91Ó°ÊÓ

The charges q₁ and q₂ are on the x-axis separated by the distance,

$r_{12}=8 cm\left(\frac{1 m}{100 cm}\right)=0.08 m$

Charge of particle 3 that is placed in between the other two particles,

$q_3=+8.00×{10}^{-19}C$

The scale of the x-axis is set at,$x_s=8 cm\left(\frac{1 m}{100cm}\right)=0.08 m$

Using the concept of Coulomb's law, we can find the net force relation of all the charges. This will determine the charge on the first particle. Again, using the same concept, we can determine the required ratio of the charges.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=\frac{1}{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_0\right)}\frac{\left|q_1\right|\left|q_2\right|}{r^2}$

(1)

According to the graph, when q₃ is very close to q₁ (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q₁ has the same sign as q₃. As is a positive-valued charge, hence, the sign of the charge q₁is positive.

Since the graph crosses zero and particle 3 is between the others, q₁ must have the same sign as q₂, which means it is also positive-valued. We note that it crosses zero at r=0.020m (which is a distance d=0.060 m from q₂). Using equation (1) at that point, we have the following relation between the particles as:

$$\begin{gathered} k\frac{\left|q_1{q}_3\right|}{r^2}=k\frac{\left|q_3{q}_2\right|}{d^2} \\ {q}_2={\left(\frac{d}{r}\right)}^2{q}_1 \\ {q}_2={\left(\frac{0.060m}{0.020m}\right)}^2{q}_1 \\ {q}_2=9.0q_1 \\ q\_2/q\_1=9.0 \end{gathered}$$

Hence, the value of the required ratio is 9.0