Particle A of charge 3.30 10 4 C is at the origin, particle B of charge 6.15 10
ID: 1427325 • Letter: P
Question
Particle A of charge 3.30 104 C is at the origin, particle B of charge 6.15 104 C is at (4.40 m, 0) and particle C of charge 1.50 104 C is at (0, 3.50 m). (a) What is the x-component of the electric force exerted by A on C?
N
(b) What is the y-component of the force exerted by A on C?
N
(c) Find the magnitude of the force exerted by B on C.
N
(d) Calculate the x-component of the force exerted by B on C.
N
(e) Calculate the y-component of the force exerted by B on C.
N
(f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C.
N
(g) Repeat part (f) for the y-component.
N
(h) Find the magnitude and direction of the resultant electric force acting on C.
Explanation / Answer
qA = 3.3*10^-4 C, qB =6.15*10^-4 C, qC = 1.5*10^-4 C
k = 9*10^9 N.m^2/C^2
F = kq1q2/r^2
(a) xAC =0
x -component of force exerted by A on C is zero
(b) Fy = kqAqC/yAC^2 = (9*10^9*3.3*10^-4*1.5*10^-4)/(3.5*3.5)
Fy = 36.37 N
(c) Fx = (9*10^9*6.15*10^-4*1.5*10^-4)/(4.4*4.4)
Fx = 42.88 N
Fy = (9*10^9*6.15*10^-4*1.5*10^-4)/(3.5*3.5)
Fy = 67.78 N
F = [(42.88)^2+(67.78)^2]^1/2 = 80.29 N
(d) from (c) Fx = 42.88 N
(e) from (c) Fy = 67.78 N
(f) Fcx = 0+ 42.88 = 42.88 N
(g) Fcy = 36.37+67.78 = 104.15 N
(h) Fc = 42.88 i+ 104.15 j
Fc = [(42.88)^2+(104.15)^2]^1/2
Fc = 112.63 N
tan(theta) = 104.15/42.88
theta = 67.6 degrees
direction is 67.6 degrees with +x axis