How would you have done this Coulomb's law problem?

[1MileCrash]

Homework Statement

In the figure, particle 1 of charge 1.0 micro Coulombs and particle 2 of charge -3.0 micro Coulombs are held at a separation of L = 10.0 cm. If particle 3, of unknown charge is to be located such that the net electrostatic force on it is 0, what must its x and y coordinates be?

(figure)

1<------L------->2

with 1 at the origin.

Homework Equations

Coulomb's law

The Attempt at a Solution

First, I noted that the particle must be to the left of 1 because 2 is of stronger net charge, so it must be closer to 1 and further from 2.

Furthermore, since the sum of the two forces is 0, they must be equal in magnitude and opposite in sign.

$kq_{1}q_{3}r^{-2}_{1}\widehat{r}_{1} = - kq_{2}q_{3}r^{-2}_{2}\widehat{r}_{2}$

$q_{1}r^{-2}_{1}\widehat{r}_{1} = - q_{2}r^{-2}_{2}\widehat{r}_{2}$

I was stumped here until I cheated and looked at the y coordinate answer. It is 0. Only knowing that was I able to solve. Can someone tell me why that is the case?


Since I know the y coordinate is 0, the vectors on both is -i, allowing me to cancel them out.


$q_{1}r^{-2}_{1} = - q_{2}r^{-2}_{2}$

$(1x10^{-6})r^{-2}_{1} = (3x10^{-6})r^{-2}_{2}$

$r^{-2}_{1} = 3r^{-2}_{2}$

$r^{2}_{2} = 3r^{2}_{1}$

$r_{2} = \sqrt{3}r_{1}$

I also know that r2 is 10 greater than r1.

$r_{2} = 10 + r_{1}$

Substituting that for r2 in the first equation,

r1 = 13.66.

Since I know it's to the left of r1, and r1 is on the origin, the x coordinate must be -13.66.

The answer in the book is -14, so I'm reasonably confident.

However, I want to know:

1.) Is this how you would have done it?
2.) Why is the y component 0?
3.) Is the y component known to be 0 through working the problem, or from examining the layout of the particles?

TY

 

[Doc Al]

Hint: Can two non-parallel vectors cancel?

 

[1MileCrash]

Nope, I see. Thanks.

does the rest look ok?

 

[Doc Al]

I didn't check your arithmetic, but your method is perfectly correct.