Electric field from a semi-infinite line charge along x-axis

[twisted079]

90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10%

Homework Statement

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 k_e as necessary.)

Homework Equations

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

The Attempt at a Solution

I used the prior equations to set up: dE = (k_eQ / 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, doesn't "L" (or x?) change, meaning I can't pull out the L? I understand k_e 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? I am kind of confused, any help would be GREATLY appreciated!

 

[SammyS]

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.

 

[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?

 

[twisted079]

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

(k_eQ) / L ₀∫^infinity dx/x²

---> 1/x² dx becomes -(1/x) which puzzles me because this leads to no solution except k_eQ/L (L or x) ... Please help!

 

[SammyS]

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.

 

[twisted079]

Here is what I have come up with so far:

dq = λdx
dE = k_eλdx / x²
E = k_eλ ₀∫^∞ dx/x²
E = k_eλ(-1/x) from 0 to ∞

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

 

[SammyS]

The integral should go from x₀ to ∞ .

 

[twisted079]

So I would have

E = keλ(-1/x₀ - 0) --> -(k_eλ / x₀) ?

 

[SammyS]

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.

 

[twisted079]

Ah okay, and the direction would be in the negative x-axis. Thank you so much for your help!

 

[SammyS]

Ah okay, and the direction would be in the negative x-axis. Thank you so much for your help!

Yes. Your previous answer is almost correct.