Given:
The charge
[tex]\begin{gathered} q1=\text{ -16.5 nC} \\ =-16.5\text{ }\times10^{-9}\text{ C} \end{gathered}[/tex]
The charge q1 is located at
[tex]x1\text{ = -1.71 m}[/tex]
The charge
[tex]\begin{gathered} q2\text{ = 36 nC} \\ =\text{ 36 }\times10^{-9}\text{ C} \end{gathered}[/tex]
The charge q2 is located at x2 = 0
The charge
[tex]\begin{gathered} q3\text{ = 52.5 nC} \\ =52.5\text{ }\times10^{-9\text{ }}C \end{gathered}[/tex]
The charge q3 is located at
[tex]x3\text{ = -1.115 m}[/tex]
To find the net force on charge q3.
Explanation:
The force on charge q3 due to q1 is
[tex]\begin{gathered} F_{13}=\text{ K}\frac{|q1q3|}{(x1-x3)^2} \\ =9\times10^9\times\frac{|(-16.5)\times10^{-9}\times52.5\times10^9|}{(-1.71+1.115)^2} \\ =2.2\times10^{-5}\text{ N} \end{gathered}[/tex]
The force on the charge q3 due to q2 is
[tex]\begin{gathered} F_{23}=K\frac{|q2q3|}{(x2-x3)^2} \\ =9\times10^9\times\frac{|36\times10^{-9}\times52.5\times10^{-9}|}{(0+1.115)^2} \\ =1.368\text{ }\times10^{-5}\text{ N} \end{gathered}[/tex]
The forces acting on q3 can be represented as
Thus, the net force will be
[tex]\begin{gathered} F_{net}=\text{ F}_{13}+F_{23} \\ =2.2\times10^{-5}+1.368\times10^{-5} \\ =3.568\times10^{-5}\text{ N} \end{gathered}[/tex]
The direction of the force will be negative as it lies on the negative x-axis.
