# Infinite lines of charge, Gauss' Law

1. Jun 26, 2011

### jumbalaya

1. The problem statement, all variables and given/known data
[PLAIN]http://smartphysics.com/images/content/EM/03/h3_lineF.png [Broken]
A cylinder of radius a = 6.1 cm and height h = 9.7 cm is aligned with its axis along the y-axis. An infinite line of charge is placed at x=3.05cm where λ1 = -2 μC/cm and λ2 = 6 μC/cm at x=-3.05 cm. Point P is located at x = 6.1 cm.
What is the value for Ex(P), the x-component of the electric field at point P?

2. Relevant equations
$\int E\cdot dA = Q_{enc}/\epsilon_o$

3. The attempt at a solution
I have calculated the Qenc which I think should just be λ1 + λ2 = 4E-4 C/m, and the radius of my gaussian surface is just .0305 m. So I try E = [(λ1 + λ2)/ (2pi(8.85E-12)(.0305)] and that didn't work. I also tried to take each charge separately and calculate it based on their individual radius' if you were to take the gaussian surface looking at just λ1 and then λ2 but this isn't working for me either. I'm confused because I thought that you just have to calculate the Qenclosed, and both charges are enclosed in the cylinder and you just have to calculate the gaussian surface with respect to the radius that you want your point to be enclosed in. A little insight and help would be much appreciated! Thanks.

Last edited by a moderator: May 5, 2017
2. Jun 26, 2011

### Staff: Mentor

The problem in applying Gauss's law is that there is not enough symmetry to allow you to calculate the field. Instead, apply superposition. Calculate the field from each line of charge separately and add them up. (You can use Gauss's law to calculate the field from a single line of charge.)

3. Jun 26, 2011

### jumbalaya

So I tried taking a gaussian surface for the first charge with a radius of (3.05+6.01)/2 = 4.575cm. Then I tried taking a gaussian surface for the second charge with a radius of 3.05 cm. However this is still not working for me... I'm not exactly sure what I'm still doing wrong

(λ1) / 2pi(8.85E-12)(((a/2))+a)/2) = (-2E-4) / 2pi(8.85E-12)(.04575) = -78616871.65

and

(λ2) / 2pi(8.85E-12)(a/2) = (6E-4) / 2pi(8.85E-12)(.0305) = 353775922.4

-78616871.65 + 353775922.4 = 275159050.8

4. Jun 26, 2011

### Staff: Mentor

I don't understand how you are calculating this radius. What's the distance between λ1 and point P? Find the coordinates of each and subtract.
OK.

It seems that λ1 and λ2 are reversed in your diagram and your problem description. Which one is closer to point P?

5. Jun 26, 2011

### jumbalaya

Oh sorry, I was calculating the radius wrong. Thanks!