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Determine electric field at x=1m

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A. The electric potential in a certain region is
    [tex]V=ax^2+bx+c[/tex]
    where [itex]a=13\frac{V}{m^2}[/itex], [itex]b=-14\frac{V}{m}[/itex], and [itex]c=50V[/itex].

    Determine the electric field at [itex]x=1m[/itex].
    Answer in units of [itex]\frac{V}{m}[/itex].

    B. Determine the position where the electric field is zero. Answer in units of m.

    2. Relevant equations
    [itex]∫Edx=V[/itex]

    3. The attempt at a solution
    Well V=49. I have no doubt about that. So if V is 49 at x=1, wouldn't the electric field also be 49 at x=1?
     
  2. jcsd
  3. Sep 16, 2013 #2

    mfb

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    There is no reason why the electric field should have the same numerical value as the potential.
    Note that this comparison has no physical meaning anyway - potential and field have different units.
     
  4. Sep 16, 2013 #3
    [itex]E=\frac{V}{r}[/itex]. [itex]\frac{49}{1}=49[/itex]. What am I doing wrong?
     
  5. Sep 16, 2013 #4

    mfb

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    E=V/r is not right.
    E is the derivative of V.
     
  6. Sep 16, 2013 #5
    Oh wow. So [itex]V=k\frac{Q}{r}[/itex] isn't correct?
     
  7. Sep 16, 2013 #6

    collinsmark

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    Not for this problem. [itex]V=k\frac{Q}{r}[/itex] only applies for situations involving spherical charge symmetry, like a point charge.
     
  8. Sep 16, 2013 #7

    collinsmark

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    Oh, and don't forget that there is a negative sign involved in the relationship between electric potential and electric field. The equation that you listed in the "Relevant Equations" section erroneously neglects this.
     
  9. Sep 16, 2013 #8
    Perfect, thank you guys
     
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