Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10
ID: 1267846 • Letter: P
Question
Particle A of charge 4.00 10-4 C is at the origin, particle B of charge -5.00 10-4 C is at (3.00 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?
(b) What is the y-component of the force exerted by A on C?
(c) Find the magnitude of the force exerted by B on C.
(d) Calculate the x-component of the force exerted by B on C.
(e) Calculate the y-component of the force exerted by B on C.
(f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C.
(g) Repeat part (f) for the y-component.
(h) Find the magnitude and direction of the resultant electric force acting on C.
PLEASE use the numbers I have, and can you also explain how you get the angle, thank you.
Explanation / Answer
(a) x component of the electric force exerted by A on C
Fx=k*QA*QC/X^2=0 N
(b) y component of the force exerted by A on C
Fy=k*QA*QC/y^2=28.8 N
(c) the magnitude of the force exerted by B on C.
F_BC=k*QB*QC/r^2=26.48 N
(d) the x component of the force exerted by B on C.
FX=F_BC*cos(3/34)=26.47N
(e) the y component of the force exerted by B on C.
Fy=F_BC sin(5/34)=0.067N
(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C.
Fnet,X=26.47NN
(g) Similarly, find the y component of the resultant force vector acting on C.
Fnet,y=28.8-0.067=28.733 N
(h) Find the magnitude and direction of the resultant electric force acting on C.
magnitude Fnet=sqrt(Fnet,x^2+Fnet.y^2)=39.06 N
direction
tan?=Fnet,y/Fnet,x=3012.65