# Coulomb's Law and Electrostatic Forces Problem

1. Aug 26, 2011

### cheerspens

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
I know that Coulomb's Law must be applied.

3. The attempt at a solution
I've drawn a force diagram with FCB pointing to the left and FAB 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.

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2. Aug 26, 2011

### tiny-tim

hi cheerspens!
i don't

how did you get that?​

3. Aug 27, 2011

### cheerspens

I solved for these vectors then added them:

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

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

4. Aug 27, 2011

### tiny-tim

hi cheerspens!

(just got up :zzz: …)

where does 198.43° come from?

5. Aug 27, 2011

### 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.

6. Aug 27, 2011

### 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?)

7. Aug 27, 2011

### cheerspens

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