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Electric field and force

  1. Mar 28, 2008 #1
    [SOLVED] Electric field and force ...

    1. The problem statement, all variables and given/known data
    We have 2 identical thin rods , each has a length of 2a and carry a charge of +Q , uniformly distributed along their lengths , and both lie on the horizontal X-axis , and the distance between their centers is "b" ...
    I need to calculate the electric field and the magnitude of the force exerted by the left rod on the right one.

    2. Relevant equations
    I know that F=Q*E
    and the electric field done by a charged rod on a point in its axis is E= K*Q/d(d+L) where L is the length of the rod , and d is the distance between the rod and the point.

    3. The attempt at a solution
    Well I didn't know how to start :S , no need for a full solution , I just need a push and tell me how to start to solve it .. thanks.
  2. jcsd
  3. Mar 28, 2008 #2

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    You have to integrate along the length of the right rod.

    If [itex]\lambda[/itex] is the linear charge density of the right rod, then find the force due to left rod on an element of length dx on the right rod which is at a distance of x from the left rod. Now integrate, putting the proper limits of x, that is, the values of x on the left and right extremities of the rod on the right respectively.
  4. Mar 28, 2008 #3
    can someone explain a litle bit more .. didn't really understand ... my english isnt that good
  5. Mar 28, 2008 #4

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    So you don’t need a push, but a slightly bigger shove. Here goes!

    Draw a diagram first of the two rods. Take a small length dx on the right rod, at a distance x from the right tip of the left rod. This dx length has got a charge of dq = [itex]

    What is the field at this point due to the charge of the left rod? This is at a distance of x from the right end of the left rod. Apply the formula you have written. (I have not checked it, but looks all right.) The field should be then KQ/[x(x+2a)].

    So the force on this dq on the right rod should be given by:

    dF = Field*Charge = KQdq/[x(x+a)] = K[itex]\lambda[/itex]dx/[x(x+a)] => the total force F should be given by:

    [tex]F = \int^{b}_{b-2a}\frac{KQ\lambda dx}{x(x+2a)}[/tex].
    Last edited: Mar 28, 2008
  6. Mar 28, 2008 #5
    thanks mate , thats very helpful ... :)
  7. Mar 29, 2008 #6
    I continued the integration .. and I got as final answer : F = (KQ^2/4a^2)ln(b^2/b^2 - 4a^2)
    I think it is right , can you confirm this ?
    thanks alot mate , I really appreciate your help ^_^ , you are the best.
  8. Mar 29, 2008 #7

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    Bracket is missing: (KQ^2/4a^2)ln(b^2/(b^2 - 4a^2)).
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