Question

19 36.90 is sin 36.9。 Fra cos 36.9° 4.00 m Fo F23 42 93 3.00 m 5.00 m 91

Using this triangle, find the vector components of the electric force on q₁ and the vector's magnitude and direction. (Use the charges given in the Practice It section.)

	Fx=	-8.94e-9 N (CORRECT)
	Fy=	??? N
magnitude		??? N
direction		??? counterclockwise from the x-axis

Answer 1

From picture F_a = F_12 = kq_1 q_2/r_12^2 = 9 times 10^9 times 6.18 times 10^7 times 1.90 times 10^-9/3^2 = 11.742 times 10^-9 N F_b = F_13 = kq_1 q_3/r_13^2 = 9 times 10^9 times 6.18 times 10^-9 times 5.03 times 10^-9/5^2 = 11.19 times 10^-9 N a) F_x = F_ax + F_bx F_x = F_a middot cos 90 degree + F_b cos 37 degree F_x = 11.742 times 10^-9 cos 90 degree + 11.19 times 10^-9 cos 37 degree F_x = 8.93 times 10^-9 N b) F_y = F_a y + F_b y F_y = F_a middot sin 90 degree + F_b sin 37 degree F_y = 11.742 times 10^-9 sin 90 degree + 11.19 times 10^-9 sin 37 degree F_y = 18.47 times 10^-9 N c) R = squareroot F_x^2 + F_y^2 R = 20.51 times 10^-9 N d) tan theta = F_y/F_x theta = tan^-1 (F_y/F_x) = tan^-1 (18.47 times 10^-9/8.93 times 10^-9) counterclockwise from x-axis. theta = 64.19 degree