Question

find the electric force vector (x and y component) on a charge of Q4=-1 microcoulombs located at the origin due to the presence of Q1= 2.7 microcoulombs AT (2,4m) Q2= 2 microcoloumbs at (-2m, 0) and Q3= 2.2 microcoulombs at (0, -4m) (Apologize about the confusion it was microcoulomb not micrometers)

Answer 1

magnitude of force exerted by Q1 on Q4,
F1 = k*Q1*Q4/r1^2

= 9*10^9*2.7*10^-6*10^-6/(2^2 +4^2)

= 1.21*10^-3 N

angle made by F1 with +x axis, theta1 = tan^-1(4/2) = 63.4 degrees

magnitude of force exerted by Q2 on Q4,
F2 = k*Q2*Q4/r1^2

= 9*10^9*2*10^-6*10^-6/2^2

= 4.5*10^-3 N

direction of F2: towards -x axis

magnitude of force exerted by Q3 on Q4,
F3 = k*Q3*Q4/r1^2

= 9*10^9*2.2*10^-6*10^-6/4^2

= 1.24*10^-3 N

direction of F2: towards -y axis

Fnetx = F1x + F2x + F3x

= F1*cos(63.4) - F2 + 0

= 1.21*10^-3*cos(63.4) - 4.5*10^-3 + 0

= -3.96*10^-3 N

Fnety = F1y + F2y + F3y

= F1*sin(63.4) + 0 - F3

= 1.21*10^-3*sin(63.4) + 0 - 1.24*10^-3

= -0.16*10^-3 N

Net force in vector notation,
Fnet = Fnetx i + Fnety j

= -3.93*10^-3 i - 0.16*10^-3 j N