# Two lines of charge, net electric field

1. Oct 4, 2007

1. The problem statement, all variables and given/known data
Short sections of two very long parallel lines of charge are shown, fixed in place, separated by L = 8.0cm. The uniform linear charge densities are $$+6.0\mu$$C/m for line 1 and $$-2.0\mu$$C/m for line 2. Where along the x axis shown is the net electric field from the two lines zero?

The known data is:
$$\lambda_{1} = 6 \times 10^{-6} C$$
$$\lambda_{2} = -2 \times 10^{-6} C$$
$$L = 0.08m$$

http://www.clan-dm.net/members/jen/netfield.jpg [Broken]
(sorry, scanner doesn't like big books)

2. Relevant equations
line of infinite charge: $$\frac{\lambda}{2\pi \epsilon_{0}r}$$
permittivity constant: $$\epsilon_{0} = 8.85*10^{-12}$$

3. The attempt at a solution
I didn't get very far with this one. From what I can tell, I need to sum the electric fields, and figure out when it's zero.

I started out like this:
0 = E1 + E2
E1 = -E2

Obviously, at this point substituting E for the line of infinite charge equation proved fruitless. I don't know if I'm overcomplicating, undercomplicating, or just plain clueless. Any help is appreciated. :)

Also, the given answer makes no sense to me:
$$x = \frac{\lambda_{1} - \lambda_{2}}{\lambda_{1} + \lambda_{2}}\left( \frac{L}{2} \right)$$

Last edited by a moderator: May 3, 2017
2. Oct 4, 2007

### Staff: Mentor

Call the coordinate of the zero-field point x. How would you write the distance to each line charge (in terms of x and L) so that you could use the infinite line charge equation?

3. Oct 4, 2007

I'm assuming that the 0 point is somewhere in the positive x region (because the first line has a larger charge - please let me know if my thinking is off).

With that assumption, line 1 would be L + x away from the point, and line 2 would be L/2 + x away?

Is this on the right track?

$$\frac{\lambda}{2\pi\epsilon_{0}(L + x)} = -\frac{\lambda}{2\pi\epsilon_{0}(L/2 + x)}$$

4. Oct 4, 2007

### Staff: Mentor

Good! Keep going.

Edit: Oops, looks like your equation is a bit off. See comment in next post.

Last edited: Oct 4, 2007
5. Oct 4, 2007

Ok, I algebra'd it out and got x = -3L/4.. which should give me x = -6cm. I guess that means my original assumption of 0 occurring in the positive side was incorrect?

The given answer is very confusing - why would it be in that form? I never actually came across it while finding x.

6. Oct 4, 2007

### Staff: Mentor

I think there's an error in your distances in your equation. Assuming you measure x from the origin, then the distance to line 1 will be x + L/2 and the distance to line 2 will be x - L/2.
To get that answer, solve the problem symbolically. Don't plug in numbers for L, $\lambda_1$, and $\lambda_2$.

7. Oct 4, 2007

Phew, got it. I had measured my distances in a weird way, but I fixed it now. :)

The answer should have been 8cm, correct?

Thanks so much! I have one more question about the etiquette on here. Is it bad form to post more than one question in a day? There's one other problem I'm banging my head against, but will hopefully figure out on my own.. I'm asking just in case. :)

Last edited: Oct 4, 2007
8. Oct 4, 2007

### Staff: Mentor

Yes. Good work.
Of course not! Post as many as you want. As long as you're showing your work, why not? (Better to post them in separate threads, of course.)