Particle A of charge 3.50 10 4 C is at the origin, particle B of charge 6.05 10
ID: 1567329 • Letter: P
Question
Particle A of charge 3.50 104 C is at the origin, particle B of charge 6.05 104 C is at (3.80 m, 0) and particle C of charge 1.40 104 C is at (0, 3.36 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
a) FACx = 0
b) FACy = k*qA*qB/y^2
= 9*10^9*3.5*10^-4*1.4*10^-4/3.36^2
= 39.1 N
c) FBC = k*qB*qC/(x^2 + y^2)
= 9*10^9*6.05*10^-4*1.4*10^-4/(3.8^2 + 3.36^2)
= 29.6 N
FBCx = FBC*cos(theta)
= 29.6*x/sqrt(x^2 + y^2)
= 29.6*3.8/sqrt(3.8^2 + 3.36^2)
= 22.2 N
d) FBCy = FBC*sin(theta)
= 29.6*y/sqrt(x^2 + y^2)
= 29.6*3.36/sqrt(3.8^2 + 3.36^2)
= 19.6 N