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What was the electron's initial speed?

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.4 mm diameter sphere is charged to -4.4 nC. An electron fired directly at the sphere from far away comes to within 0.31 mm of the surface of the target before being reflected.
    a) What was the electron's initial speed?
    b) At what distance from the surface of the sphere is the electron's speed half of its initial value?
    c) What is the acceleration of the electron at its turning point?

    2. Relevant equations



    3. The attempt at a solution
    how do i start this problem?
     
  2. jcsd
  3. Mar 25, 2009 #2

    LowlyPion

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    When the electron is fired at the sphere it will have kinetic energy right?

    How much work will you need to do to slow it down and reverse?

    Isn't Work = q*ΔV ?

    V at ∞ is = 0. So what is the Voltage Potential at .31 mm?
     
  4. Mar 25, 2009 #3
    so the kinetic energy = 1/2 mv^2
    what is the kinetic energy i use to slove for v?
     
  5. Mar 25, 2009 #4

    LowlyPion

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    That's the wrong question.

    What is the V you have at .31mm to let you solve for the ½mv²?
     
  6. Mar 25, 2009 #5
    can you plz explain it a little more.
    its due at 11 and im totally lost
     
  7. Mar 25, 2009 #6
    how do you do part b if you get a?
     
  8. Mar 25, 2009 #7

    LowlyPion

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    What is the formula for determining Voltage?

    V = kq/r perhaps?

    q is your charged ball. And r = .31 mm.

    That helps you determine the Voltage and then you can get the kinetic energy.

    That gives you the Velocity at ∞.

    So what voltage will it be moving through when the Velocity is half the Velocity at ∞ ?

    The acceleration back out at .31mm is a little easier.

    You can determine that from F = q*E = m*a
     
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