Question

A 5nC charge is placed at the origin of our coordinate system. There are three other charges; 5C located 3m in the positive i direction, 2C located 2m in the negative i direction, and -4C located 5m in the positive j direction. What is the net force acting on the 5nC charge?

Answer 1

Step-by-step explanation:

F1 = (Kq1 q2)/r^2

Where K = 9x10^9NmC-2

F1 = (9×10〗^9×2×5×〖10〗^(-9))/2^2 =22.5N

F2= (Kq_2 q_3)/r^2

F2 = (9*〖10〗^9×5×〖10〗^(-9)×5)/2^2 =25N

F3= (Kq_3 q_4)/r^2

F3 = (9*〖10〗^9×5*〖10〗^(-9)×-4)/2^2 =-7.2N

Net horizontal force F = F2 – F1

F = 25 – 22.5 = 2.5N

Net force in 5nC = √((〖7.2〗^2 )+〖2.5〗^(2 ))=7.62N