A) Find the x -component of the total force exerted on the third charge by the o
ID: 1422046 • Letter: A
Question
A) Find the x-component of the total force exerted on the third charge by the other two.
B) Find the y-component of the total force exerted on the third charge by the other two.
C) Find the magnitude of the total force acting on the third charge.
D) Find the direction of the total force acting on the third charge.
A charge 4.98 nC is placed at the origin of an xy-coordinate system, and a charge -2.04 nC is placed on the positive x-axis at x = 4.02 cm . A third particle, of charge 5.99 nC is now placed at the point x 4.02 cm , y 2.98 cmExplanation / Answer
let
q1 = 4.98 nC at (0,0)
q2 = -2.04 nC at (4.02 cm, 0)
q3 = 5.99 nc at (4.02 cm, 2.98 cm)
F13 = k*q1*q3/d13^2
= 9*10^9*4.98*10^-9*5.99*10^-9/(0.0402^2 + 0.0298^2)
= 1.07*10^-4 N
F23 = k*q2*q3/d13^2
= 9*10^9*2.04*10^-9*5.99*10^-9/(0.0298^2)
= 1.24*10^-4 N
a) Fnetx = F13x + F23x
= F13*cos(theta) + 0
= 1.07*10^-4*4.02/sqrt(4.02^2 + 2.98^2)
= 8.6*10^-5 N <<<<<<<-------------------------Answer
b) Fnety = F13y + F23y
= F23*sin(theta) + F23y
= 1.07*10^-4*2.98/sqrt(4.02^2 + 2.98^2) - 1.24*10^-4
= -6.03*10^-5 N <<<<<<<-------------------------Answer
c) |Fnet| = sqrt(Fnetx^2 + Fnety^2)
= sqrt(8.6^2 + 6.03^2)*10^-5
= 1.05*10^-4 N <<<<<<<-------------------------Answer
d) theta = tan^-1(Fnety/Fnetx)
= tan^-1(-6.03/8.6)
= 35 degrees below +x axis <<<<<<<-------------------------Answer
or 325 degrees with +x axis in counter clockwise direction.