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Electric field due to a charged rod

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    https://gyazo.com/d502fb408d6224ffa70700cf047bad20 (link to problem: #4).
    upload_2016-9-22_22-4-21.png
    [moderator edit: Image inserted for clarity]
    A total charge Q is distributed uniformly over rod length L. The rod is aligned on the x-axis, with one end at the origin and the other at point x=L.
    a) calculate electric field at a point (0,D)
    b) field at point (L/2,D)
    2. Relevant equations
    E=Kq/r^2
    q = lambda*length

    3. The attempt at a solution
    I was able to find the answer to letter a. However, I deduce that for b), the answer will only be the y-component of that found in a) as the x one cancel out. However, this is not the case as the answer key shows.

    Thus, I tried doing the calculations and I still arrive at the y-component found in a).

    Thanks
     
    Last edited by a moderator: Sep 22, 2016
  2. jcsd
  3. Sep 22, 2016 #2

    gneill

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    Hi Jabbar, Welcome to Physics Forums!

    Please show the details of your calculations.
     
  4. Sep 23, 2016 #3

    andrevdh

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    Part b is very much like part a if you consider sliding the rod halfway to the left
     
  5. Sep 23, 2016 #4

    SammyS

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    How do your results for the x component compare in parts (a) and (b) ?
     
  6. Sep 24, 2016 #5
     
  7. Sep 24, 2016 #6

    gneill

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    Can you explain what you used to substitute for sin(θ)? I'd expect to see a 'D' in the numerator as a result of the substitution.

    Since the x-components cancel by symmetry, if you slide the rod to the left so that its center point is at the origin and half the rod is to the left of the origin and half to the right (this was suggested previously by @andrevdh) then you can take further advantage of symmetry and integrate from x = 0 to x = L/2. Just multiply the result by two.

    Does the answer key solution keep the charge density as λ or does it insert Q/L for it?
     
  8. Sep 24, 2016 #7
    This is my theta.

    For the second part, yeah I see that if I do that I can integrate quite easily from either -L/2 to L/2 or "x = 0 to x = L/2. Just multiply the result by two". Thanks for the insight.

    I just don't know why what I'm doing is wrong. Moreover, why can't I say that its electric field will be equal to the y-component found in a). It feels logical to me. I'd really like to understand the concept :( .

    Thanks
     

    Attached Files:

  9. Sep 24, 2016 #8
    This is clearer image to compared to what I attached previously
    @gneill
     

    Attached Files:

  10. Sep 24, 2016 #9

    gneill

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    Consider that the field strength due to some dq on the rod varies inversely with the square of the distance between the point of interest and the dq. Then consider the average distance of the dq's in each case. Which case should have a larger sum of contributions?
     
  11. Sep 24, 2016 #10
    It's going to stay the same no? As the distance the two points have to dq are the same.
     
  12. Sep 24, 2016 #11

    gneill

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    Consider the two cases:
    upload_2016-9-24_17-57-12.png
    The average line distances up to the midpoint in the first case are the same as those on both sides of the point of interest in the second case. After the midpoint in the first case every distance is longer than any distance in the second case.
     
  13. Sep 24, 2016 #12
    Oh alright, so the Electric field at that point in a) will be smaller than in b) since for dqs' on the rod... the first case will have dqs' with a greater distance r. right?
     
  14. Sep 24, 2016 #13

    gneill

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    Right. More of the charge is further away.
     
  15. Sep 24, 2016 #14
    Thanks you very much!!!! and just for my calculation mistake. Is it just that it is better to have the point of interest along one axis to do a problem. Because won't changing the axis affect my results? I can't possible change my axis mid problem no?

    Thanks
     
  16. Sep 24, 2016 #15

    gneill

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    In this problem they're asking for the electric field which is a vector quantity: As long as you don't rotate the axes it won't change the result.
     
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