91Ó°ÊÓ

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

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

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,

$F_1=\frac{1}{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_0\right)}\left(\frac{\left|q_1\right|\left|q_2\right|cos\mathrm{θ}}{r^2}\hat{i}+\frac{\left|q_1\right|\left|q_2\right|sin\mathrm{θ}}{r^2}\hat{j}\right)$ (1)

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

F₃=F₃₁+F₃₂+F₃₄

$=\frac{1}{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_0\right)}\left(-\frac{\left|q_1\right|\left|q_3\right|}{a^2}\hat{j}+\frac{\left|q_2\right|\left|q_3\right|}{{\sqrt{2a}}^2}(\cos {45}^{∘}\hat{i}+\sin {45}^{∘}\hat{j})+\frac{\left|q_4\right|\left|q_3\right|}{a^2}\hat{i}\right)$

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

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

.

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

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

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

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