Respuesta :
Answer:
0.00019665625 N towards +x axis
0.0008091 N
towards +x axis
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
[tex]q_1[/tex] = 3 nC
[tex]q_2[/tex] = -7 nC
[tex]q_3[/tex] = 5 nC
From Coulomb's law we have
[tex]F_{13}=\dfrac{kq_1q_2}{r_{13}^2}\\\Rightarrow F_{13}=\dfrac{8.99\times 10^9\times 3\times 10^{-9}\times 5\times 10^{-9}}{0.02^2}\\\Rightarrow F_{13}=0.000337125\ N[/tex]
The force is 0.000337125 N towards -x axis
[tex]F_{23}=\dfrac{kq_3q_2}{r_{23}^2}\\\Rightarrow F_{13}=\dfrac{8.99\times 10^9\times -7\times 10^{-9}\times 5\times 10^{-9}}{0.04^2}\\\Rightarrow F_{23}=-0.00019665625\ N[/tex]
The force is 0.00019665625 N towards +x axis
Net force on [tex]q_3[/tex]
[tex]F=F_{13}+F_{23}\\\Rightarrow F=0.000337125-0.00019665625\\\Rightarrow F=0.00014046875\ N[/tex]
[tex]F_{31}=0.000337125\ N[/tex]
The force is 0.000337125 N towards +x axis
[tex]F_{21}=\dfrac{kq_1q_2}{r_{21}^2}\\\Rightarrow F_{13}=\dfrac{8.99\times 10^9\times -7\times 10^{-9}\times 3\times 10^{-9}}{0.02^2}\\\Rightarrow F_{23}=-0.000471975\ N[/tex]
The force is 0.000471975 N towards +x axis
Net force on [tex]q_1[/tex]
[tex]F=F_{31}-F_{21}\\\Rightarrow F=0.000337125-(-0.000471975)\\\Rightarrow F=0.0008091\ N[/tex]
The force is 0.0008091 N towards +x axis