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Question about electrial potential

  1. Feb 12, 2005 #1
    The work done by an external force to move a [tex]-8.50{\mu}C[/tex] charge from a point a to point b is [tex]15.0*10^{-4}J[/tex]. If the charge was started from rest and had [tex] 4.82 * 10^{-4}J[/tex] of kinetic energy when it reached point b, what must be the potential difference between a and b.

    I am not sure if this is suppose to be a trick question or not, but if the final kinetic energy is 4.82 * 10^ -4, then the initial potential difference has to be 4.82 * 10^-4 right?
     
  2. jcsd
  3. Feb 12, 2005 #2
    Consider the units of electric potention (V or J/C) and work (J)
     
  4. Feb 12, 2005 #3

    dextercioby

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    What is the unit of potential difference...?

    Daniel.
     
  5. Feb 12, 2005 #4
    so
    [tex]{\Delta} KE = -{\Delta}PE}[/tex]
    [tex] \frac{1}{2}mv^2 = -qV_{ba}[/tex]
    [tex]4.82*10^{-4}J=8.5*10^{-6} V_{ba}[/tex]
    So V_ba = 5.6 * 10 V

    But when I use
    [tex] W = -qV_{ba}[/tex]
    [tex] 15 * 10^{-4} = 8.5 * 10^{-6} V_{ba}[/tex]
    [tex] V_{ba} = 1.76 * 10 ^2 [/tex]

    I am kind of lost here, can somebody help
     
  6. Feb 12, 2005 #5

    xanthym

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    You're not properly using the energy information provided. What's the relationship between the 2 energy values given in the problem?

    ~~
     
    Last edited: Feb 12, 2005
  7. Feb 12, 2005 #6
    [tex] 4.82 * 10^{-4}J[/tex], which is the kinetic energy
     
  8. Feb 12, 2005 #7

    xanthym

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    What about:

    [tex] Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]


    ~~
     
  9. Feb 12, 2005 #8

    xanthym

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    HINT:

    [tex] :(1): \ \ \ \ Work_{ext} = \Delta K.E. + \Delta P.E. [/tex]

    [tex] :(2): \ \ \ \ (15x10^{-4} Joules) \ \ = \ \ (4.82x10^{-4} Joules) \ \ + \ \ (Charge)*(PotentialDifference) [/tex]

    Do you see the technique?


    ~~
     
    Last edited: Feb 12, 2005
  10. Feb 12, 2005 #9
    I have the answer -1.19 x 10 ^ -2 V
     
  11. Feb 12, 2005 #10

    dextercioby

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    I've said that b4.It's the OP-s job to do the "dirty" work.Namely the simple calculations...We assume that the person coming in for advice/help has minimum knowledge of arithmetics...

    Daniel.

    P.S.Your result is 10.000 times smaler...
    P.P.S.xanthym,the latex code for multiplication is either "\times" or "\cdot" or leaving a space...
     
  12. Feb 12, 2005 #11
    I didnt know this forumla before because I started from the middle of the book (since that i am taking second half of the course), so i skipped all the Mechnical energy stuff

    But once i have this, it makes everything easier
     
  13. Feb 12, 2005 #12
    you mean it is -1.2 x 10 ^3?
     
  14. Feb 12, 2005 #13

    xanthym

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    LEOLAW --
    Don't guess. Repeat the calculation. (10^3 is still wrong.)
     
  15. Feb 12, 2005 #14
    [tex] (15*10^{-4} Joules) = (4.82x10^{-4} Joules) + (Charge)*(PotentialDifference) [/tex]

    [tex]1.018 * 10 ^{-3} = 8.5*10^{-6} * (PotentialDifference)[/tex]

    and i still get the result -1.19 x 10 ^2

    now I get the idea.... but not the answer?/
     
  16. Feb 12, 2005 #15

    xanthym

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    CORRECT!! :smile:
     
  17. Feb 12, 2005 #16
    finally..wholala
     
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