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Potential difference and electric fields

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data
    An electric field is given by E_x = (5 kN/C) x^3. Find the potential difference between the points on the x-axis at x= 1 m and x = 5 m. Answer in units of kV.


    2. Relevant equations
    V = -Ed


    3. The attempt at a solution
    V1 = -5*10^6 N/C * 1 = -5*10^6 volts
    V2 = -5*10^6 * 5 = -25*10^6 volts = -2.5*10^7 volts

    V2 - V1 = -25*10^6 -(-5*10^6) = -20*10^6 = -2.0*10^4 kV

    I also tried cubing x and got -6.2e+5 but that is also wrong.
     
    Last edited: Jul 5, 2009
  2. jcsd
  3. Jul 5, 2009 #2

    gabbagabbahey

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    [itex]V=-Ed[/itex] only when [itex]E[/itex] is uniform/constant along the length of [itex]d[/itex]...is that the case here?

    If not, you will need to use the more general equation:

    [tex]V_b-V_a=-\int_a^b\vec{E}\cdot\vec{dl}[/tex]
     
  4. Jul 5, 2009 #3
    I would think that the x^3 would mean that E is not uniform/constant...

    -5*10^6 times the integral of x^3 from 5 m to 1 m.... which is the same as -5*10^6*[(1/4)x^4] from 5 m to 1 m... which is equal to -7.8*10^5.

    Is that it?
     
  5. Jul 5, 2009 #4

    gabbagabbahey

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    Where is the [itex]10^6[/itex] coming from?
     
  6. Jul 5, 2009 #5
    I meant 10^4 since it's in kN/C, so the answer would be -7.8*10^3 kN?
     
    Last edited: Jul 5, 2009
  7. Jul 5, 2009 #6
    That apparently isn't the answer... what did I do wrong?
     
  8. Jul 5, 2009 #7

    diazona

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    [tex]1\,\mathrm{kN} = 10^3\,\mathrm{N}[/tex]
    and
    [tex]1\,\mathrm{kV} = 10^3\,\mathrm{V}[/tex]

    Could it be that?
     
  9. Jul 5, 2009 #8
    Oh wow... yet another stupid mistake. Thanks so much everyone!
     
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