# Electric Field in a point

1. Sep 4, 2007

### yevi

I want to find an electric field in a point (x,y,z) generated by 2 charges:
q1 = q, and q2=2q, q1’s position = (1,2,3) and q2’s position = (4,5,6),

I tried to solve it by using superposition principle, adding the E of each charge to point x,y,z, I am doing something wrong probably with vectors calculation.

2. Sep 4, 2007

### G01

Where in the calculation are you getting stuck? I'm not going to be able to give you any decent help if you don't show some work.

3. Sep 4, 2007

### yevi

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1 + q/r2^2 * r2)
r1 and r2 vectors

4. Sep 4, 2007

### yevi

r1=(x-1,y-2,z-3)
r2=(x-4,y-5,z-6)

after that and calculating all constants i get stuck.

5. Sep 4, 2007

### learningphysics

The vectors should be unit vectors...

6. Sep 4, 2007

### yevi

yes unit vector,
I get: r1/|r1| but this doesn’t give me anything

7. Sep 4, 2007

### rootX

If it should be like this:

E[x] = E1*(1/sqrt(14),0,0)+E2..

?
Thanks.

8. Sep 4, 2007

9. Sep 4, 2007

### yevi

I don't know the answer, but my answer has form with a lot of x,y and z variables.

10. Sep 4, 2007

### learningphysics

There should be 2q in the numerator or r2^2.

11. Sep 4, 2007

### rootX

well,
E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1/|r1| + q/r2^2 * r2/|r2|)
r1 and r2 vectors

did you solve this one?
You would get lots of x,y,z vars, if you do

yea, changing to q2=2q should give the answer..

12. Sep 4, 2007

### yevi

Yes, 2q.

I have a follow up question, to find field in (10,10,10) where q1=10^-6 q2=2*10^-6
and the answer for that is 10^-9K(21.09x’+17.76y’+14.43z’)

My answer and the point (10,10,10) doesn’t give this result.

13. Sep 4, 2007

### yevi

rootX,
yes I get a mess of x,y and z.

14. Sep 4, 2007

### yevi

I am doing something wrong with vectors, it's shouldn't be complicated...

15. Sep 4, 2007

### learningphysics

Just leave it as the sum of the two fields... also you can factor out $$\frac{q}{({r_1^2})^{3/2}}$$ for the first part... and $$\frac{q}{({r_2^2})^{3/2}}$$ for the second part...

so all the messy stuff in the denominator can get factored out of the vectors... don't actually multiply out (x-a)^2 + (y-b)^2.... etc.

Last edited: Sep 4, 2007
16. Sep 4, 2007

### rootX

I get
kq (0.123,1.17798..)
I guess I also messed up something..

17. Sep 4, 2007

### rootX

ok, I am 55% sure that this answer is wrong: 10^-9K(21.09x’+17.76y’+14.43z’)

here's what I got from maxima:
kq(0.14227258171877 , 0.20662308551546 , 0.270973589312)

18. Sep 4, 2007

### learningphysics

I'm getting the answer that is given: 10^-9K(21.09x’+17.76y’+14.43z’)

Last edited: Sep 4, 2007
19. Sep 4, 2007

### yevi

learningphysics,

20. Sep 4, 2007

### learningphysics

Basically I got:

$$kq*[\frac{(9,8,7)}{{194}^{3/2}} + 2*\frac{(6,5,4)}{{77}^{3/2}}]$$

this evaluates to the given answer.

Last edited: Sep 4, 2007