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Electric Potential Zero Point

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Two point charges are placed on the x axis. The charge + 5 q is at x =1.50m , and the charge −q
    is at x= -1.50m .
    There is a point on the x axis between the two charges where the electric potential is zero. Where is this point?

    The electric potential also vanishes in one of the following regions: Region 1, 10.0m>x>1.50m ;
    region 2, -1.50m >x> - 10.0m. Identify the appropriate region.
    Find the value here.


    2. Relevant equations

    V= kq/r

    3. The attempt at a solution

    For part a)
    0 = -1/x + 5/(3-x)
    Solve for x, i got 0.5, which is from -1.50

    I don't understand how to get the second region.
     
  2. jcsd
  3. Feb 21, 2014 #2

    SammyS

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    Hello Peetah. Welcome to PF !


    How did you solve the equation to get a distance of 0.5m from x = -1.50 m ?

    What is the x coordinate of that point?
     
  4. Feb 21, 2014 #3
    Thanks,

    1/x = 5/(3-x)
    3-x = 5x
    x = 0.5,
    Since x is from -1.50m, for the first part, it is -1.0m
     
  5. Feb 21, 2014 #4

    SammyS

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    Don't you really want

    ##\displaystyle 0=\frac{-1}{\left| x \right|} + \frac{5}{\left| 3-x \right|} \ \ ? ##
     
  6. Feb 22, 2014 #5
    I figured it had something to do with absolute values. would the other situation be:

    0 = -1/-x + 5/(3-x)?

    Sorry, I have no idea how to add an equation to make it look nice
     
  7. Feb 22, 2014 #6

    SammyS

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    Yes, that should work. Why does it work?


    Actually, if you want to have x be the location on the x-axis at which you are evaluating the electric potential (That would be the usual thing to do.) then the electric potential is given by the following.

    ## \displaystyle V(x)=\frac{-1}{\left| x-(-1.5) \right|} + \frac{5}{\left| x-1.5\right|} \ \ ##
    ##\displaystyle =\frac{-1}{\left| x+1.5 \right|} + \frac{5}{\left| x-1.5 \right|} \ \ ##​

    Then consider the 3 cases individually:
    x > 1.5

    -1.5 < x < 1.5

    x < -1.5​

    You will find some x that makes V(x) = 0 for two of these cases.
     
  8. Feb 22, 2014 #7
    Awesome, thanks a lot!
     
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