# Gauss' Law

1. Mar 19, 2007

### suspenc3

1. The problem statement, all variables and given/known data
A long, straight wire has a linear charge density of magnitude 3.6nC/m. The wire is to be enclosed by a thin, no-conducting cylinder of ouside radius 1.5cm, coaxil witht he wire. The cylinder is to have positive charge on its outside surface with a surface charge density $$\sigma$$ such that the net external electric field is zero. Calculate the required $$\sigma$$.

2. Relevant equations
$$E=\frac{\lambda}{2\pi \epsilon_0 r}$$

3. The attempt at a solution

The electric field can be found, but then how do I go about finding the required $$\sigma$$ value?

Last edited: Mar 19, 2007
2. Mar 19, 2007

### Gyroscope

Last edited by a moderator: Mar 19, 2007
3. Mar 19, 2007

### suspenc3

Thats what it sayss....

4. Mar 19, 2007

### Gyroscope

The wire has obviously negative charge. Hasn't it?
Apply Gauss's Law, to a gaussian surface, I recommend you to use another cylinder, and put the condition that:

$$\oint E\cdot dS=0$$

5. Mar 19, 2007

### suspenc3

Im still confused

6. Mar 19, 2007

### robb_

The total field is zero right? It is due to a superposition of the field from the wire and the field from the cylindrical shell.

7. Mar 19, 2007

### suspenc3

So Do I just find an electric field for the gaussian surface that when summed with the electric field already found wll equal zero?

8. Mar 19, 2007

### robb_

not sure what exactly you mean there

9. Mar 19, 2007

### suspenc3

Since the total electrical field is 0, and I can find $$E=\frac{\lambda}{2\pi \epsilon_0 r}$$, cant I just find $$E=\frac{\sigma}{\epsilon_0}$$

and then $$\frac{\lambda}{2\pi \epsilon_0 r} = -\frac{\sigma}{\epsilon_0}$$ and then solve for $$\sigma$$

Sorry if this make no sense, I dont really understand this stuff.