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Homework Help: Electric fields

  1. Feb 7, 2005 #1
    I'm not sure if this has been asked, but i am in major need of help. i have 2 questions. first is determining the speed of a particle when the speed is parallel to the electric field. i talked to my prof. and he suggested finding the energy, U, and by integrating E with respect to z, by doing so, that left me with E*z from Zinitial (1-cos(theta)*L) to Zfinal(0). after doing this, i st this equal to 1/2 mv^2. to solve for the velocity, however, i am not getting the correct answer. :mad: and i'm getting frustrated. does anyone know where i am going wrong?

    my second question is finding finite value(s) where an electric field is zero. by having a +q charge at the orgin and a -2q charge at x=5.8m. my professor told me to use kq/r^2 and where q is and q or -2q and r is x and x-d respectively where d is 5.8 m. i solving and put the equation = 0, and used the quadratic formula to get the answer. However, i am not getting this correct, when i solve for x i get x^2+2xd-d^2. i have used various other ways of trying to get this answer correct, by changing the signs and and i am still getting incorrect answers. I need to do the same with electric potential, but i do not understand how i can get 2 answers from a linear equation. i am getting d-3x where d =5.8 m. :confused:
    if anyone could help, i would appreciate it so much
     
  2. jcsd
  3. Feb 7, 2005 #2

    dextercioby

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    Homework Helper


    Who's "Z"...?

    The equation is:
    [tex] m\frac{d\vec{v}}{dt}=q\vec{E} [/tex]

    If the acceleration has the sense with the velocity (the sense of the acc.is the same with the one of the electric field for a positively cherged particle),then project the previous relation on an axis and integrate wrt to corresponding limits...

    No,on normal basis it should reduce to a quadratic...

    Daniel.
     
  4. Feb 7, 2005 #3
    Z is the limit bounds....initial to final. plus, i was using it as a variable to integrate on
     
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