Coulomb's Law and Electrostatic Forces Problem

[cheerspens]

Homework Statement

Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.

Homework Equations

I know that Coulomb's Law must be applied.

The Attempt at a Solution

I've drawn a force diagram with F_CB pointing to the left and F_AB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 1.42 x 10^-5 $\hat{x}$ + 8.54 $\hat{y}$. From here what do I do to get an answer?

I know the answer is supposed to be 1.38 x 10^-5 at 77.5 degrees so I need help with how to get there.

 

[tiny-tim]

hi cheerspens!

… I got 1.42 x 10^-5 $\hat{x}$ + 8.54 $\hat{y}$.

i don't

how did you get that?​

 

[cheerspens]

I solved for these vectors then added them:

F_CB=[(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][cos 180]$\widehat{x}$ + [(9x10⁹|(6x10^-9)(5x10^-9)|) / (.3²)][sin 180]$\widehat{y}$

F_AB=[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][cos 198.43]$\widehat{x}$ +[(9x10⁹|(-3x10^-9)(5x10^-9)|) / (.1²)][sin 198.43]$\widehat{y}$

 

[tiny-tim]

hi cheerspens!

(just got up :zzz: …)

where does 198.43° come from?

 

[cheerspens]

Good morning!

So I plugged the 198.43 degrees in for some reason but that was the angle at which the ball is being pushed by the other balls.

So then in the vectors I had before, I fixed the degree in the second one to 270 degrees but then I get 2.7x10^-5$\widehat{x}$ + 0 $\widehat{y}$ and that still does not match the correct answer.

 

[tiny-tim]

let's see …

it's |-9*10^-9 (6*5/.09y + 3*5/.01x)|

= |-9*10^-7 (2*5/3y + 3*5/1x)|

= 9*10^-7 √(100/9 + 225)

= 9*10^-7 √(111.11 + 225)

… hmm, that's not the answer in the book

(though the angle is correct, isn't it?)

 

[cheerspens]

The angle I got isn't even correct. According to the book it should be 77.5 degrees below the -x axis.