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If a charged particle is released from rest...

  1. Feb 26, 2016 #1
    1. The problem statement, all variables and given/known data

    F9D0D1C5-9B48-4538-BF46-BC96603529CC_zpsrjlbhk6a.jpg

    A) Calculate the electric potential at point A.

    B) If a small charged particle with a mass m=5.0 mg and charge q=7.0 nC is released from rest at point A, what will be its final speed vf

    2. Relevant equations
    V=Kq/r
    Uelec=qV
    Uelec=1/2mv2

    3. The attempt at a solution
    A) I calculated V for each point.

    V=(Kq1/r1)+(Kq2/r2)+(Kq3/r3)
    V=[(9x109)(2x10-9)/0.03]+[(9x109)(2x10-9)/0.04]+[(9x109)(2x10-9)/0.05]
    V=600+450+360
    V=1410

    For part A, I converted all my nanocoulombs (nC) to Coulombs (C), as well as my centimeters (cm) to meters (m) before plugging them into my equation.

    ________________________________________________________________________________________

    B) In order to calculate velocity (v), I need to calculate potential energy (Uelec).
    Uelec=qV
    Uelec=(7x10-9)(1410)
    Uelec=9.87x10-6

    Again, I convert the 7.0 nC into Coulombs for the 1st half of part B

    Now I solved for velocity (v) after converting mass from mg to kg:
    Uelec=1/2mv2
    9.87x10-6=1/2(5x10-6)(v2)
    v2=1/2(5x10-6)(9.87x10-6)
    v2=2.4675x10-11
    I then took the square root of both sides
    v=5x10-6 m/s

    It's asking for an assessment of the problem. "How does the velocity compare with a typical bullet's velocity, vbullet=120 m/s?"

    I'm not sure there is even anything there to compare. With the velocity of the particle at v=0.000005 m/s, and the velocity of a typical bullet at v=120 m/s, the particle is substantially slower than the average bullet. So I must be doing something wrong when there is nothing there to compare.
     
  2. jcsd
  3. Feb 26, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    There is an error going from the first line to the second.
     
  4. Feb 26, 2016 #3
    ah. This is where my math skills are starting to show.

    It appears I went the hard route and attempted to put the v2 on the left side the "=" when it could've remained on the right and rather move the 2.5x10-6 to the left.

    If I were to manipulate the equation in this fashion:
    9.87x10-6=1/2(5x10-6)(v2)
    9.87x10-6=2.5x10-6(v2)

    then divide both sides by 2.5x10-6:
    9.87x10-6/2.5x10-6=v2
    3.948=v2

    Take the square root of both sides:
    v=1.99 m/s

    That still seems like an extremely small number to be comparing to a much larger number. Does it not?
     
  5. Feb 26, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    I think your answer is correct. Note that the value of the initial potential energy is small.
     
  6. Feb 28, 2016 #5
    I apologize for the late reply. My truck's oil cooler o rings started leaking profusely and I had to rebuild my oil cooler.

    I really appreciate both your help. TSny, when you say the value of the potential energy is small, are you reffering to the 7.0 nC charge for the charged particle moving from point A?
     
  7. Feb 28, 2016 #6
    I apologize for the late reply. My truck's oil cooler o rings started leaking profusely and I had to rebuild my oil cooler.

    Never mind. I answered my own question by going back and looking through my answer, realizing I calculated the potential energy.

    Anyways, thanks again everyone. I really appreciate the guidance.
     
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