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Potential Difference between two points

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A bird stands on a DC electric transmission line carrying 2800A. The line has 2.5 x 10^-5 ohms resistance per meter, and the bird's feet are 4.0 cm apart. What is the potential difference between the bird's feet?


    2. Relevant equations

    I wish I knew

    3. The attempt at a solution
    Alright so since the resistance is 2.5 x 10^-5 per meter, I found that the resistance is .000001 ohms between the bird's feet. Since the current is 2800A, I used the equation V=IR and got .0028V. Is this correct or is the problem asking for something else?

    Also, there is another problem I have

    1. The problem statement, all variables and given/known data

    Part A:
    An electron in the picture tube of a television is accelerated from rest through a potential difference Vb-Va= 3000v. What is the change in electric Potential energy of the electron?

    Part B:
    What is the speed of the electron as a result of this acceleration?

    2. Relevant equations

    W=q(Vb-Va)

    3. The attempt at a solution
    I solved part A using the equation stated above and got 4.8 x10^-16
    The part that I need help with is Part B. What equation should I use to find the speed?
     
    Last edited: Jul 18, 2009
  2. jcsd
  3. Jul 18, 2009 #2

    Dick

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    Looks fine to me. Well done.
     
  4. Jul 18, 2009 #3
    Thanks Dick :) what about the second problem? any idea how to go about doing that one?
     
  5. Jul 18, 2009 #4

    Dick

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    If you put units on 4.8x10^(-16) then you've got the kinetic energy of the electron. It's Joules, right? So what's the relation between speed and kinetic energy?
     
  6. Jul 18, 2009 #5
    Well, KE= 1/2mV^2, right? I have the KE, so all I need to do is put in the mass of the electron and solve for velocity? But I thought the mass of an electron is negligible in calculations?
     
  7. Jul 18, 2009 #6

    Dick

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    Depends on the calculation. You certainly can't ignore it in this one!
     
  8. Jul 18, 2009 #7
    alright so using this equation:
    [tex]
    v = \sqrt { 2 \frac {\mbox{KE}} m }
    [/tex]

    and 9.10938188 × 10^(-31) kilograms as the mass of an electron i get
    [tex]
    v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.10938188 × 10^{-31}Kg }
    [/tex]

    32463868.12 (assuming my math is correct)

    and that would be in m/s?
     
  9. Jul 18, 2009 #8

    Dick

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    You should round those numbers off to the precision implied by the problem. But yes, that looks ok. You should be able to figure out the units yourself. J=kg*m^2/s^2. What's sqrt(J/kg)? It's a good habit to get into carrying units around and checking the result has the correct dimension. On a more interesting side note, which you may ignore until you study relativity, that's about 10% of the speed of light. If the answer had been much closer to the speed of light or exceeding it, that would have been a clue that (1/2)*mv^2 is not a good enough approximation and you need to use the relativistic form for kinetic energy.
     
  10. Jul 18, 2009 #9
    Well, When I plugged it into my calculator I acually used [tex]

    v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.109x10^{-31}Kg }

    [/tex]

    and as far as the units go, That was a good call.... it Should have been m^2/s^2. Thank You for that. I really appreciate your help.
     
  11. Jul 18, 2009 #10

    Dick

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    Uh, J/kg is m^2/s^2. Sqrt(J/kg)=m/s. You were right, it's m/s. m^2/s^2 isn't a even a velocity.
     
  12. Jul 18, 2009 #11
    my apologies, I misunderstood. I thought you meant that J=kg*m^2/s^2.

    Your help is much appreciated :)
     
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