Net electric field-whats wromg with this?

[nothingatall]

Homework Statement

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?

Homework Equations

E=kqcos(theta)/r^2

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.

 

[tiny-tim]

Hi nothingatall!

(have a theta: θ and try using the X² tag just above the Reply box )

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?
…
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?

Perhpas I'm misunderstanding what you did, but don't you have to double it for two charges?

 

[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?

 

[tiny-tim]

(just got up :zzz: …)

… Then the tutorial mentions adding individual vectors to get the net field but should the 7.58e-11 value be the one I double?

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?

 

[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:(

 

[tiny-tim]

Hi nothingatall!

(please use the X² tag just above the Reply box )

For the y-comp the answer was 0 because they cancel out.

Yup! (I was just testing, of course )

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:(

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.

Yes, 5.22 m is correct.

What values did you use for k and for cosθ ?

 

[nothingatall]

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

 

[nothingatall]

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

Thank you for your help!