Answer:
[tex]-180.38\ \text{N}[/tex]
Explanation:
[tex]q_1=-53\ \mu\text{C}[/tex]
[tex]q_2=105\ \mu\text{C}[/tex]
[tex]q_3=-88\ \mu\text{C}[/tex]
r = Distance between the charges
[tex]r_{12}=0.5\ \text{m}[/tex]
[tex]r_{23}=0.95\ \text{m}[/tex]
[tex]r_{13}=1.45\ \text{m}[/tex]
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Net force is given by
[tex]F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}[/tex]
The force on the particle [tex]q_1[/tex] is [tex]-180.38\ \text{N}[/tex].
