91Ó°ÊÓ

The charge value of particles 1, 2, and 3 in $20.0\mathrm{μ}C$each.

The separation between particles 1 and 2, $r_{12}=1.50m$

The three in the second figure are positioned as vertices of the equilateral triangle.

Using the concept of Coulomb's law, we can get the magnitude of the force on particle 1 due to particle 2. Similarly, for more than one charge, the net force is the sum of all the forces.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

Substituting the given data in the equation (1), we can get the magnitude of the force acting on particle 1 due to particle 2 as given:

$$\begin{gathered} F_{12}=\left(9×{10}^9\frac{N.m^2}{C^2}\right)\left(\frac{20.0×{10}^{-6}C}{{(1.50m)}^2}\right) \\ =1.60N \end{gathered}$$

Hence, the value of the magnitude of the force is 1.60N.

The Force diagram is shown as well as our choice of the y-axis (the dashed line)

The y axis is meant to bisect the line between q₂ and q₃ in order to make use of the symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges q₁ = q₂= q₃ =q). We see that the resultant force is along this symmetry of the y-axis, and we obtain the net force on particle 1 by summing the force due to particle 2 and particle 3 and thus, using equation (1), it is given as:

$F_{\mathrm{γ}}=2k\frac{\left|q^2\right|}{r^2}\cos \left({30}^{∘}\right)$

$=2(9×{10}^9\frac{N.m^2}{C^2})\frac{{(20.0×{10}^{-6}C)}^2}{{\left(1.50m\right)}^2}\cos \left({30}^{∘}\right)$

=2.77N

Hence, the net value of the force is 2.77N$$\begin{gathered} F_{\mathrm{γ}}=2k\frac{\left|q^2\right|}{r^2}\cos \left({30}^{∘}\right) \\ =2\left(9×{10}^9\frac{N.m^2}{C^2}\right)\frac{{(20.0×{10}^{-6}C)}^2}{{\left(1.50m\right)}^2}\cos \left({30}^{∘}\right) \\ =2.77N \end{gathered}$$