Particle A of charge 3.21 x 10^-4 C is at the origin, particle B of charge -6.48
ID: 1773890 • Letter: P
Question
Particle A of charge 3.21 x 10^-4 C is at the origin, particle B of charge -6.48 x 10^-4 C is at (4.00 m, 0), and particle C of charge 1.05 x 10^-4 C is at (0, 3.00 m). We wish to find the net electric force on C.
(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 from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. ______ N
(g) Similarly, find the y component of the resultant force vector acting on C. ______ N
(h) Find the magnitude and direction of the resultant electric force acting on C.
magnitude ______ N
direction ______ ° counterclockwise from the +x-axis
Explanation / Answer
a] FAx = 0 N
b] FAy = kqAqC/r^2 = 9e9*3.21e-4*1.05e-4/3^2 = 33.705 N
c] FB = kqCqB/r^2 = 9e9*6.48e-4*1.05e-4/[3^2+4^2] = 24.5 N
d] FBx = kqBqC/r^2 cos theta = 9e9*6.48e-4*1.05e-4/(3^2+4^2) * 4/5 = 19.6 N
e] FBy = kqBqC/r^2 sin theta = 9e9*6.48e-4*1.05e-4/(3^2+4^2) * -3/5 = -14.7 N