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A proton is placed in an electric field of intensity 700 N/C

  1. Jun 17, 2012 #1
    A proton is placed in an electric field of intensity 700 N/C. What is the magnitude and direction of the acceleration of this proton due to this field?

    in a google I found that
    1 proton= 1.6022 x 10 ^ -19
    and 1.6726 x 10 ^ -27 kg

    so I first get the force :
    F= q*E
    f= (1.6022 x 10 ^ -19 )(700 N/C)
    f=1.1215 * 10^ -16N

    f=ma

    f/m=a
    1.1215 * 10^-16 / 1.6726 * 10^-27 = a
    6.71* 10^10 m/s^2 = a

    am I right?

    Thanks!
     
  2. jcsd
  3. Jun 17, 2012 #2

    SammyS

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    Looks like the correct magnitude.

    What's the direction of the acceleration ?
     
  4. Jun 17, 2012 #3
    since it a positive charge and a positive field, i guess opposite. am I right??
     
  5. Jun 17, 2012 #4
    In second thought, it goes with the elctric field... because is positive and the proton is positive, so they go in the same direction... right?
     
  6. Jun 17, 2012 #5
    Yes, the proton is going in the direction of the electric field.
     
  7. Jun 17, 2012 #6
    thank you a lot SammyS and Xisune!!!
     
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