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Thin rod symbolic questions based on Image.

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of -Q is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location ‹ 0, y, 0 › due to the rod.

    16-050-Erod_setup.jpg

    Use the following as necessary: x, y, dx, A, Q. Remember that the rod has charge -Q.

    (a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?
    λ = ?

    (b) What is the amount of charge dQ on the small piece of length dx?
    dQ = ?

    2. Relevant equations

    Electric field of a uniformly charged Rod: 1/4πε0 * 2(Q/L) / r

    3. The attempt at a solution
    PartA: -ΔQ: I tried this because since i'm dividing the rods by a based amount, the charge Q, should be written by -ΔQ.
    -ΔQ/L: I input this wrong, because it's suppose to be A, since that's the lenght of the rod. I was thinking of -ΔQ/A, because since i'm dividing the charge by a certain amount, I'm also dividing the length of the rod as well. Haven't tried it yet.

    PartB: -dx/A: Well I think similarly like part a, we're dividing the rod again, should be divided by A. Was thinking of Q (dx/A) if i'm trying to find the dQ

    Just need help on these parts. From there I'll attempt the rest of the parts on my own.
     
  2. jcsd
  3. Feb 16, 2013 #2

    tms

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    The rod goes from x = -A to x = A. How long is it?
    You've got the right idea, just the wrong length.
     
  4. Feb 17, 2013 #3
    Ok thanks , er dq = λ*dx? :P. I'll figure it out eventually.
     
  5. Feb 17, 2013 #4

    tms

    User Avatar

    In the first part, the charge density is the total charge divided by the total length. The total length is not A.

    Yes, [itex]dq = \lambda \:dx[/itex].
     
  6. Feb 17, 2013 #5
    Ok got part A I believe now. It's -Q/2*A.

    Thanks for your help. I got the rest of the parts on this problem right as well.
     
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