# Homework Help: 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
21. Sep 4, 2007

### yevi

lol
I was doing [sqrt(194)]^3/2 and [sqrt(77)]^3/2
Thanks

22. Sep 4, 2007

### rootX

and I was using (1,2,3) for (9,8,4) and (4,5,6) for..

23. Sep 5, 2007

### yevi

Another related question:

There are 2 linear charges distributed on y and x axis, segments lengths 0<=x<=l and 0<=y<=l.
Density of the charges is not uniformed: gamma(x)=bx and gamma(y)=by.
Need to fine E(0,0,z)

I tried to do following:

Because those 2 segments lay on axis x and axis y the needed Electric field is on axis z.
And because of symmetric I can calculate the contribution of one segments and contribution of the second will be the same.

So I need to find dE and then dE_z.

dE= $$\frac{kdq}{r^2}$$

dE_z= $$\frac{kdq}{r^2}$$ sin ($$\alpha$$)

r^2 = $$\sqrt{z^2+l^2}$$

Am I doing right so far?

24. Sep 5, 2007

### learningphysics

You meant:

$$r=\sqrt{z^2+l^2}$$

Yes, what you've described will give you the field for one segment at the point z along the z-axis... then you can double it to get the field along the z-axis...

But you also need another component of the field at (0,0,z)

25. Sep 5, 2007

### yevi

What do you mean by another component?