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?

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Physics

Guest

18 February, 23:59

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?

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Answers (1)

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Koen French · Answer 1

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