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Electric field problem

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data
    2 positive charges are placed on the why axis. one is at y=+a the other at y=-a. show that the e field where x is much smaller than a is approximately equal to (2kqx)/a^3. The 1st part of the question was to show that the e field = (2kqx/(x^2+a^2)^1.5) I did that part already so we can just use that equation.



    3. The attempt at a solution
    okay so, using the equation (2kqx/(x^2+a^2)^1.5) to do this part where x is much smaller than a means that x is approaching 0. so shouldn't the whole e field be approx = to 0 instead of like the equation they gave (2kqx)/a^3. where they only got rid of the x at the bottom and not the one at the top?
     
  2. jcsd
  3. Aug 31, 2011 #2

    ehild

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    Write the expression x2+a2 in the form a2[1+(x/a)2] If x is much smaller than a its square is even much smaller. For example, x/a = 0.01, then (x/a)2=0.0001. Ignoring it causes 0.01% error only. What you can call "small" depends on the accuracy of a. If it is given with 3 digits, the error of a is greater than the error the approximation brought in.

    ehild
     
    Last edited: Aug 31, 2011
  4. Aug 31, 2011 #3
    i don't understand how you got a[1+(x/a)^2] out of the denominator.
     
  5. Aug 31, 2011 #4

    ehild

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    My bad... I did not take the power 1.5 into account, and forgot a square. Now I edited my original post. The denominator becomes

    [a2(1+(x/a)2]1.5=a3(1+(x/a)2)1.5,
    and the relative error when ignoring x/a is 1.5 (x/a)2.

    ehild
     
    Last edited: Aug 31, 2011
  6. Aug 31, 2011 #5
    sorry but I'm still confused why that gives the answer. when you're doing the denominator, you're assuming x approaches 0 right? which gives you the denominator with only a in it. But the answer STILL has x in the numerator. shouldn't what was done in the denominator be also applied to the numerator which gives an answer of 0?
     
  7. Sep 1, 2011 #6

    ehild

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    Yes there is x in the numerator. As it stands alone (nothing added) you can not compare it to anything and ignoring with respect to. x approaches zero, but is not zero, only very small with respect to a. Assume a=1 and evaluate x/[1+x2]1.5 for x=0.1, 0.01, 0.001. You will get what the approximation means.


    ehild
     
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