# Finding Electric Field

1. Feb 11, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
The charge per unit length on a long, straight filament is -91.9 µC/m.

(a) Find the electric field 10.0 cm from the filament, where distances are measured perpendicular to the length of the filament. (Take radially inward toward the filament as the positive direction.)
MN/C

(b) Find the electric field 50.0 cm from the filament, where distances are measured perpendicular to the length of the filament.
MN/C

(c) Find the electric field 150 cm from the filament, where distances are measured perpendicular to the length of the filament.

2. Relevant equations
$\lambda_q = \frac{dq}{dl} \rightarrow dq= \lambda_q dl$

$\vec{E} = k_e \frac{q}{r^2}\hat{r}$

3. The attempt at a solution

$||\vec{E}||= \frac{\lambda_q dl}{r^2}$ How can I solve this without knowing the length of the filament? Supposedly this is the suggested way of solving this problem--and I'd to understand this method very much. Could someone help me?

EDIT: Also, in addition to solving it by this expedient, is it possible to solve this problem by employing a Gaussian surface?

Last edited: Feb 11, 2013
2. Feb 11, 2013

### BruceW

You can assume that the filament is infinitely long, unless they specify otherwise. But using
$$\int^{\infty}_{-\infty} \lambda_q dl$$
as the value of q in Coulomb's law will not work - clearly the line of charge is not a point charge.

As you say, the easiest way to solve the problem is by using Gaussian surface. The other way is to use Coulomb's law, in the limit of infinitely many point charges, all along the filament. Gaussian surface way is easier. Have you used that method before? By using symmetry, what kind of Gaussian surface would be the most useful?

3. Feb 11, 2013

### Bashyboy

Well, the equation I was advised to use is: $E=\frac{2K_e\lambda}{r}$. How did the author procure this equation?

As for the Gaussian surface, I know it is most wise to choose one that is highly symmetric; and the first one that pops into my head is a cylinder. However, I am having a difficult time figuring out why it is symmetric with respect to the filament.

4. Feb 11, 2013

### Bashyboy

It appears that the only difference between the equation I derived and the one derived by the author is that there is a factor of 2, and for some reason they have cancelled out $dl$ with $r$. Why did they do that?

5. Feb 11, 2013

### Saitama

Nope that's not the reason.

The idea of selecting a cylinder is correct. As to why its the best choice, try to sketch the field lines.

6. Feb 11, 2013

### BruceW

The easiest way is to use a cylindrical Gaussian surface. But since you are given this equation, then you can just use the equation given. I would guess that the teacher does not expect you to derive the equation, if it is given to you. Or maybe you are supposed to derive it. This depends on what the teacher asked you to do. Have you done this kind of thing in class before? If not, then it is not a bad thing to get ahead. Just keep in mind that there might be steps that are not completely intuitive, since it has not been shown to you in class.