Question

In Fig. 22-66, particle 1 (of charge $\mathbf{+}\mathbf{2}\mathbf{.00}\mathbf{\text{pC}}$), particle 2 (of charge$\mathbf{-}\mathbf{2}\mathbf{.00}\mathbf{\text{pC}}$), and particle 3 (of charge$\mathbf{+}\mathbf{5}\mathbf{.00}\mathbf{\text{pC}}$) form an equilateral triangle of edge length $\mathit{a}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{.50}\mathbf{\text{cm}}$.

(a) Relative to the positive direction of the x-axisdetermines the direction of the force $\overrightarrow{{\mathbf{F}}_{\mathbf{3}}}$on particle 3 due to the other particles by sketchingelectric field lines of the other particles.

(b) Calculate the magnitude of$\overrightarrow{{\mathbf{F}}_{\mathbf{3}}}$

Short answer

1. Relative to the positive direction of the x-axis, the direction of the force, $\overrightarrow{F_3}$on particle 3 due to the other particles is zero in the form of electric field lines.
2. The magnitude of the force $\overrightarrow{F_3}$is.$9.96\times {10}^{-12}\text{N}$

Step-by-step solution

The given data

As given in Fig., particle 1 $(q_1=+2.00\text{pC})$, particle 2$(q_2=-2.00\text{pC})$, and particle 3 $(q_3=+5.00\text{pC})$form an equilateral triangle of edge length.$a=9.50\text{\hspace{0.17em}cm}$

Understanding the concept of electrostatic force

Using the concept of Coulomb's law of electrostatics, the value of the net force on the third particle can be calculated. Similarly, using a similar concept, we can see that the electric field lines from two equal and oppositely charged bodies cancel each other, resulting in a net-zero force component of the third particle.

Formula:

The magnitude of the electrostatic force acting on a particle 1 due to particle 2 that is making an angle with each other, $F_1=\frac{q_1{q}_{2}cos\theta }{4\pi {\epsilon }_o{a}^2}$ (i)

a) Calculations for sketching the electric field lines of the force

From symmetry, we see the net force component along the y-axis for two charges of equal magnitude and opposite direction is zero.

b) Calculation of the magnitude of the force

The net force component along the x axis points rightward. With θ = 60^o, the magnitude of the net force acting on particle 3 due to particle 1 and particle 2 using equation (i), is given as:

$\begin{array}{c}{F}_3=2\frac{q_3{q}_{1}cos\theta }{4\pi {\epsilon }_o{a}^2}\text{(}\because \left|q_1\right|=\left|q_2\right|\text{)}\\ F_3=\frac{k{q}_3{q}_1}{a^2}\text{(}\because {\text{cos(60}}^0\text{)=1/2)}\\ =\frac{(8.99\times {10}^9\text{N}\cdot \frac{\text{m}}{{\text{C}}^{\text{2}}})(5.00\times {10}^{-12}\text{C})(2.00\times {10}^{-12}\text{C})}{{\left(0.0950\text{m}\right)}^2}\\ =9.96\times {10}^{-12}\text{N}\end{array}$

Hence, the value of the net force is.$9.96\times {10}^{-12}\text{N}$