# Net electric field-whats wromg with this?

1. Jan 23, 2010

### nothingatall

1. The problem statement, all variables and given/known data
The figure shows two charged particles on an x axis: -q = -8.00 × 10-19 C at x = -1.50 m and q = 8.00 × 10-19 C at x = 1.50 m. What are the (a)x- and (b)y- components of the net electric field produced at point P at y = 5.00 m?

2. Relevant equations
E=kqcos(theta)/r^2

3. The attempt at a solution

I went through the tutorial to solve the problem and i got:
What is the magnitude of the x component of the field set up by particle 1 (or particle 2)?

Number 7.58e-11 Units N/C or V/m

-- that was correct but the question is what are the net components of the net field and i put in that number and its wrong. What step am i missing?
thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 23, 2010

### tiny-tim

Hi nothingatall!

(have a theta: θ and try using the X2 tag just above the Reply box )
Perhpas I'm misunderstanding what you did, but don't you have to double it for two charges?

3. Jan 23, 2010

### nothingatall

Well i got 7.58e-11 using E=kqcos(theta)/r^2; r=5.22 m distance bewteen particle 1 and point P and the field set up by particle 1 and point P was 2.64e-10. Then the tutorial mentions adding individual vectors to get the net field but should the 7.58e-11 value be the one I double?

4. Jan 24, 2010

### tiny-tim

(just got up :zzz: …)
Yes, you add the vectors to get the total field.

Alternatively, you can add the x components to get the x component of the total field (and the same for the y components).

If the x components of the two fields are the same, of course you can just double one of them.

If one's minus the other, then of course they add to zero.

Which is it in this case for the x components? And for the y components?

5. Jan 24, 2010

### nothingatall

For the y-comp the answer was 0 because they cancel out. However I just doubled 7.58e-11 and i got 1.52e-10 and its wrong... Oh wait magnitude means do the Sqrt(7.58e-11^2+0^2) right?

Never mind its the same:(

6. Jan 24, 2010

### tiny-tim

Hi nothingatall!

Yup! (I was just testing, of course )
Yes, 5.22 m is correct.

What values did you use for k and for cosθ ?

7. Jan 24, 2010

### nothingatall

I used 8.99e9 for k and my theta= 73.3.

8. Jan 24, 2010

### nothingatall

Ah I just add a negative to my 1.52e-10!