# Infinite lines of charge, Gauss' Law

• jumbalaya
In summary, the problem involves a cylinder with a given radius and height that is aligned along the y-axis. An infinite line of charge is placed at specific points with given values for electric charge. The task is to find the value of the x-component of the electric field at a given point using Gauss's law. However, due to lack of symmetry, superposition must be applied by calculating the field from each line of charge separately and adding them up. The attempt at a solution involved calculating the Qenclosed for each charge, but the calculation was incorrect due to incorrect calculation of the radius. After correcting the radius calculation, the correct values for Qenclosed were obtained and the final result for the x-component of the electric field at the given
jumbalaya

## Homework Statement

[PLAIN]http://smartphysics.com/images/content/EM/03/h3_lineF.png
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?

## Homework Equations

$\int E\cdot dA = Q_{enc}/\epsilon_o$

## 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:
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.)

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

jumbalaya said:
So I tried taking a gaussian surface for the first charge with a radius of (3.05+6.01)/2 = 4.575cm.
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.
Then I tried taking a gaussian surface for the second charge with a radius of 3.05 cm.
OK.

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

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

## 1. What is Gauss' Law and how does it relate to infinite lines of charge?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. For infinite lines of charge, it can be used to calculate the electric field at any point in space.

## 2. How is an infinite line of charge different from a finite line of charge?

An infinite line of charge has an infinite length and a uniform charge distribution, whereas a finite line of charge has a defined length and a non-uniform charge distribution. This difference affects the mathematical calculations and application of Gauss' Law.

## 3. Can Gauss' Law be used to calculate the electric field at any point around an infinite line of charge?

Yes, Gauss' Law can be used to calculate the electric field at any point in space around an infinite line of charge. The law states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface, regardless of the shape or size of the surface.

## 4. What is the mathematical equation for Gauss' Law and how is it applied to infinite lines of charge?

The mathematical equation for Gauss' Law is ∫E⃗ · dA⃗ = Q/ε0, where E is the electric field, dA is the differential area, Q is the total charge enclosed, and ε0 is the permittivity of free space. For infinite lines of charge, this equation can be simplified to E = λ/2πε0r, where λ is the linear charge density and r is the distance from the line of charge.

## 5. How does the electric field vary around an infinite line of charge?

The electric field around an infinite line of charge varies inversely with the distance from the line, according to E = λ/2πε0r. This means that the electric field is strongest closest to the line, and decreases as the distance from the line increases. The electric field lines are also perpendicular to the line of charge and form concentric circles around it.

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