrm{L}}_{23}\text{)}$

We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on ${\mathrm{q}}_3$ by $q_1$) is negative if they are unlike charges, indicating that ${\mathrm{q}}_3$is being pulled toward $q_1$, and it is positive if they are like charges (so $q_3$would be repelled from $q_1$). From, the given data, the net force on particle 3 is zero, thus, the above equation becomes

$\begin{array}{c}\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_3\right|}{{({\mathrm{L}}_{12}+{\mathrm{L}}_{23})}^2}+\mathrm{k}\frac{\left|{\mathrm{q}}_2\right|\left|{\mathrm{q}}_3\right|}{{\left({\mathrm{L}}_{23}\right)}^2}=0\\ \frac{{\mathrm{q}}_1}{4.00}+{\mathrm{q}}_2=0\\ \frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=-4.00\end{array}$

Hence, the value of the ratio is $-4.00$