In Fig. 21-26, particle 1 of charge +qand particle 2 of charge +4.00q are held at separation L=9.00cmon an x-axis. If particle 3 of charge q₃ is to be located such that the three particles remain in place when released, what must be the (a) xand (b) ycoordinates of particle 3, and (c) the ratio q₃/q?

A particle of charge +q and particle 2 of charge +4.00q are separate at distance on the x-axis,

$L=9.00\hspace{0.33em}cm\left(\frac{1m}{100cm}\right)=0.09\hspace{0.33em}m$

The third particle of charge is placed such that all remain in place.

Using the same attraction and repulsion of the charges due to the forces is used to find the position coordinates of the third particle. Using the same Coulomb's law, we can get the value of the required ratio of the charges.

Formula:

The value of the electrostatic force between any two particles,

$F_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_1{q}_2}{r^2}\right)$ (1)

If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and could not be in equilibrium. Suppose q₃ is at a distance x from q, and (L-x) from +4.00q. The force acting on it is using equation (1) is given by:

$F_3=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q{q}_3}{x^2}-\frac{4q{q}_3}{{(L-x)}^2}\right).....................\left(2\right)$

where the positive direction is rightward. For all charges to be at their place, we need: F₃=0.

Thus, canceling all the common terms for net force as zero, the equation (2) becomes

$$\begin{gathered} \frac{1}{x^2}=\frac{4}{{\left(L-x\right)}^2} \\ \frac{1}{x}=\frac{2}{L-x} \\ x=\frac{L}{3} \\ x=3.00\hspace{0.33em}cm \end{gathered}$$

Hence, the value of the x-coordinate is 3.00cm.

Similarly, to have a balanced system in order to have particles in their original position, the y coordinate of q₃ is 0.

The force on q using equation (1) is given as:

$F_q=\frac{-1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q{q}_3}{x^2}-\frac{4q^2}{L^2}\right)$..........................(3)

The signs are chosen so that a negative force value would cause q to move leftward. Now, for the particles to remain in their position, the net force is F_q =0 Thus, using this in equation (3) and canceling the common terms, we get that

$$\begin{gathered} q_3=-\frac{4q{x}^2}{L^2} \\ {q}_3=\frac{-4q}{9} \\ \frac{q_3}{q}=-\frac{4}{9} \\ \frac{q_3}{q}=-0.444 \end{gathered}$$

We may also verify the ratio value for the force on 4.00q that also vanishes:

$$\begin{gathered} F_{4q}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4q^2}{L^2}+\frac{4q{q}_0}{{\left(L-x\right)}^2}\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4q^2}{L^2}+\frac{4(-4/9)q^2}{\left(4/9\right)L^2}\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4q^2}{L^2}-\frac{4q^2}{L^2}\right) \\ =0 \end{gathered}$$