# Gauss's law problem

1. Feb 1, 2009

### winterwind

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

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.

2. Relevant equations

I suppose net electric flux = ErA = Q/e0

3. 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.

2. Feb 1, 2009

### Kurdt

Staff Emeritus
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. Feb 1, 2009

### winterwind

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. Feb 1, 2009

### Kurdt

Staff Emeritus
Yes, the gaussian surfeace would be a cylinder that is bigger than the cylinder of charge. The surface area of a cylinder is not $4\pir^2$, 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. Feb 1, 2009

### winterwind

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. Feb 1, 2009

### Kurdt

Staff Emeritus
You can ignore the ends thats 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. Feb 3, 2009

### winterwind

I just σ/E0.

Is this correct?

8. Feb 4, 2009

### Kurdt

Staff Emeritus
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.