Two lines of charge, net electric field

[jendead]

Homework Statement

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\muC/m for line 1 and -2.0\muC/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
(sorry, scanner doesn't like big books)

Homework Equations

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

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)

 

[Doc Al]

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?

 

[jendead]

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)}

 

[Doc Al]

Good! Keep going.

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

 

[jendead]

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.

 

[Doc Al]

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?

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.

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

To get that answer, solve the problem symbolically. Don't plug in numbers for L, \lambda_1, and \lambda_2.

 

[jendead]

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. :)

 

[Doc Al]

The answer should have been 8cm, correct?

Yes. Good work.

Is it bad form to post more than one question in a day?

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.)