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Homework Help: 2 solutions to the same problem but which is the correct one?

  1. Sep 23, 2004 #1
    You are given a positively charged rod of length L and of univform charge Q. TYhere is a positive test charge qo placed alongside this rod like the figure

    +++++++++++++++++++++++++++++++++<-------distance x------>Q

    find the net force on Q

    First of all there is no y or z component so F refers to the x axis only

    One solution :

    Let dq = YdL where Y is charge density
    then dF = k dq Q/(L+x)^2

    dF = kqYdL/(L+x)^2

    integrating you get

    F = kqY dL/L+x)^2 from 0 to L

    then F = kqYL/x(L+x)

    then resub for Y = Q /L

    F = KqQ/x(x+L)

    is there anything wrong with this solutioin please point out a mistake i don't seem to ppick up on.

    My prof suggested this way

    Divide the rod like so

    ++++++++++++++++<------distance x--------->Q
    <----L/2---><---------(x - L/2)------------------->
    then dq = Y dL

    dF = k dq Q / (x-L/2)^2

    dF = kYdLQ/ (x-L/2)^2

    F = kYQ (integrate from 0 to L) dL / (x-L/2)^2

    solving gives kqQ/x(2x-L) why this inconsistency of solutions i know one is wrong, but which one??
    i have a feeling that the integration for the second one is not done properly...
    Last edited: Sep 23, 2004
  2. jcsd
  3. Sep 23, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Please describe where q0 is placed. For example, if the charged rod extends from x = 0 to x = L, where is q0?
  4. Sep 23, 2004 #3
    it i sm y typing mistake qo = Q

    and Q is a distance x from the tip of the rod
  5. Sep 23, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Your notation is a bit confusing, but your answer is correct. I believe you probably solved it correctly. I'd write it like this:
    Let an element of charge on the rod be: [itex]dQ = \lambda dx[/itex], where I use "x" as the variable specifying the distance of the charge element from [itex]q_0[/itex]. (I'll call the distance of the test charge from the end of the rod [itex]x_0[/itex].)

    So [itex]dF = k \lambda q_0/x^2 dx[/itex].
    Integrate from x = [itex]x_0[/itex] to [itex]x_0 + L[/itex].
    The answer is identical to yours.

    I have no idea what your prof was telling you to do.
    Last edited: Sep 24, 2004
  6. Sep 24, 2004 #5
    The method is correct but something is wrong.
    We take an element 'dl' at a distance 'l' and not 'L'.then apply the same formulae and integrate from x to x+L.this gives u the correct result.

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