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Electric quadropole

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    four point charges +q,-q,+q,-q are placed at the four ends of a horizontal square of side 'a' .no of neutral points (where the electric field vanishes) is......
    2. Relevant equations


    3. The attempt at a solution
    i think that there two dipole .. there is no null point due to dipole .but there to like charges ...due to that a neutral point exist in middle of square ...there fore i think that there exist only one neutral point...but ans provide in the key is infinity......how it will be..pls explain...this is a question from tifr 2009
     
  2. jcsd
  3. Jul 22, 2015 #2

    You have two dipoles placed in opposite direction . For any point d which is very far from these two dipoles (d >> distances between dipoles' monopoles) , the two will exert opposite electric field , equal in magnitude . Thus the net electric field at that point would be zero . This would be true for any far enough point and thus there would be ∞ null points , plus the center of course .

    *I have assumed you know what the electric field is at any general point . Also I have not considered close points for I have already answered your question . I however think that electric field at a close by point would not be zero .

    I hope this helps .
     
  4. Jul 22, 2015 #3

    BvU

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    Hello sivadas, welcome to PF :smile: !

    Is this a two-dimensional problem ? :wink:

    PS I don't agree with qwerty2 :rolleyes:
     
  5. Jul 22, 2015 #4

    My mistake ?
     
  6. Jul 22, 2015 #5

    BvU

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    Yes.

    The fact that the field falls off faster than ##1/r^2## and also faster than ##1/r^3## doesn't mean it becomes zero at bigger distances: there is no ##r<\infty## for which ##|{\bf \vec E}| < \epsilon ## for all ##\epsilon > 0##. Except the r = 0 that was found already in the OP.

    Sivadas' reasoning is correct - in the horizontal plane :wink:
     
  7. Jul 22, 2015 #6
    Have I said any of that ?

    Suppose there is a dipole placed along the x-axis at the origin . Electric field at a point at a large distance along the y-axis is (magnitude) kp/r3 - where p is the dipole moment .

    Now you place a similar dipole which is aligned in an opposite sense (direction) to the original at the same place as the first . Are you saying that the resultant of the two will not be zero at the far away on the y-axis (/or any other far away point for that matter) ?
     
  8. Jul 22, 2015 #7

    BvU

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    That is equivalent to removing the first dipole. I call that cheating :smile: . In the OP the charges are clearly not all in the origin.
     
  9. Jul 22, 2015 #8
    No , my point is this - if you take a far enough point , it won't matter whether the dipoles are in the exact same position or not - for a really far point , they will equivalently be at the same position , and then - what I said in my previous post .
     
  10. Jul 25, 2015 #9
    this is the actual problem- A17
    look through that..........
     

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  11. Jul 25, 2015 #10
    As BvU has said , it will be zero along all points on the z-axis , already giving ∞ points .

    As I have said , it should be zero at far away points all over the plane and all over the three dimensional volume too - however , you need to use your discretion for my point , i.e. , it is upto you whether you consider Limitd→∞ a/d = 0 . :cool:

     
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