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Electrostatics-Taylor expansion

  1. Oct 15, 2009 #1
    Hi everyone

    First, I’m sorry if this topic is not in the right place here (I did hesitate with introductory physics).

    My questions are the following:

    In a Cartesian coordinate system:
    Consider that you have four charges q placed in (a,0,0);(-a,0,0);(0,a,0);(0,-a,0).
    The electrostatic field in O(0,0,0) will be null(because of the symmetries). But what is the form of the Taylor expansion of the potential near the origin O?

    The answer is V(M)=V0+b*x+c*y+d*z+e*x^2+f*y^2+g*z^2+h*x*y+i*x*z+j*y*z
    But I can’t understand WHY?

    Second question:

    Consider the electrostatic field E(M)=(x,y,-2*z) (in a Cartesian coordinate system).
    “If you put a particle of charge q in M, then if it’s movement is stable in xOy, then this movement won’t be stable in (Oz)” Again, I can’t understand why.

    Any help would be welcome!
  2. jcsd
  3. Oct 15, 2009 #2


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    Hi penguin007, you've been here long enough that you should know you need to show some attempt at a solution, or at least explain where you are stuck!

    What is the expression for the potential due to a collection of point charges?

    Read the section of your textbbook on Earnshaw's theorem.
  4. Oct 16, 2009 #3
    Hi, gabbagabbahey, sorry if my questions were not clear...
    actually I guessed that I had to do a Taylor expansion at the order 2 to get the first result but didn't know its expression (I got it:https://www.physicsforums.com/showthread.php?t=346001).
    For my second question, I understand it is linked to Earnshaw's theorem, but I still don't get the idea:how can this field be stable in xoy and not in (oz)? Any explanation would be welcome...
  5. Oct 16, 2009 #4


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    Well, Earnshaw's theorem tells you it can't be stable in three dimensions...If its stable in two dimensions, it must therefor be unstable in the third dimension.
  6. Oct 16, 2009 #5
    Therefore, it could also be stabe in (xoz) and not in (oy)?
  7. Oct 16, 2009 #6


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