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Electrostatics, parallel plates

  1. Jun 11, 2013 #1
    1. A proton with kinetic energy of 2.1 x 10-17 J is moving into a region of charged parallel plates. The proton will be stopped momentarily in what region (attached diagram)?

    a) Region K
    b) Region L
    c) Region N
    d) Region M


    2. Relevant equations

    [itex]\Delta[/itex]Ek + [itex]\Delta[/itex]Ep = 0

    3. The attempt at a solution

    [itex]\Delta[/itex]Ek + [itex]\Delta[/itex]Ep = 0
    1/2mv2= -2×Q[itex]\Delta[/itex]v

    v=√[itex]\frac{-2Q\Delta v}{m}[/itex]
    [itex]\Delta[/itex]v = 2.1×10-17 / 1.6×10-19 = 131.25

    I have an answer, but all the [itex]\Delta[/itex]v in the regions are 100, so I don't get how I am supposed to know what region it's supposed to stop in.
     

    Attached Files:

  2. jcsd
  3. Jun 11, 2013 #2

    gneill

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    Staff: Mentor

    Each region presents a separate ΔV to the moving proton. When the proton moves from one region to the next, it only sees the ΔV of that region, but it's experienced the energy change due to the previous one. So it loses some energy climbing the potential of the previous region before entering the next. At some point the proton won't have enough energy to completely traverse a region. In fact, it will stop and reverse direction.

    You've calculated a total ΔV that will reduce the proton's KE to zero. Within which region will the total ΔV traversed reach this value?
     
  4. Jun 11, 2013 #3
    Is it the second region (L) because 131 is between 100 and 200?
     
  5. Jun 12, 2013 #4

    gneill

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    Staff: Mentor

    That's a good reason :smile:
     
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