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Homework Help: Electromagnetism: Finding Electric Field of non Uniform Charged Rod

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Here is the question:

    Thin rod AB has length l=100 cm and total charge q=37 nC that is distributed in such a way that its line density ? is proportional to the square of the distance from the end A, i.e. ?(x) =kx^2. Determine electric field E at the end A of the rod.

    ? stands for Lambda, sorry I do not understand TeX.

    I know that I need to integrate SOMETHING. How I do that I am not sure. Here are a couple of questions:

    2. Relevant equations

    - Can I assume that at 1m: (37nC)/1m=k(1^2) to find k? (or is k just 8.97E9?)
    - Do I use the formula E=lambda/((2)(pi)(epsilson-nought)(x)) because this is a cylinder?

    3. The attempt at a solution

    This is what I have so far:

    solve for k at 1m and see that it equals 37 nC (should I just convert that to C?)

    input lambda into the equation and do an integral so you get: k(x^2)/((2)(pi)(epsilon-nought)(x)) from A (0) to B (1 m). From this I got that the integral is k(x^2)/((4)(pi)(epsilon-nought)). This is where would end the problem and put in my values.

    I really just want to understand the problem, obviously there is something I am not getting.

    Thanks in advance.

    PS I know there is this exact question (I did my research on this problem before I came here). I didn't understand the explanation there so please help!
  2. jcsd
  3. Jun 27, 2011 #2
    I did the math and I got that the electric field equals 9.99E2 (this is converting nC to C)

    what would my units be? C/m?
  4. Jun 27, 2011 #3

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    Welcome to PF, olivia.belli3!

    (Have a λ :wink:)

    Let's start here and try to find k.

    Your line density λ defines how much charge is between x and x+dx measured from point A, where dx is a very small distance.
    That is, the charge between x and x+dx is:
    [tex]\lambda(x)dx = k x^2 dx[/tex]

    To find the total charge you need to calculate the integral of these charges over the length of the rod.
  5. Jun 27, 2011 #4
    I set up my integral that q= the integral of kx^2 dx to get that q=(kx^3)/3
    from A=0 to B=1.

    this means that k=3Q or k=111 nC (Do I convert to C?)

    Thanks for your reply :)
  6. Jun 27, 2011 #5

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    Right! :smile:

    No need to convert from nC to C unless you really want to. ;)
    However, your unit is not quite right, since you have that the charge q (coulomb) equals your k times x3 divided by 3. So what should the unit of k be?

    Now for the electric field at point A.
    I'm not sure where you got your formula, but you should use the formula from Coulomb's law (http://en.wikipedia.org/wiki/Coulomb%27s_law" [Broken]).

    That is:
    [tex]dE = \frac 1 {4 \pi \epsilon_0} \frac {dQ} {r^2}[/tex]
    where dE is the small contribution of a small charge [itex]dQ=\lambda(x)dx[/itex] at a distance r.

    Can you apply this to your problem?
    And integrate it over the length of the rod?
    Last edited by a moderator: May 5, 2017
  7. Jun 27, 2011 #6
    well in that case my K= 111 nC/m^3.

    So I have Q, of course I can integrate over the length of the rod using dr from 0 to 1, but I'm stuck. Where does k come in? I'm a little lost . I'm assuming that you meant to put QQ on the top instead of dQ? In that case I would taket he charge from A (which is zero) then take the charge from B which would lead me to to the answer, but how do you do that?
  8. Jun 27, 2011 #7

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    [tex]E_A = \int_0^{AB} dE = \int_0^{AB} \frac 1 {4 \pi \epsilon_0} \frac {\lambda(x)dx} {x^2}=\ ...[/tex]
    Last edited: Jun 27, 2011
  9. Jun 27, 2011 #8
    Okay, so I integrated that and put in that lambda=kx^2 (since we know K I am assuming this is okay). I'm assuming that AB=1m?

    I got that the integral equals k(x)/((4)(pi)(epsilon-nought)) from 0 to AB which I would think would equal K/((4)(pi)(epsilon-nought)).
  10. Jun 27, 2011 #9

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  11. Jun 27, 2011 #10
    so would my final answer be:

    (111nC/m^3)(1m)/((4)(pi)(epsilon-nought)) to yield the electric field as: 9.99E10^11?
  12. Jun 27, 2011 #11
    sorry 9.98E10^11
  13. Jun 27, 2011 #12
    units would be nC/m^2
  14. Jun 27, 2011 #13

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    Yep! :smile:

    (Now would be a good time to convert nC to C. :wink:)

    Uhh, no.
    Did you take the unit of [itex]\epsilon_0[/itex] into account?

    Btw, do you know how the electric field is related to the force it would exert on a charged object?
    That might help you deduce the unit of the electric field.
    Last edited: Jun 27, 2011
  15. Jun 27, 2011 #14
    Eek sorry! forgot about epsilon nought. The fnal units is N/C!

    Thank you SO much!
    I really appreciate you!
  16. Jun 27, 2011 #15

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    Any time! :smile:
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