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Help finding distance d for particle to travel

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    a beam of charged particles is emitted from a very small opening called an "orifice" from the source. Above the source, a set of parallel plates is used to create a uniform electric field E. The distance between the two plates is D = 20 cm. The charged particles enter the electric field with the same kinetic energy Ek = 7.84e-18 J through a small hole (or "pin hole") at the center of the lower parallel plane. The particles are identical and do not interact with each other. The motion of the particles is along the electric field lines. You can ignore gravity.

    1. The charge of the particles is q=-0.0004 C. If the electric field is used to prevent the particles from hitting the upper parallel plane, then the direction of the electric field should be ___.

    2. If the magnitude of the applied electric field is |E|=4.41×10-13 N/C and the mass of the particles is m = 3×10−18 kg, then what is the time t for one particle to be stopped by the electric field after it enters the area? (We call it "time of flight", technically.)

    3. What is the distance d for one particle to travel between the plates before it is stopped by the electric field? (We call this "penetration depth", technically.)

    2. Relevant equations
    qE=change in momentum/time
    initial momentum=change in momentum=mv

    3. The attempt at a solution
    i have answers to both #1 and #2 i just need to know how to do #3. reason i included them is if u need answers to them to calculate #3.

    1.up
    2.using kinetic energy given in problem i got velocity=2.29. then i used qE=change in momentum/time to get answer of 3.90×10^-2 sec. (this is correct i checked).

    i just need #3 to find distance d.
     
  2. jcsd
  3. Feb 19, 2013 #2

    tms

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    You have a particle with a known initial velocity subject to a known constant force for a known time. How do you find out how far it goes? (I didn't do the calculation, so I'm assuming this is not relativistic.)
     
  4. Feb 19, 2013 #3
    sorry just got the answer a while ago. cnt delete this post :(
     
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