magnitude N direction Particle A of charge 2.97 x 10 - 4 degree C is at the orig
ID: 2144873 • Letter: M
Question
magnitude N direction Particle A of charge 2.97 x 10 - 4 degree C is at the origin, particle B of charge - 5.70 x 10 - 4 degree C is at(4.00 m, 0), and particle C of charge 1.07 x 10 - 4 degree C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force exerted by A on C? What is the y component of the force exerted by A on C? Find the magnitude of the force exerted by B on C. Calculate the x component of the force exerted by B on C. Calculate the y component of the force exerted by B on C. Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. Similarly, find the y component of the resultant force vector acting on C. Find the magnitude and direction of the resultant electric force acting on C.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=31.779 N
(c) the magnitude of the force exerted by B on C.
F_BC=k*QB*QC/5^2=21.95 N
(d) the x component of the force exerted by B on C.
FX=F_BC*cos(4/5)=21.947N
(e) the y component of the force exerted by B on C.
Fy=F_BC sin(4/5)=0.306N
(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=21.947 NN
(g) Similarly, find the y component of the resultant force vector acting on C.
Fnet,y=31.779-0.306=31.472 N
(h) Find the magnitude and direction of the resultant electric force acting on C.
magnitude Fnet=sqrt(Fnet,x^2+Fnet.y^2)=38.369 N
direction
tan?=Fnet,y/Fnet,x=304.88
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