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Homework Help: Electric field - point of zero field

  1. Sep 14, 2010 #1


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    1. The problem statement, all variables and given/known data
    Find a point on the line connecting two charges [tex]q_{1}[/tex] and [tex]q_{1}[/tex]
    where the electric field strength is zero. These charges [tex]q_{1}[/tex] and [tex]q_{1}[/tex] at a distance of [tex]d[/tex].
    NB - the charges can be of the same charge or different charge.

    3. The attempt at a solution
    I know the answer to be (say x is the distance):

    [tex]x=\frac{\sqrt{q_{1}}}{\sqrt{q_{1}}\pm\sqrt{q_{2}}}d[/tex] (1)

    Right, for starters I assume that these charges are both positive? and I place a negative test charge to the right of the two charges, at a distance of [tex]x[/tex]. This means that the test charge is at a distance [tex]d+x[/tex] from the first charge.
    Right, without revealing the in-between I reach a point where, by superposition


    and dividing that by k


    from where I reach the following by taking a square root of the above expression (by which I lose one solution?)


    but that does not match up with the answer in the book. (1)
    Especially the fact that [tex]q_{2}[/tex] is in the top of the division, not [tex]q_{1}[/tex], the plus-minus part I don't mind.
  2. jcsd
  3. Sep 14, 2010 #2


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    Right, for starters I assume that these charges are both positive?

    If both the charges are positive, the point of zero electric field will in between the two charges, i.e. if x is the distance from one charge, d-x will be the distance from the other charge. The value of x depends on the charge from which the distance is measured.
  4. Sep 14, 2010 #3


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    If the charges are positive yes, even simply by using a mind experiment, the zero point is in between them. Depending on the charge of either [tex]q_{1}[/tex] or [tex]q_{2}[/tex], the distance of the zero point from the first charge (say it is the starting point of the coordinate system) will vary by d-x...

    And I don't know where I went wrong... I actually did the calculation before, but I assumed [tex]q_{1}=q_{2}[/tex] and that gave me [tex]x= \frac{d}{2}[/tex], which makes sense when keeping my assumption in mind.

  5. Sep 14, 2010 #4


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    Homework Helper

    At neutral point

    q1/x^2 = q2/(d-x)^2

    Simplify and find the value of x.
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