In Fig. 21-26, particle 1 of charge$\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{q}$and particle 2 of charge $\mathbf{+}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{q}$ are held at separation Lon anx-axis. If particle 3 of unknown charge q₃is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the (a) x and (b) y coordinates of particle 3?

Charge of particle 1,$q_1=-5.00q$

Charge of particle 2,$q_2=+2.00q$

The separation between particle 1 and particle 2,

Charge of particle 3 is$q_3$.

Net electrostatic force on particle 3 from particle 1 and particle 2 is zero.

When more than one force is acting on a body, you follow the concept of the superposition principle to calculate the net force acting on the given charge or body. Our system consists of two charges along a straight line. The position of the third charge so that the net force on it due to charges 1 and 2 vanishes can be calculated using Coulomb's law.

Formula:

The electrostatic force of attraction or repulsion on a body by another charged body according to Coulomb’s law,

$F_{ij}=\frac{k\left|q_i\right|\left|q_j\right|}{{r_{ji}}^2}$…..(i)

In order to have net force as zero, particle 3 must be on the x-axis and be attracted by one and repelled by another as both the particles are on the x-axis. As the result, it cannot be between particles 1 and 2, but instead either to the left of particle 1 or to the right of particle 2.

Let $q_3$be placed to the right of the first particle. Then its attraction to the first charge will be exactly balanced by its repulsion from second charge.

Now, the net electrostatic force on particle 3 by particle 1 and 2 can be given using equation (i), and the x-coordinate of the particle 3 can be calculated as follows:

$\begin{array}{rcl}{F}_3& =& F_{13}+F_{23}\\ 0& =& \frac{k{q}_1{q}_3}{r_{13}^2}+\frac{k{q}_2{q}_3}{r_{23}^2}\\ & & \end{array}$

Substitute known valued in th above equation.

$\begin{array}{rcl}0& =& \frac{k\left(-5.00q\right)\left(q_3\right)}{{\left(L+x\right)}^2}+\frac{k\left(+2.00q\right)\left(q_3\right)}{{\left(x\right)}^2}\\ \frac{\left(5.00\right)}{{\left(L+x\right)}^2}& =& \frac{\left(2.00\right)}{{\left(x\right)}^2}\\ \frac{x+L}{x}& =& \sqrt{\frac{5}{2}}\end{array}$

$\begin{array}{rcl}\text{1.58x}& =& x+L\\ 0.58x& =& L\\ x& =& \frac{L}{0.58}\\ x& =& \text{1.72}L\end{array}$

Therefore, the change $q_3$ is placed at a distance of $\text{1.72}L$ from the charge $q_2$.

The distance to the chargefrom is given as,

$\begin{array}{c}{r}_{13}=L+x\\ =L+1.72L\\ =2.72L\end{array}$

Hence, the value of x-coordinate of the charge$q_3$is $2.72L$.

As per the calculations of part (a), the y-coordinate of particle q₁ and q₂ is zero. Hence, the y-coordinate of the charge q₃ is,

y = 0