Particle A of charge 5.00 10-4 C is at the origin, particle B of charge -5.00 10
ID: 2158669 • Letter: P
Question
Particle A of charge 5.00 10-4 C is at the origin, particle B of charge -5.00 10-4 C is at (5.00 m, 0) and particle C of charge 2.00 10-4 C is at (0, 5.00 m).(a) What is the x-component of the electric force exerted by A on C?
1 N
(b) What is the y-component of the force exerted by A on C?
2 N
(c) Find the magnitude of the force exerted by B on C.
3 N
(d) Calculate the x-component of the force exerted by B on C.
4 N
(e) Calculate the y-component of the force exerted by B on C.
5 N
(f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C.
6 N
(g) Repeat part (f) for the y-component.
7 N
(h) Find the magnitude and direction of the resultant electric force acting on C. magnitude 8 N
direction 9
Explanation / Answer
QA = 5*10-4 C ; QB = -5*10-4 C ; QC = 4*10-4 C
a) Both A and C lie on the y axis. Thus the x component of FCA = 0
b) Y component of Force exerted by A on C : KQAQC/22 = 9*109*5*10-4*4*10-4/4 = 450 N
c) Force exerted by B on C = FBC = KQBQC/(22+22) = -9*109*5*10-4*4*10-4/(22) = 636.396 N
d) X component of FCB = FBC cos(45) = 450 N
e) Y component of FCB = -FBC sin(45) = - 450 N
f) Net x component of force on C = 450+0 = 450 N
g) Net y component of force on C = 450-450 = 0 N
h) Resultant force on C = 450 N in x direction