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Coulomb Forces

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the force on a point of charge of 30μC at (0,0,5)m due to a 4m square int he z=0 plane between x=±2m and y=±2m with a total of 500μC, distributed uniformly.

    2. Relevant equations


    3. The attempt at a solution
    R=-x,-y,+5z
    dQ=ρdydx=500/4 dydx

    dE=dQ(5az) / (4πε0((x2+y2+25)3/2 )
    x and y axis cancel

    How to do integral with x2 and y2 in the denominator?

    Ans. 4.66azN
     
    Last edited: Oct 25, 2015
  2. jcsd
  3. Oct 25, 2015 #2

    Simon Bridge

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    ... use the relationship of dQ to dx and dy to make a double integral.
     
  4. Oct 28, 2015 #3
     
  5. Oct 28, 2015 #4
    Actually I really don't understand the question. Area enclosed by x=±2 and y=±2 will be 16m square. What does it means by" due to a 4m square "?
     
  6. Oct 28, 2015 #5

    gneill

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    Draw a square 4 m per side. Distribute 500 μC uniformly on the edges (so each side of the square is a line segment with 500/4 μC on it.
     
  7. Oct 28, 2015 #6
    F1d=3x5x9/16 ∫-22 dy (√29d -yay)/(29+y2)3/2
    F1d=8.43 x 0.129 =1.087
    The integral value using app integral calculator.
    F1z=1.087 x 5 /√29 = 1.009.

    Total force= 4Naz less than the answer..
     
  8. Oct 28, 2015 #7

    gneill

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    I don't understand what your last line is trying to convey. Can you explain in more detail?
     
  9. Oct 28, 2015 #8
    4 segments, so I added all that z-axis component. X and y axis cancelled. The answer is 4.66azN
     
  10. Oct 28, 2015 #9

    gneill

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    So your result is about 4 N and the answer given in the text is 4.66 N ?

    I'm inclined to agree with your result over that of the book.
     
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