1. Oct 15, 2011

### twisted079

1. The problem statement, all variables and given/known data
A continuous line of charge lies along the x-axis, extending from x=+x0 to positive infinity. The line carries positive charge with a uniform linear charge density λ0.
What is the magnitude of the electric field at the origin? (Use λ0, x0 and ke as necessary.)

2. Relevant equations

1) dE = (kedq) / r2
2) dq = λdx = (Q/L)dx

3. The attempt at a solution

I used the prior equations to set up: dE = (keQ / L) * dx/x2
Now time to integrate, but first a few questions. I understand "L" is the length of the entire rod? But so is x? Am I using L and x the right way or should I put everything in terms of x? Second, what are the constants that I pull out of the integral? Since its to infinity, doesnt "L" (or x?) change, meaning I cant pull out the L? I understand ke and Q are constant so I pull them out, is that all? Also I am integrating from 0 to infinity correct? Depending on what the set integral is, I understand that there is a possibility that when I integrate, infinity might be a denominator, making that part go to 0? Im kind of confused, any help would be GREATLY appreciated!

2. Oct 15, 2011

### SammyS

Staff Emeritus

x is not the length. It's just a general position along the x-axis.

You don't need L. The answer should have λ0 in it dq = λ0 dx .

You're about 50% done, at best.

Last edited: Oct 15, 2011
3. Oct 16, 2011

### twisted079

But λ = Q/L, or in this case would it be Q/x?

Edit: What about the constants that I could pull out of the integral?

4. Oct 16, 2011

### twisted079

Ok, so this is my idea of what the solution would look like:

(keQ) / L 0infinity dx/x2

5. Oct 16, 2011

### SammyS

Staff Emeritus

Maybe I wasn't clear. Wherever you have Q/L, use λ0 instead.

If you had such a line of charge, then if you cut a length, L, of that line and measured the amount of charge, Q, for that length, L, you would find the ratio of charge to length to be λ0 .

In other words, Q/L = λ0 , anywhere on the line.

6. Oct 16, 2011

### twisted079

Here is what I have come up with so far:

dq = λdx
dE = keλdx / x2
E = keλ 0 dx/x2
E = keλ(-1/x) from 0 to ∞

Now, when I plus 0 in, I run into a problem

7. Oct 16, 2011

### SammyS

Staff Emeritus

The integral should go from x0 to ∞ .

8. Oct 16, 2011

### twisted079

So I would have

E = keλ(-1/x0 - 0) --> -(keλ / x0) ?

9. Oct 16, 2011

### SammyS

Staff Emeritus

Yes, Except you're finding the magnitude. Drop the "-" sign .

I edited this post, but you must have seen it before I made the changes.

10. Oct 16, 2011