Question

Particle A of charge 3.50 times 10^-4 C is at the origin, particle B of charge 6.20 times 10^-4 C is at (3.74 m, 0) and particle C of charge 1.40 times 10^-4 C is at (0, 3.46 m). What is the x-component of the electric force exerted by A on C? N What is the y-component of the force exerted by A on C? N Find the magnitude of the force exerted by B on C. N Calculate the x-component of the force exerted by B on C. N Calculate the y-component of the force exerted by B on C. N Sum the two x-components to obtain the resultant x-component of the electric force acting on C. N Repeat part (f) for the x-component. N Find the magnitude and direction of the resultant electric force acting on C. N

Answer 1

d)
Angle of BC with the x axis 180 – θ and tan θ = ¾ and hence 180 – θ = 143.13°
the x component of the force exerted by B on C. is
30.15 cos 143.13 = - 24.12 minus shows that it is toward left

e).
the y component of the force exerted by B on C.
30.15 sin 143.13 = 18.10 N

f)
Sum the two x components from part (a) and (d) = the resultant x component of the electric force acting along x -axis is - 24.12
g)
Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis is 36.83 + 18.10 = 54.93 N

h)

Resultant of - 24.12 and 54.93 is
√ [(- 24.12)^2 + (54.93)^2] = 60 N
Direction is at angle (90 + φ) from the x axis.

60 sin φ = 54.93 => φ = 66.27°

(90 + φ) = 90 + 66.27 = 156.27° from the x axis.