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Why is the electric field intensity zero at only one point?

  1. Feb 4, 2007 #1
    Suppose you have a Cartesian coordinate system with:

    a 2uC charge at (0,0) and a -8uC charge at (4,0), why is there only one point in which the EFI is zero (-4,0)? Shouldn't that be the first point where the EFI is zero (so (-5,0),(-6,0)... all have Electric Fields of zero)

    The reason I think this is because if you drew the Electric field lines from one charge to the other, they wouldn't go either way infinitely.

    Thanks
     
  2. jcsd
  3. Feb 4, 2007 #2

    mjsd

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    Electric field strength is zero only at (-4,0) because field strength are given by (up to a multiplicative constant)
    [tex]\displaymath{\frac{q}{r^2}}[/tex]
    so at (-4,0)
    you have
    [tex]\displaymath{\frac{2}{4^2}+\frac{-8}{(4+4)^2}=
    \frac{2}{16}-\frac{8}{64}=0}[/tex]
    whereas at (-5,0) you get
    [tex]\displaymath{\frac{2}{5^2}+\frac{-8}{(5+4)^2}=
    \frac{2}{25}-\frac{8}{81} \neq 0}[/tex]
     
  4. Feb 4, 2007 #3
    I know that, but can you explain the field itself, since drawing the electric field lines doesn't seem to explain this
     
  5. Feb 4, 2007 #4

    mjsd

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    it is very hard to draw field lines to explain this, you will need a computer to help to do this properly. mind you an E-field is a vector field and at each spacetime point it has a magnitude and direction, so strictly speaking, to draw this properly you need to insert at each point a little vector, do this for each charge separately and you will find that only at (-4,0) will the magnitude be the same AND direction is opposite (ie cancel after doing the vector sum)
     
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