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Surface charge

  1. Mar 8, 2005 #1
    a square of side 2a is uniformly charged with a surface density [tex]\rho[/tex]
    find the potential at a point distance a directly above the midpoint of one of the sides, express your answer as a numerical multiple of [tex]k_e \rho a[/tex].

    wow ok soo... i got this going for me
    k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + z^2}}dy' dx' [/tex] where (x,y,z) is the point simple enough..how do i evaulate this...i know it goes all natural log on me...but damn what do i do then!? suggestions anyone?
  2. jcsd
  3. Mar 8, 2005 #2
    Find the potential at the point due to a small rod of length 2a and then integrate from a to a+2a
  4. Mar 8, 2005 #3
    how will that be different? other then a change of limits?
  5. Mar 8, 2005 #4
    Is the solution
    2 ln[4+(2)^1/2]* (KP2a)^2 ?
  6. Mar 8, 2005 #5
    probley not considering it asks for a numerical multiple of kpa...
    no solution is given i just have to figure it out

    make it easier..sorta whats this?

    k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(-x')^2 + (-a-y')^2 + a^2}}dy' dx' [/tex]
    Last edited: Mar 8, 2005
  7. Mar 8, 2005 #6
    how come everytime i do one of these no one answers lol...
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