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A proton accelerates

  1. Jan 26, 2005 #1
    Hello, and thank you for your existance. I have been reading posts since the beginning of last semester. I must take physics as a degree requirement, and I have in the past had a lot of trouble with understanding the concepts of physics, or really I have more trouble trying to understand how the ideas are applied. This is my first post.
    Here is my current struggle:
    A proton accelerates from rest in a uniform electric feild of 640 N/C. At some later time, its speed is 1.20 x 10^6 m/s. a)find the magnitude of the acceleration of the proton. b) how long does it take the proton to reach this speed? c) How far has it moved in this interval? d) What is its kinetic energy at the later time?

    I read the definition of an electric feild: it exists in a region of space around a charged object.
    I know what they look like, and I was able to draw them in lab. I was given equations regarding an electric feild, q=-Ed (I think that's right) where q is the point charge, E is the electric feild, and d is the distance between the point charge, and the measured particle, again, I think that's right, (left my notes in the other room). I am not sure how to relate this info or any of the other info given in class to the questions. I even went back to the beginning of the book, where it defines acceleration as change in velocity over time. I have the answer from the back of the book, which is (for a) 6.12 x 10^10 m/s^2. I tried setting it up to get the unknown time, a=deltaV/deltaT, having the change in speed (velocity) (even though my instructor says speed and velocity arent the same, and even though the words are used interchangeable throught the book) and there must be some effect the field has on the acceleration of the proton, or else I would have gotten the answer for b when I worked it out, which the answer for b is 19.6 (mu)s. Honestly, I have so much trouble with all of it. This is the tip of the iceberg. There are 6 of us left this semester, for the second half of the class, and no one got an A last semester. My instructor says we can come to him with any problem and ask any question, but he belittles our questions, and is intimidating. I explain to him that we are beginners, and he is an old pro, and ask him to try to soften his blows, but here I am again, out of class, and without a clue. sorry to cry on ya like that. I hope you will help me. Thank you again. (And if you are my teacher, and you know who I am, I am sure you have heard these very words from me at the college.) :confused:
     
  2. jcsd
  3. Jan 26, 2005 #2
    Wow, that's a lot of typing.

    There's a quick way and a slower way. I'll walk you through the latter because it is easier to grasp what is going on.

    F=ma

    F=e where B is the field strength and e is the electron charge (use correct sign)

    Be=ma

    a=dv/dt thus:

    (Be)dt=mdv

    integrate this:

    [tex]Be \int_0^t dt=m \int_0^{1.2\times 10^6}dv[/tex]

    which gives [itex]Be(t)=1.2\times 10^6 M_p[/itex]

    [tex]t=\frac{1.2\times 10^6 M_p}{Be}[/tex]

    You now know the time it take to accelerate from zero to 1.2e6 m/s.

    Now use a=dv/dt to solve for the acceleration because you know the change in velocity and the change in time.

    Distance can be found by using [tex]a=\frac{d^2s}{dt^2}[/tex] where s is the displacement.

    Kinetic energy is 1/2mv^2 is it not?

    That should cover all of the steps I think.

    Good luck.
     
    Last edited: Jan 26, 2005
  4. Jan 26, 2005 #3
    Thank you so much!

    I appriciate your help very much. After I posted my sad song, I phoned one of the other students, who gave me :
    F=ma and F= q E(-->) and the ma could be set equal to the q E(-->)
    and I said "wow that is a good start" I hadn't even thought of it. Now with your help and my friend's help, I will surely do well on this problem.
    Again, I say Thank you!
    :rofl:
     
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