Question

How do you do #4? I need a single number with a direction

en Consider three point charges located at the corners of a triangle as shown in the accompanying figure. lfg.-6.0 nC, 9,--1.0nC, and q3-5,0 nC, what is the net force on q3? 935.0 nC 0.10 m 0.30 m , 6.0C 91.0 nC

Answer 1

q_= 6.0 mc q_2 = -1.0 mc q_3 = 5.0 mc Net force on q_3 F_3 = f^rightarrow_3, 1 + F^rightarrow_3, 2 therefore f^rightarrow_3, 2 therefore F^rightarrow_3 = -k q_1 q_3/r^2_13 i^cap - k q_1 q_2/r^2_23 j^cap therefore F^rightarrow_3 = [- 9 times 10^9 times 6 times 10^-9 times 5 times 10^-9/(0.3)^2 i^cap - 9 times 10^9 times 1 times 10^-9 times 5 times 10^-9/(0.10)^2 j^cap] N F^rightarrow_3 = (- 3 times 10^-6 i^cap - 4.5 times 10^-6 j^cap) N - Magnitude of Vertical-bar F_3 Vertical-bar = squareroot (3)^2 + (4.5)^2 times 10^-6 N = 5.41 times 10^-6 N therefore Vertical-bar F_3 Vertical-bar = 5.41 times 10^-6 N Direction phi = tan^-1 (4.5 times 10^-6/3 times 10^-6) = 56.3 degree therefore theta = 180 + phi = 180 + 56.3 = 236.3 degree therefore theta = 236.3 degree leftarrow counter clockwise direction from + x - axis