Net electric force of multiple charged particles in 3-d space

[Edel Crine]

Homework Statement

An xyz coordinate system contains three charged particles: particle 1, q1=-5.0 muC, at (4.0m,-2.0m,0); particle 2, q2=12 muC, at (1.0m, 2.0m, 0); and an electron at (-1.0m,0,0).
(a) Draw a diagram showing the vectors you need in order to determine the direction of the separate electric forces exerted on the electron by particles 1 and 2.
(b) Calculate the vector sum of these two forces.

Relevant Equations

f=k((q1*q2)/r^2)*unit vector r

I draw the graph like this:

For (b), I divided each force vector to e from p1 and p2 as x and y parts.

I computed them and got
Fx=-4.608*10^(-15)N
Fy=-2.52*10^(-15)N

However, I am not sure whether I did it correctly or not...
I appreciate every help from all of you!
Thank you!

 

[haruspex]

You missed out a chunk of your working, and I don't have time to check your final result right now, but I notice the directions of the arrows in your diagram. Are these intended to represent the directions in which the forces will act? Note that the two particles are oppositely charged.

 

[Edel Crine]

Just redraw the graph,

Then I got,
Fx=4.032*10^(-15)N
Fy=6.12*10^(-15)N

 

[haruspex]

Just redraw the graph,
View attachment 269790
Then I got,
Fx=4.032*10^(-15)N
Fy=6.12*10^(-15)N

I get somewhat smaller numbers. Please post your detailed working.

 

[Edel Crine]

I get somewhat smaller numbers. Please post your detailed working.

Just got new values:
Fx=1.297*10^(-15)N
Fy=1.620*10^(-15)N

 

[haruspex]

Just got new values:
Fx=1.297*10^(-15)N
Fy=1.620*10^(-15)N
View attachment 269793

Those numbers look about the right magnitude. What about direction?

 

[haruspex]

Those numbers look about the right magnitude. What about direction?

You may have missed my edit re direction.

 

[Edel Crine]

Those numbers look about right.

Your are the best teacher always! I appreciate all your help in my questions! I really hope you have a nice day!

 

[Edel Crine]

You may have missed my edit re direction.

You mean the arctan(Fy/Fx) maybe...?

 

[haruspex]

You mean the arctan(Fy/Fx) maybe...?

No, I mean the signs on Fx and Fy.

 

[Edel Crine]

I think I got
Fx= +1.297*10^(-15)N
Fy= +1.620*10^(-15)N

 

[haruspex]

I think I got
Fx= +1.297*10^(-15)N
Fy= +1.620*10^(-15)N

Good.

 

[Edel Crine]

Good.

Again, thank you sooooooooo much!