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Uniformly charged distributions (Electricity)

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Positive charge +Q is distributed uniformly along the +x axis from x=0 to x=a. Negative charge - Q is distributed uniformly along the +x axis from x=0 to x=-a. A positive point q lies on the positive y axis, a distance y from the origin. Find the force (mag and dir.) that the +ve and -ve charge distributions together exert on q.

    2. Relevant equations

    F= Eq

    3. The attempt at a solution

    I am so fine with the understanding and setting up the answer to this question. Initially, due to symmetry, I know that the resultant force would be just like the double of the field produced by the negative rod (or positive) in the negative x axis. So the answer should be - i direction. That's fine! ... My problem now is the final answer. Now let us consider the negative rod. I know originally that my answer is negative and what i usually did with point charges too if the direction is negative is that i take the absolute value of every single thing including q and then i add a negative sign at the end. Now by taking the negative rod to examine, I reached to the formula of KQ ( -1/(x^2+y^2)^.5) with the integral limits of -a to 0. Is this right... Is it right to take the integral from the small to the negative?? ... Now after doing the calculation I end up with a negative sign which is ( - K Q ( 1/y - 1/(y^2+a^2)^.5) .. The final answer is in NEGATIVE.... I am confused what to do next... Should I multiply it again by negative i restor in every question.. I need general rules for this and not just an answer to this..
    Questions.
    1) Should my answer of the integral be inline with my physics asumptions... I mean I know that the answer should be + i and the integral comes out in negative.. WHat should I do?
    2) Should the integral always to be from the smallest to the largest

    THANKS A LOT FOR YOUR PATIENCE
     
  2. jcsd
  3. Feb 22, 2013 #2

    haruspex

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    A couple of issues there.
    1. How does a field vary as a function of distance?
    2. As you noted, you only want the component parallel to the x axis.
     
  4. Feb 23, 2013 #3
    Can you kindly clarify your points you are trying to explain\?
     
  5. Feb 23, 2013 #4

    haruspex

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    If I've understood your post, you're saying that the force on a charge q at relative position (x, y) from a charge element δQ is KδQ(x2+y2)-1/2. That would mean force varies inversely as the distance, instead of as the square of the distance.
    Also, as you noted beforehand, you don't want the whole force. You know that the forces in the y-direction will cancel, so you want to integrate only the x-component of the force.
     
  6. Feb 23, 2013 #5
    Nope not at all... In this question, I meant that the answer comes in negative where i know from the physics of the charges that it should point to the positive x axis
     
  7. Feb 23, 2013 #6

    haruspex

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    Ok, I think I see where I had misunderstood the OP. The formula you quoted is after integrating but before applying the limits.
    ##\int_{x=-a}^0\frac{-x dx}{(x^2+y^2)^{\frac32}} = \int_{x=-a}^0\frac{- d\left(x^2\right)}{2(x^2+y^2)^{\frac32}} = \int_{x^2=a^2}^0\frac{- d\left(x^2\right)}{2(x^2+y^2)^{\frac32}}##
    The problem is that over the range x = -a to 0 d(x2) is negative. To put it another way, had your initial range been x=+a to 0 you would still have got a2 to 0 for x2. Whenever there are square roots involved there's the potential for ambiguity.

    Edit: Ignore the above. I'm back to not understanding the issue. The answer should be negative, and it is. In my algebra above, I should have started with +xdx, not -xdx.
     
    Last edited: Feb 23, 2013
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