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Help understanding E fields for line of charge and ring.

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    I am going through the example problems in the book and it shows me how to find the electric field of a line of charge, which is this

    http://imageshack.us/a/img24/8932/img0303fq.jpg [Broken]

    I get to this part where they change ΔQ to (Q/L)ΔY
    http://imageshack.us/a/img132/170/img0304lm.jpg [Broken]

    Ok so next question tells me we can treat this thin circle as a line of charge with λ=Q/2∏r
    http://imageshack.us/a/img543/9461/img0305wxd.jpg [Broken]

    Then i get to this part again where the ΔQ is involved
    http://imageshack.us/a/img27/927/img0306rk.jpg [Broken]

    So why is it that for the ring, ƩΔQ= Q, and for the line of charge, ΔQ must first be converted into (Q/L)Δy? Why didn't we plug in ΔQ= (Q/2∏r)Δy for the ring?

    my guess is that for the line of charge, y = L and y was changing for each segment of the portion of E_field we were taking. While for the ring, (Q/2∏r) is constant no matter what segment of E_field we were measuring. Can someone please explain why we substituted ΔQ for the line and not the ring?

    Thank you very much in advance


    2. Relevant equations
    λ=Q/L
    ΔQ=λL = (Q/L)Δy



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 2, 2013 #2

    tiny-tim

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    hi johnnyboy53! :smile:
    yes, that's correct …

    you can use ∆Q = Q∆θ/2∏ for the ring, but then you'll have to integrate over θ, and that will just give you Q anyway! :wink:
     
  4. Mar 2, 2013 #3
    Your intuition is right; notice how for the line charge, the sum actually depends on y (that's why the y has subscript i). And that makes sense--look at the problem, as you get farther and farther up the y-axis, the cosine of theta becomes smaller and smaller. So, even though the rod is uniformly charged, charge farther away from the x-axis makes a smaller contribution to the E-field at the point (the smaller contribution is also due to the inverse-square fall off of the E-field). And the problem hints at why you have to substitute (Q/L) dy for dQ: (Q/L) dy is descriptive of how the charge is configured in space, dQ is not.

    I actually think their use of (Q/L) is confusing! Usually, we call this quantity lambda--linear charge density. As you get into more advanced physics, you'll encounter charge distributions that are a function of your position! So the value changes based upon your y-position (or, once you get to 3D-space, it will depend on x,y,z--we call that volume charge density and use rho instead of lambda).

    For the ring, you're equidistant from all charge on the ring and all charge makes the same contribution to the E-field at the point. That's why the sum is independent of z. But if you're unhappy with that--just redo the derivation and keep everything inside the sum. I'll use integrals instead of sums, but it looks like this:

    [tex]E= \frac{1}{4\pi \epsilon_{0}} \int \frac{zdQ}{(z^2+R^2)^{3/2}}[/tex]

    which is just

    [tex]= \frac{1}{4\pi \epsilon_{0}} \int \frac{z(Q/(2 \pi R))}{(z^2+R^2)^{3/2}} dl[/tex]

    [tex]= \frac{1}{4\pi \epsilon_{0}} \frac{z(Q/(2 \pi R))}{(z^2+R^2)^{3/2}} \int dl[/tex].

    So now you might be wondering, what the hell is dl (note that Q/(2 \pi R) dl= dQ). It's the length of a little segment of the ring. Since this isn't too easy to describe in Cartesian coordinates, let's just think about what

    [tex] \int dl [/tex]

    might be. Well. It's the circumference of the ring! So that integral is just

    [tex] \int dl = 2\pi R[/tex]

    and the E-field becomes


    [tex]E = \frac{1}{4\pi \epsilon_{0}} \frac{zQ}{(z^2+R^2)^{3/2}} [/tex].


    I also recommend you maybe check out a different physics book (I'd recommend Halliday & Resnick--I think the newer editions aren't as good as the older ones...but you should be able to find the 5th or 6th edition for like $5 on Amazon). The explanations aren't great in your book, and their use of sums seems kind of silly to me. If you're up for a bit of a challenge, they just started reprinting Purcell's book. It's a great book (a lot of people think it's one of the best physics books ever written).
     
    Last edited: Mar 2, 2013
  5. Mar 4, 2013 #4
    One more question, For the line of charge equation, can we take the r in the numerator and put it on the outside of the summation/integral since it is constant?

    Thank you both for your input and also thank you rob for the great explanations. They made it much more clear. I have looked into the books you have suggested and will be supplementing it with my textbook. Thanks!
     
  6. Mar 5, 2013 #5

    tiny-tim

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    hi johnnyboy53! :wink:
    that's correct … it's a constant, and you can always take a constant outside the ∫ :smile:
     
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