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Homework Help: Finding Electric Field on rod or ring?

  1. Dec 13, 2005 #1
    Its from an example in the book, and it doesnt seem to make sense,

    A rod of length l has a uniform positive charge per unit length (lambda) and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end.

    So then the example takes a small part of the rod, dx which has charge of dq, and is distance x from point P.

    dq = (lambda)dx and dE = ke dq/x^2 or ke lambda dx / x^2

    fine so far.

    Now the example says we must sum up the contributions of all the segments.

    it becomes an integral E = (integral) from a to l+a of ke lambda dx/x^2

    the example breaks the dq component out into ke lambda [ - 1/x ] from a to l+a Im somewhat confused now.

    then it goes on, ke lambda(1/a - 1/l+a) = keQ/a(l+a) ??? what??!!

    okay it kind of makes sense, the total charge divided by length, but the last part there is a divide by zero error to my thought process, multiplying an item with example values: Lambda(1/a - 1/b) or Q/l(1/a - 1/b) might give (Q/l * 1/a) - (Q/l * 1/b) if doing the same thing for the actual values, should give Q/la - Q/la + l^2 ?? no?

    reducing it down to Q/a(a+l) ? the book doesnt explain how it arrived at this.

    Can anyone give an example of calculating the Field from a charged rod? apparently its the same in a ring from the x axis but using vectors, but this concept seems tough, thanks for any explanations
     
  2. jcsd
  3. Dec 13, 2005 #2

    Doc Al

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    Staff: Mentor

    I assume you understand and agree that that's the total field at the point in question.

    What's the anti-derivative of [itex]1/x^2[/itex]? That's where the [itex]1/x[/itex] comes from.

    [tex]k\lambda (\frac{1}{a} - \frac{1}{(l + a)}) = k\lambda (\frac{l+a}{a(l+a)} - \frac{a}{a(l+a)}) = k\lambda l \frac{1}{a(l+a)} = \frac{k Q}{a(l+a)}[/tex]
     
    Last edited: Dec 13, 2005
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