Finding Magnitude and Direction of Force on a Charge

1. Aug 27, 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 2.7 x 10-5 $\hat{x}$ + 0 $\hat{y}$. 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 27, 2011

Staff: Mentor

You must have made a mistake, since the resultant must have a component in the y direction. Show how you computed FCB and FAB.

3. Aug 27, 2011

cheerspens

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 270]$\widehat{x}$ +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 270]$\widehat{y}$

4. Aug 27, 2011

Staff: Mentor

OK, now simplify these results by evaluating those trig functions.

5. Aug 27, 2011

cheerspens

I did and thats how I got 2.7x10-9$\hat{x}$ + 0$\hat{y}$
Was my set up right and I went wrong somewhere in the math? I was always thinking I set it up wrong or wasn't using the right values.

6. Aug 27, 2011

Staff: Mentor

Give me your values for sin(180), cos(180), sin(270), and cos(270).

7. Aug 27, 2011

cheerspens

sin(180) = 0
cos(180) = -1
sin(270) = -1
cos(270)=0

8. Aug 27, 2011

Staff: Mentor

Good. Now simplify your equations in post #3 accordingly.

9. Aug 27, 2011

Lobezno

What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y axis and one along the x axis. Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.

Last edited by a moderator: Aug 27, 2011
10. Aug 27, 2011

Staff: Mentor

While you don't need to use trig functions, there's nothing wrong with using them.
Charges A and B are oppositely charged, thus the force is attractive.

11. Aug 28, 2011

Lobezno

C and B, sorry. They'll produce an opposite charge.