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Potential difference given velocity of electron at 2 points

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data

    An electron moves through an electric field, and its speed drops from 2900 {\rm m/s} to 1500 {\rm m/s} .

    What's the potential difference between the two points at which the speed was measured?

    2. Relevant equations
    U=qV
    K=mv^2/2

    q=electron
    m=mass of electron

    3. The attempt at a solution

    Ki=.5 * 9.109*10^-31 * 2900^2
    = 3.83*10^-24J
    Kf=.5 * 9.109*10^-31 * 1500^2
    =1.0247*10^-24J
    deltaK=2.80557*10^-24

    p.d= deltaK/q
    = 2.80557*10^-24/1.602*10^-19C
    =1.8*10^-5V

    Is it supposed to be negative 1.602*10^-19C for the charge on the electron and is that what is stuffing me up? Onto my final go of this on the online homework so trying to get it :)

    Thanks for help in advance.
     
  2. jcsd
  3. Jul 23, 2010 #2

    rock.freak667

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    I believe your method looks correct. But the negative sign does not really matter, as your change in KE would be negative so your pd would still be positive.
     
  4. Jul 24, 2010 #3
    Damn the homework already said no to the answer 1.8*10^-5V and 1.7*10^-5V :'(

    Maybe it doesn't like V and wants J/C I dunno :(
     
  5. Jul 24, 2010 #4

    tiny-tim

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    hi louza8! :smile:

    (try using the X2 icon just above the Reply box :wink:)
    oooh … it's not 24 :redface:

    EDIT: oops! yes it is 24 :rolleyes:
     
    Last edited: Jul 24, 2010
  6. Jul 24, 2010 #5
    Perhaps you need a more precise answer like 1.75. If the homework system you are using is something like UTexas, 3 decimals are needed as they only tolerate 2% error.
     
    Last edited: Jul 24, 2010
  7. Jul 24, 2010 #6

    vela

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    Energy conservation gives you ΔU=-ΔK, so ΔU>0. ΔV=ΔU/q should be negative.

    The OP should also verify he or she has the right number of sig figs.
     
  8. Jul 24, 2010 #7
    To verify, the correct number of sig figs are 2, deduced from the question. or are they 3? whoops? confused.
     
    Last edited: Jul 24, 2010
  9. Jul 24, 2010 #8

    ehild

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    The speed drops so ΔK is negative. The work of the field is negative.

    ehild

    Edit: I left out one sentence before, saying that the potential difference was postive. :blushing: Work is negative p.d. times charge. It is electron, so the p.d. is negative. -1.751 *10-5 V
     
    Last edited: Jul 24, 2010
  10. Jul 24, 2010 #9

    vela

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    The potential difference is not the work per unit charge done by the field; it's the work per unit charge that would have to be done against the field to move it between two points.

    Negative charges are a bit confusing because they want to roll up potential hills. But here, the electron is slowing down, so it's ending up at a lower potential (but higher potential energy) than from where it started. The potential difference is therefore negative.
     
  11. Jul 24, 2010 #10
    thanks guys, the answer given from mastering physics was 1.75*10^-5V. they were asking for the deltaV.

    I don't know why they answered in 3 significant figures when only 2 are provided in the question? am I missing something here?
     
  12. Jul 24, 2010 #11
    No. You are not missing anything. When carrying out sig figs, it really should be two but many systems use three automatically.
     
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