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Homework Help: Electric force on between 2 rods

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Identical thin rods of length (2a) carry equal charges +Q uniformly distributed along their lengths.

    The rods lie on (along) the x axis with their *centers* separated by a distance b > 2a.

    (Left rod from x=-a to x=a, right rod from x=b-a to x=b+a)

    Show that the magnitude of the force exerted by the left rod on the right one is
    [tex]F = k_e\frac{Q^2}{4a^2} \ln \frac{b^2}{b^2-4a^2}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I think i should find the electric field at a point and then integrate it over the length of the rod to find the force But i even can't find the electric field . I see this question in the archieve but i just don't understand .
    I know that we first find the E by integrating -a to a .After that we find the force by integrating from b-a to b+a right ? But the answer is integrating both from -a to a . Why not from b-a to b+a for the second times integration ?
    Can someone help me to solve the questions?
    Last edited: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2


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    Staff Emeritus
    Science Advisor

    Your approach is correct.

    Here is a similar problem - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c1

    The integration and form of integrand depend on the coordinate system.

    If the one used the same coordinate system with x=0 at the center of the left rod, then one would integrate the right rod from b-a to b+a, but then one has to be careful about using x which is the position rather than distance.

    By treating the second rod with another independent variable e.g. y it is less confusing.

    If y is the position with respect to the center of the second (right) rod, then one integrates the second rod from -a < y < +a, just as one would integrate the first (left) rod from -a < x < a, and the distance between any two points is (b+y-x).

    Also, the charge density is [itex]\lambda\,=\,\frac{Q}{2a}[/itex], and one can see that the force is proportional to [itex]\lambda^2[/itex].
  4. Oct 28, 2007 #3

    Doc Al

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    Staff: Mentor

    Why don't you first find the field from the first rod as a function of distance from its center?

    As far as the second integration goes, its range depends on how you define the variable of integration. If you use distance (x) from the center of the first rod, then the integration will be from x = b -a to x = b + a. But you can also define the distance with respect to the center of the second rod (x'): x = b + x'. In which case the integration will be from x' = -a to x' = +a.
  5. Oct 28, 2007 #4
    I don't really understand what you mean . Can you give the step by step integration so that i can understand more or give some picture to visualise what you mean?
  6. Oct 29, 2007 #5
    someone please help me !!!
  7. Oct 29, 2007 #6

    Doc Al

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    Staff: Mentor

    Please do as I suggest in post #3: Find the field from the first rod as a function of distance from its center.
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