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Electric Potential Question

  1. May 5, 2010 #1
    A proton's speed as it passes point A is 53,000 m/s. It follows the trajectory shown in the figure below, in which V1 = 15 V and V3 = 5 V. What is the proton's speed at point B?

    Figure: http://www.webassign.net/knight/p29-44alt.gif

    1. The problem statement, all variables and given/known data
    Mass of proton: 1.67e-27
    Charge of proton: 1.6e-19
    Difference in Electric Potential
    velocity at point A: 53000m/s
    V1 = 15V
    V2 = 10V
    V3 = 5V

    2. Relevant equations
    KE=1/2mv^2
    V=(kq)/r
    V=Ed

    3. The attempt at a solution
    10q=1/2mv^2
    v=sqrt(20*1.6e-19/1.67e-27)
     
  2. jcsd
  3. May 5, 2010 #2
    [removed]
     
    Last edited: May 5, 2010
  4. May 5, 2010 #3
    never mind I got the solution:

    KE(B) + PE(B) = KE(A) + PE(A)

    (1/2)mvB^2 + qV(B) = (1/2)mvA^2 + qV(A)\

    vB^2 = vA^2 + (2q/m)[V(A) - V(B)]

    Answer: 68739.85m/s

    I should have known it was a simple conservation of energy problem
     
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