Question

In Fig. 21-25, the particles have charges q₁ = -q₂=100nCand q₃ = -q₄=200nC and distance a=5.0cm. What are the (a) xand (b) ycomponents of the net electrostatic force on particle 3?

Short answer

1. The x-component of the net electrostatic force on particle 3 is 0.17N.
2. The y-component of the net electrostatic force on particle 3 is -0.046N.

Step-by-step solution

The given data

The values of the charges, $q_1(=-q_2)=100nC$and$q_3(=-q_4)=200nC$

The distance between the particles, $r=5.0cm\left(\frac{1m}{100cm}\right)=0.05\hspace{0.33em}m$

Understanding the concept of Coulomb’s law

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the dot product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=\frac{1}{\left(4\pi {\epsilon }_0\right)}\left(\frac{\left|q_1\right|\left|q_2\right|cos\theta }{r^2}\hat{i}+\frac{\left|q_1\right|\left|q_2\right|sin\theta }{r^2}\hat{j}\right)$ (1)

a) Calculation of x-component of the net electrostatic force on particle 3

The net force experienced by q₃ using equation (1) is given as:

F₃=F₃₁+F₃₂+F₃₄

$=\frac{1}{\left(4\pi {\epsilon }_0\right)}\left(-\frac{\left|q_1\right|\left|q_3\right|}{a^2}\hat{j}+\frac{\left|q_2\right|\left|q_3\right|}{{\sqrt{2a}}^2}(\cos {45}^{\circ }\hat{i}+\sin {45}^{\circ }\hat{j})+\frac{\left|q_4\right|\left|q_3\right|}{a^2}\hat{i}\right)$

The x-component of the resultant force on q₃ is given as:

$$\begin{gathered} F_{3x}=\frac{\left|q_3\right|}{\left(4\pi {\epsilon }_0\right)}\left(\frac{\left|q_2\right|}{2\sqrt{2}}+\left|q_4\right|\right) \\ =(9\times {10}^{9}N.m^2/C^2)\times \frac{2{(1\times {10}^{-7}C)}^2}{{(0.050m)}^2}\left(\frac{1}{2\sqrt{2}}+2\right) \\ =0.17N \end{gathered}$$

.

Hence, the value of the x-component of particle 3 is 0.17N

b) Calculation of y-component of the net electrostatic force on particle 3

Similarly, the component of the net force is given as:

$$\begin{gathered} F_{3Y}=\frac{\left|q_3\right|}{\left(4\pi {\epsilon }_0\right)}\left(-\left|q_1\right|+\frac{\left|q_2\right|}{2\sqrt{2}}\right) \\ =(9\times {10}^{9}N.m^2/C^2)\times \frac{2{(1\times {10}^{-7}C)}^2}{{(0.050m)}^2}\left(-1+\frac{1}{2\sqrt{2}}\right) \\ =-0.046N \end{gathered}$$

Hence, the value of the y-component of the net electrostatic force of particle 3 is -0.046N