# Homework Help: Force between two lines of charge?

1. Sep 14, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"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?"

2. Relevant equations
$\int E⋅dA=\frac{q_{enc}}{ε_0}$

3. 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?

2. Sep 14, 2017

### Orodruin

Staff Emeritus
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?

3. Sep 14, 2017

### Eclair_de_XII

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

4. Sep 14, 2017

### Orodruin

Staff Emeritus
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?

5. Sep 14, 2017

### Eclair_de_XII

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

So $r=x$?

6. Sep 14, 2017

### Orodruin

Staff Emeritus
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.

7. Sep 14, 2017

### 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?

8. Sep 14, 2017

### Orodruin

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

9. Sep 14, 2017

### 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.

10. Sep 14, 2017

### Orodruin

Staff Emeritus
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.