Electric Field Calculation given three point charges?

[P1nkButt3rflys]

Homework Statement

Three point charges are located in an equilateral triangle with side 0.330 cm, where q1 = +8.30 µC, q2 = −8.20 µC, and q3 = +1.90 nC.

A) What is the electric field at the location of q3? Enter the x component, then the y component.

B) What is the force on q3? Enter the x component, then the y component.

DIAGRAM:
http://www.learning.physics.dal.ca/dalphysicslib/Graphics/Gtype51/threechrg1.gif

Homework Equations

|Fe|= (k*|q1|*|q2|)/r^2
|E|=(k*|Q|)/r^2

k = 8.99*10^9 (N*m^2)/C^2
Fe = Electrostatic force between a pair of charges
r = charge separation
q1, q2 = charge magnitudes
E = electric field
Q = charged object

The Attempt at a Solution

I've attempted this many ways and have been incorrect each time. I know the angles at the three points of the triangle will be 60deg. Knowing this and that the sides of the triangle are 0.330cm in length I calculated "r" -> the distance in respect to X and Y from q1 to q3 and from q2 to q3.

I got r(1-3x) = 0.165cm and r(1-3y) = 0.286cm and r(2-3x) = 0.165cm and r(2-3y) = 0.286cm.

Any help would be greatly appreciated. Even to just the first portion would help a lot.

 

[jackarms]

Careful -- to calculate the components of the force, you can't use the components of the separation. This is a pretty common mistake, I think, and not very intuitive, but you have to calculate the field vector using the full separation between the charges, and then find the components from that vector.

The difference comes from the formula for electric field. Say for finding the y component of the field, using just the y component of the separation looks like this:
$$E_{y} = k\frac{q_{1}q_{2}}{(rsin\theta)^{2}}$$
Whereas finding the component of the whole field vector looks like this:
$$E_{y} = k\frac{q_{1}q_{2}}{r^{2}}sin\theta$$

 

[P1nkButt3rflys]

Hi jackarms! Thanks for the help but I'm still very confused. I attempted another solution, again it's wrong.. I'll post my attempt and maybe someone can give me some guidance. I'm not sure if I'm on the right track or completely off in the wrong direction.

[IMAGE REMOVED]

Thank you!

 

[P1nkButt3rflys]

So not only did I use some wrong values in my previous calculation, after reading through my textbook further I approached it differently and got the correct answer for Part A :)

I'm not sure what to do for Part B, but am going to attempt it and post my work.

 

[P1nkButt3rflys]

Attempted B, to find the force on q3 and it is incorrect.

Any suggestions:

 

[jackarms]

All the calculations look right for both the field and the force, so I think one must be wrong. For the field, I'm not sure about the 7.15 x 10⁷ value - are you sure that's right?

Also, an easy way to go from field to force is to just use the definition of field: $E = \frac{F}{q}$
It's just nice to save some work.

 

[P1nkButt3rflys]

Thanks for the suggestion! I used your equation and got the same result, well, almost. The force for x was the same, the force for y was 0.134 N, so I must have had an issue with my sig figs. I entered both answers and they are correct!

Thank you for your help!