Electric potential/potential energy

  1. Ok I have a couple questions here.

    1. Calculate the speed of a proton and an electron after each particle accelerates from rest through a potential difference of 190V.

    I think I got the part of the electron because I randomly came across the energy value of an electron. The equation I'm using is 1/2mv^2 = q(190V)

    2. A hydrogen electron orbits its proton in a circular orbit of radius 0.529X10^-10 meters. What is the electric potential due to the proton at the electron's orbit?

    V=kq/r right? But where do I find the q??

    3. A uniform electric field E=7500 N/C points in the negative x direction. What is the distance between the +14-V and +16-V equipotentials?

    I have no idea on this one.

    Any help on any of these woulod be greatly appreciated. Thanks a ton in advance. :confused:

    Eric
     
  2. jcsd
  3. part 1 looks right... you can do the same for a proton just don't forget to change the values of m and q.

    part 2: thats the right equ'n... the q is referring to the source charge which in this case I believe to be the charge of the proton (should be in your book / same as e but opposite sign)

    part 3: the equ'n for potential for a uniform electrical field is ED
     
  4. For part 1 I just don't see how to get the q for the proton. I just stumbled upon it for the other part.

    FOr part 2 the q value is definitely not in my book. I looked in all the appendices and everything.

    For part 3 you're saying that V = ED?
     
  5. Realize that the charge of the proton is equal to that of an electron but has the opposite sign (that is its positive)... I believe I said this in an earlier post
    If you don't believe me remember that
    [tex]\Delta V = \int\vec{E}\cdot\vec{dl}[/tex]
    since [itex]\vec{E} [/itex] is constant , this can be taken out of the integral
    and this becomes
    [tex]\Delta V = \vec{E}\int\vec{dl} [/tex]
    And since [tex]\int\vec{dl} [/tex] is just equal to the distance traveled by the particle
    [tex]\Delta V = ED [/tex]
    (where the above integrals are definite line integrals whose lower limit is the starting point and upper limit is the end point of the path )
     
    Last edited: Jan 19, 2005

  6. Sorry if it sounded like I didn't believe you. I just didn't fully understand you. :smile:
     
  7. Thats fine.... you should question everything rather than take it at face value, its part of the learning process :)

    PS: It should show a "Delta v" before each equals sign,,, is it showing up for you?
     
  8. It doesn't show, but it makes more sense now. Thanks a bunch.
     
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