Given :
Three charges :
[tex]q_1=-1.95\ nC\\\\q_2=3.60\ nC\\\\q_3=4.90\ nC[/tex]
There location on y-axis :
[tex]y_1=-0.550\ m\\\\y_2=0\ m\\\\y_3=-0.435\ m[/tex]
To Find :
Total force acting on [tex]q_3[/tex] .
Solution :
Total force acting on [tex]q_3[/tex] is :
[tex]F_3=\dfrac{kq_1}{r_1^2}+\dfrac{kq_2}{r_2^2}\\\\F_3=k[(\dfrac{1.95\times 10^{-9}}{0.550-0.435})+(\dfrac{3.6\times 10^{-9}}{0.550-0})]\\\\F_3=9.0 \times 10^9 \times [(\dfrac{1.95\times 10^{-9}}{0.550-0.435})+(\dfrac{3.6\times 10^{-9}}{0.550-0})]\\\\F_3=211.52\ N[/tex]
Therefore , force applied on [tex]q_3[/tex] is 211.52 N .
