Figure 21-14 shows two charged particles on an axis. The charges are free to move. However, a third charged particle can be placed at a certain point such that all three particles are then in equilibrium. (a) Is that point to the left of the first two particles, to their right, or between them? (b) Should the third particle be positively or negatively charged? (c) Is the equilibrium stable or unstable?

Figure 21-14 represents two charged particles on an axis. However, a third particle is placed at a point to get equilibrium condition for the three particles.

The condition of equilibrium due to any two charges on a third particle can be achieved if the forces due to the two charges are equal in magnitudes and opposite in direction. For a given set of equal polar charges, the third particle should be of opposite polarity and placed between the charges to get the force direction to be opposite. Now, the equilibrium condition can be considered unstable if the movement of one charge will affect the movements of other charges and the state cannot be reversed, if not, then it is a stable one.

Formula:

The magnitude of the electrostatic force due to the two charges, $\mathbf{\left|}\mathbf{F}\mathbf{\right|}\mathbf{=}\frac{\mathbf{k}\mathbf{\left|}{\mathbf{q}}_{\mathbf{1}}\mathbf{\right|}\mathbf{\left|}{\mathbf{q}}_{\mathbf{2}}\mathbf{\right|}}{{\mathbf{r}}^{\mathbf{2}}}$ (i)

Let us consider the separation between both the given charges to beand the point is placed between them at a distance$\mathrm{x}$from the charge $-3\mathrm{q}$as for equal polarities charges, the forces due to the two forces cancels out if the point is between them with opposite polarities. Thus, consider that the charge is $+\mathrm{q}\text{'}$.

Then, the condition of equilibrium at this point can be given using equation (i) as follows:

$\begin{array}{c}{\mathrm{F}}_{-3\mathrm{q}}={\mathrm{F}}_{-\mathrm{q}}\\ \frac{\mathrm{k}|-3\mathrm{q}||+\mathrm{q}\text{'}|}{{\mathrm{x}}^2}=\frac{\mathrm{k}|-\mathrm{q}||+\mathrm{q}\text{'}|}{{(\mathrm{d}-\mathrm{x})}^2}\\ \frac{\mathrm{d}-\mathrm{x}}{\mathrm{x}}=\frac{1}{\sqrt{3}}\end{array}$

As this condition is justified, thus, the assumptions are correct.

Hence, the third particle is placed between the given charges for equilibrium condition.

From the above calculations and assumptions found to be correct, it can be said that the third particle is positively charged.

The found condition of equilibrium is unstable because if$+\mathrm{q}\text{'}$ is moved slightly to the left, the attraction due to$-3\mathrm{q}$ will be more comparatively to attraction due to$-\mathrm{q}$ and it will not come back to its original form.

Hence, the equilibrium is unstable one.