Particle A of charge 2.79 *10 -4 C is at the origin, particle B of charge -5.70
ID: 2097414 • Letter: P
Question
Particle A of charge 2.79 *10-4C is at the origin, particle B of charge -5.70 * 10-4C is at (4.00 m, 0), and particle C of charge 1.02 *10-4C 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?
?
(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 from parts (a) and (d) to obtain the resultant x component of the electric force acting on C.
(g) Similarly, find the y component of the resultant force vector acting on C.
(h) Find the magnitude and direction of the resultant electric force acting on C.
magnitu
N
direction
magnitu
N
direction
Explanation / Answer
a) 0
b) F = 8.987*10^9*2.79*1.02*10^-8/9 = 28.416894 N
c)-8.897*10^9*5.7*1.01*10^-8/25 = 20.4880116 N
d)20.4880116*4/5 = 16.39040928 N
e)20.4880116*3/5= 12.29280696 N
f) 0-16.39040928 = 16.39040928 N
g)28.416894-12.29280696 = 16.12408704 N
h) F= sqrt((16.12408704*16.12408704) + (16.39040928*16.39040928)) = 22.9919921 N ANSWER
angle = tan-(16.12408704/16.39040928) = 44.53071 degrees ANSWER