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

  1. Oct 15, 2011 #1
    90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10%

    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!
     
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  3. Oct 15, 2011 #2

    SammyS

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

    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
  4. Oct 16, 2011 #3
    Re: 90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10

    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?
     
  5. Oct 16, 2011 #4
    Re: 90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10

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

    (keQ) / L 0infinity dx/x2

    ---> 1/x2 dx becomes -(1/x) which puzzles me because this leads to no solution except keQ/L (L or x) ... Please help!!
     
  6. Oct 16, 2011 #5

    SammyS

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

    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.
     
  7. Oct 16, 2011 #6
    Re: 90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10

    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
     
  8. Oct 16, 2011 #7

    SammyS

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

    The integral should go from x0 to ∞ .
     
  9. Oct 16, 2011 #8
    Re: 90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10

    So I would have

    E = keλ(-1/x0 - 0) --> -(keλ / x0) ?
     
  10. Oct 16, 2011 #9

    SammyS

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

    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.
     
  11. Oct 16, 2011 #10
    Re: 90% done with this "evenly dstrbtd chrg on infinite line" problem. PLEASE help 10

    Ah okay, and the direction would be in the negative x-axis. Thank you so much for your help!
     
  12. Oct 16, 2011 #11

    SammyS

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

    Yes. Your previous answer is almost correct.
     
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