Gauss's Law: Electric Field from Long Cylinder w/ Charge Density

In summary, the Gaussian surface of charge will have a total flux through it of EA, where A is the surface area of the Gaussian surface multiplied by the charge density.
  • #1
winterwind
29
0

Homework Statement



Use Gauss’s law to determine a formula for the electric field outside (and far from the ends) of a long, uniformly charged cylinder of radius R and surface charge density σ C/m2.


Homework Equations



I suppose net electric flux = ErA = Q/e0


The Attempt at a Solution



I solved for E, and then plugged in 4pi^2R^3h for A (surface area of cylinder). Just a blind stab in the dark.
 
Physics news on Phys.org
  • #2
You need to set up a Gaussian surface of radius r around the cylinder and then use Gauss' law. What will the total flux through the Gaussian surface be? What is the total charge inside the surface?
 
  • #3
Kurdt said:
You need to set up a Gaussian surface of radius r around the cylinder and then use Gauss' law. What will the total flux through the Gaussian surface be? What is the total charge inside the surface?

Could the Gaussian surface be a cylinder with a greater radius than the original cylinder?

Would the total flux be EA, where A is 4pir^2?

Total charge inside the surface...not sure. =(
 
  • #4
Yes, the gaussian surfeace would be a cylinder that is bigger than the cylinder of charge. The surface area of a cylinder is not [itex]4\pir^2[/itex], you're thinking of a sphere.

For the toatal charge, you know the charge per unit area so what is the surface area of a cylinder?
 
  • #5
Wouldn't the flux just go through the two ends, and not through the side of the cylinder for a Gaussian surface? So it would be 2piR^2 + 2piR^2?

Charge per unit area is σ C/m2. So (surface area)(σ)??
 
  • #6
You can ignore the ends that's why the question mentions the point is far away from them. To work out the surface area of the cylinder without the ends imagine unfolding it into a rectangle the same length of the cylinder and the width of the circumference of the cylinder.

You are correct for the total charge. It is simply the surface area multiplied by the charge density. Remember we're ignoring the ends.
 
  • #7
I just σ/E0.

Is this correct?
 
  • #8
Not quite. You were at the point where you had the E-field multiplied by the area of the Gaussian cylinder was equal to the total charge (i.e. the charge density multiplied by the surface area of the cylinder of charge) divided by epsilon nought. Remember the radius of each of those cylinders is different.
 

Related to Gauss's Law: Electric Field from Long Cylinder w/ Charge Density

What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

How is Gauss's Law applied to a long cylinder with charge density?

In the case of a long cylinder with uniform charge density, Gauss's Law states that the electric field at any point outside the cylinder is directly proportional to the charge density and inversely proportional to the distance from the center of the cylinder.

What is the formula for calculating the electric field from a long cylinder with charge density using Gauss's Law?

The formula is E = λ/2πε0r, where E is the electric field, λ is the charge density, ε0 is the permittivity of free space, and r is the distance from the center of the cylinder.

Can Gauss's Law be used to calculate the electric field inside the cylinder?

No, Gauss's Law can only be used to calculate the electric field outside the cylinder. This is because the charge enclosed by any surface inside the cylinder is equal to zero, resulting in an electric field of zero according to Gauss's Law.

What is the significance of Gauss's Law in electromagnetism?

Gauss's Law is one of the four Maxwell's equations, which form the basis of classical electromagnetism. It is a powerful tool for understanding and analyzing the behavior of electric fields and charges, and has many applications in physics, engineering, and technology.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
771
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
894
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
231
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
870
  • Introductory Physics Homework Help
Replies
4
Views
599
Back
Top