Force between two lines of charge?

[Eclair_de_XII]

Homework Statement

"Two very long uniform lines of charge are parallel and are separated by $x=\frac{3}{10}m$. Each line of charge has charge per unit length $\lambda = 5.2_{10^{-6}}\frac{C}{m}$. What magnitude of force does one line of charge exert on a $y=\frac{1}{20} m$ section of the other line of charge?"

Homework Equations

$\int E⋅dA=\frac{q_{enc}}{ε_0}$

The Attempt at a Solution

Basically, the only use I make out of Gauss' law is that the only electric field that has any effect on the $\frac{1}{20}m$ segment of charge is directly aligned with it. I don't think any other parts of the line interfere with the electric field of this segment. Anyway, this is what I did:

$q_{enc}=\lambda y$
$\int E⋅dA=\frac{q_{enc}}{ε_0}=\frac{\lambda y}{ε_0}$
$E⋅(2\pi ry)=\frac{\lambda y}{ε_0}$

And I don't feel I'd be right in trying to continue because I still have an arbitrary r that I don't think I will be able to eliminate through algebra. I'm so confused. Can anyone tell me if I should flip my Gaussian surface ninety degrees, or perhaps use some other one? I'm leaning on the former. But I'm just trying to figure out how to get the force on one segment of the line to the segment directly across from it, using flux and electric field. I'm so confused. Can anyone help me?

 

[Orodruin]

How is $r$ arbitrary? What you are doing is computing the field a distance $r$ from the line charge. At what distance is your field relevant for the problem?

 

[Eclair_de_XII]

I'm guessing $\frac{1}{2}x$?

 

[Orodruin]

I'm guessing the halfway point?

The electric field at the half-way point is going to determine the force on the other conductor?

How is the force on a charge related to the electric field?

 

[Eclair_de_XII]

An electric field is basically the force per charge, I think.

So $r=x$?

 

[Orodruin]

An electric field is basically the force per charge, I think.

So $r=x$?

Yea, the electric field at the position of the charge. In your case, this is the electric field of the other conductor at the position of the conductor you are computing the force on.

 

[Eclair_de_XII]

Oh, so the electric field at the other conductor is: $E=\frac{2}{4\pi \epsilon_0}(\frac{\lambda}{r})$?

So how would I translate that to force?

 

[Orodruin]

Oh, so the electric field at the other conductor is: $E=\frac{2}{4\pi \epsilon_0}(\frac{\lambda}{r})$?

So how would I translate that to force?

You tell me. What is the charge in the region you are interested in?

 

[Eclair_de_XII]

The charge is... $q=\lambda y$?

So... $E=\frac{2}{4\pi \epsilon_o}\frac{\lambda}{x}=\frac{F}{\lambda y}$ or $F=\frac{2}{4\pi \epsilon_0}\frac{\lambda ^2 y}{x}$. Got it. Thanks.

 

[Orodruin]

Right, actually the height of the Gaussian cylinder you chose to compute the field is irrelevant (as came out from your formula). The only place where y enters is in the computation of the charge the field is acting on.