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Coulomb: Electrostatic forces

  1. May 15, 2012 #1
    I am working on the following.
    A point charge of √3Q is at point (3a, a, 2a)
    Find charge at (2a, 2a, 3a)

    I'm using Coulomb's Law
    E(r) = q / 4∏ε0 (r-r0)3 (lr-r0l
    & can get through most of it, but looking at my book I can't see why the last step occurs. See attachment.
    I can find r-ro & its magnitude lr-r0l as well, so can sub into the equation.

    r = (2a, 2a, 3a)
    r0 = (2a, 2a, 3a)
    r-r0 = (-a, a, a)
    lr-r0l = √ -1)2 +1)2 +1)2 = √3

    E(r) = √3q / 4∏ε0 (3a)3 (-a, a, a)
    Then the answer is
    E(r) = q / 4∏ε0 (√3)2 (-?a, ?a, ?a) How do you change this, mathematically, from the line above?

    Can't see the wood for trees again, I expect, but I need a bit of a memory jog, please.
     

    Attached Files:

    Last edited: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2
    Hmm :confused:

    There is no such thing as charge at (2a, 2a, 3a) due to (3a, a, 2a) From the question in the attachment, you're probably trying to find out the electric field at that point.

    These don't tally with the given values in the link, so probably this is a different question?

    Just mathematical simplification. As for the attachment case, they simply divided it by 9 that was outside the (bracket).
     
  4. May 15, 2012 #3
    Yes, you're competely right, it was the field strength I meant. Again, yes, the cartesian coords. were mixed from another question as I'm going through similar examples & inserted them by mistake.
    The attachment was as a solved example, & the solution you gave was for an "I've stopped thinking" moment.
    All fine now. Thank you.
     
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