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Continuous Charge

  1. Mar 4, 2004 #1
    A rod with a continuous charge extending from x=2.00m to x=3.00m has a charge density lambda=3x^2.

    a.)What is the charge on the rod?
    b.)What is the Electric potential of the rod at the origin?
    c.)What is the Electric field at the origin.

    This was a test question that I totally messed up....

    How do I find the charge of the rod? Everything I did wrong stemmed off that first part of the question.

    For part a I totally messed up, I'm sure it was extremely easy, but I ended up integrating from 2 to 3 dq/r^2 only to realize with no time left that I'm finding the E field of rod, not the charge...[b(] Thinking about it now, should I have treated it as a guassian surface and found the charge by q_in / epsilon knot ?

    When finding the electric potential I used V = Kq/r because I figured since it was the origin it was defined as V=0 and R_a = infinity.

    When finding the E field at the origin I again treated the origin as a point and used E = kq/r^2 but then after the test realized I probably should have used E=int Kdq/r^2 and integrated from 0 to 3..
    Should I of?

    Any help is greatly appreciated
  2. jcsd
  3. Mar 4, 2004 #2
    Be careful with units. You really need to mention what units the charge density is given in. I'll assume C/m.

    a)[tex]Q = \int_a^b \lambda\,dx = \int_2^3 3x^2\,dx[/tex]

    b)[tex]V = \frac{1}{4\pi\epsilon_0}\int_a^b \frac{\lambda}{r}\,dx = \frac{1}{4\pi\epsilon_0}\int_2^3 \frac{3x^2}{x}\,dx[/tex]

    you really should have stated the y-location of the rod. I took it to be 0.

    c)[tex]E = \frac{1}{4\pi\epsilon_0}\int_a^b \frac{\lambda}{r^2}\,dx = \frac{1}{4\pi\epsilon_0}\int_2^3 \frac{3x^2}{x^2}\,dx[/tex]
    obviously in the -x direction.

  4. Mar 4, 2004 #3

    Doc Al

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    Staff: Mentor

    By integrating the charge density (λ= 3x^2) over the interval x = (2,3).
    Guass's law won't help.
    I'm not sure what you're talking about. V = kq/r is the standard form for the potential at distance "r" from the charge "q". (Yes, that defines the potential as zero at r = infinity.) So if you integrated kq/x over the interval x = (2,3), where q = λdx, you got it.
    You should have integrated kq/x^2, where q = λdx, but over the interval x = (2,3), not (0,3). There is no charge in the interval x = (0,2), so what would you be integrating?

    Edit: Hey, cookiemonster beat me to it!
    Last edited: Mar 4, 2004
  5. Mar 4, 2004 #4
    I know I always just wanted to see the math after a test, so I kind of forewent the explanations. Guess now he's got both!

  6. Mar 4, 2004 #5

    Doc Al

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    Staff: Mentor

    And I was too lazy to write the equations properly using Latex like you did. We make a good tag team.
  7. Mar 4, 2004 #6
    Hmmm, well it looks like I got it wrong, which isn't good on a three question test.

    Yeah I understand what was done, but got confused when it said the origin was involved.

    Thanks for all the help.:smile:
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