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Two infinite rods with charges

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Two infinite parallel rods are separated a distance 2d. One carries a uniformly distributed positive charge (lambda), the other an opposite charge (-lambda) the rods are parallel to the axis and intersect the x-y plane at x=0 y = +/- d. Find the electric field for a point on the positive x axis. How does E behavor for x>>d?

    2. Relevant equations
    E = 1/(2pi*epsilion0) * (lambda) / r

    3. The attempt at a solution

    Ok, so my solution is that the r = (x^2 + d^2)^1/2 using Pythagoreans theorem from any point down the x axis to the two rods.

    I'm also guessing that the two electric fields don't cancel each other and that actually they add to one another since the difference between the two rods is 2 * lambda so if I add lambda to both rods my negative lambda goes to zero and my positive 1 lambda goes to two..

    So my answer for part one was

    2[1/(2pi * epsilion0) * (lambda)/(x^2 + d^2)^1/2]

    and then for x >> d I just took out the d term

    2[1/(2pi * epsilion0) * (lambda)/x]
  2. jcsd
  3. Feb 12, 2009 #2


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    Doesn't look quite right. You should have a nice diagram showing the circular cross sections for the two wires, one a distance 2d below the other. From the middle of these, off to the right a distance x, you have your point P where you will find the E field. At P, there should be an E arrow going away from the positive wire and another E arrow going toward the negative wire. You need to add these two E vectors together. There is some symmetry that will simplify things but you will definitely have a sine or cosine in your final answer (which can be expressed in terms of d and x).
  4. Feb 12, 2009 #3
    I believe after you draw the picture you will realize that the net E will point straight in the -y direction just using symmetry, if I understand the problem correctly. So I guess the sine or cosine Delphi51 was talking about will just be 1.
  5. Feb 12, 2009 #4


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    No, it won't be 1. The E's are at an angle from the y-axis that varies with x and d.
  6. Feb 12, 2009 #5
    Yes, they do vary with x and d but after an infinite number of E vectors are drawn from an infinite number of dq's on the rod, the net E vector should point in the -y hat direction.
  7. Feb 12, 2009 #6


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    I agree that the total points in the -y direction. But there are only two E vectors, one from each line charge. That formula already includes all the infinite number of dq's on the line.

    So, you only need to find the y component for each of the two E's.
    However, these E's do not point in the y direction so you must do some trig to find the y component in each case.
  8. Feb 12, 2009 #7
    Ok, I understand what you are saying now and it is correct.
  9. Feb 12, 2009 #8
    Could someone help me out with this? I can't, for the life of me, figure out where to even begin. I've got a great professor, and I understand everything he says in class, but when it comes to actually solving homework problems, I don't know what to do. Any help would be greatly, greatly appreciated.

    EDIT: Like the OP, I'm thinking that the fields don't cancel, but add. Also, I don't understand why the field is in the y-direction... I have the same equation, but I can't really think of how to describe 'r' because there are two rods producing fields...
    Last edited: Feb 12, 2009
  10. Feb 12, 2009 #9


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    Welcome to PF.

    You are right the y-components of the sum of the e-fields do add. But the x components cancel. Since the y component basically vanishes at x>>d (due to the fact that you are dealing not only with the diminishing y component proportional to sinθ, which is d/x, but also the 1/R relationship).
  11. Feb 12, 2009 #10
    I don't understand how the x-components cancel... I could see that being the case if the rods had equal charges, but opposite? Wouldn't that make the field go doubly in the direction of negative charge?
  12. Feb 13, 2009 #11


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    If they were the same charge then the y-components would cancel.

    But they are opposite. When you are on the x axis, one points toward the other away. Hence they cancel as they are equidistant.

    (Note: the arrangement is ± d on the y-axis.)
  13. Feb 13, 2009 #12
    Oh, I think I get that, then. So, if the fields add in the y-direction, how to I denote that?
  14. Feb 13, 2009 #13


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    From any point x on x, the y-component will be sinθ*|E|.

    Sinθ = d/x
  15. Feb 13, 2009 #14
    I don't mean to sound like an idiot, but that's for /both/ charges? How does one come to that conclusion?
  16. Feb 13, 2009 #15
    I'm still trying... this is for one pole.

    E = K*Lambda*Integrate[r-hat/r^2,{r,-Infinity,Infinity}]
    E = K*Lambda*Integrate[<0,y,0>/(y*r^2),{r,-Infinity,Infinity}]
    E = K*Lambda*Integrate[<y>/(((0^2+y^2+0^2)^(1/2))*r^2),{r,-Infinity,Infinity}]
    E = K*Lambda*Integrate[<y>/(y*r^2),{r,-Infinity,Infinity}]

    r = (x^2+d^2)^(1/2)

    E = K*Lambda*Integrate[<y>/(x^2+d^2),{d,-Infinity,Infinity}]

    This comes out to (K*Lambda*Pi/x^2)<y>. If I subtract the other E-field (subtract, as positives mut subtract to cancel, and I know these two add, so...), I get...

    (K*Lambda*Pi/x^2)<y> - (K*-Lambda*Pi/x^2)<y> =2K*Lambda*Pi/x^2

    Is that right? If not, where have I gone wrong, and by what terrible degree?

    EDIT: Can't be right. Y in the denominator up and flew away, it would seem.
  17. Feb 13, 2009 #16


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    Look at the problem as 2 charges, 1 + and the other -.

    Now examine a point on the ⊥ bisector and write an expression for the field contribution of both charges at once.

    E = ∑ E = ∑ i + ∑ j

    From one charge you have (+i,+j) for the other you have (-i,-[-]j) = (-i,+j)

    When you sum the pair wise contribution at that point you should recognize then that so long as you are on the ⊥ bisector, the components +,- along the direction of the bisector cancel. That means you can represent the field at that point as being

    2*Sinθ*|E| j = 2*d/r*|k*q/r| j = 2*d*k*q/r² j

    where Sinθ can be represented far away as d/r (small angle). (i.e. d << r )

    Otherwise it is d/(d² + r²)½
    Last edited: Feb 13, 2009
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