A uniform electric field of magnitude 15.0 N/C is directed parallel to the posit

Question

A uniform electric field of magnitude 15.0 N/C is directed parallel to the positive x-direction. A proton is at the origin (x=0, y=0) at time t=0 when it is moving with a velocity of 5.0 x 10^4 m/s in the positive y-direction. Ignore and non-electric forces. At some particular later time, it has an x-coordinate of 4.0m.

a.) What is the acceleration (magnitude and direction) of the proton?
b.) What is the y-component of the velocity of the proton at the later time?

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Explanation / Answer

             Given:     Applied   uniform electric field of magnitude = E = 15.0 N/C             (directed parallel to the + x-direction )                                     As it moving with a velocity of = V = 5.0 x 10^4 m/s                    ( in the positive y-direction)              charge on the proton = q = 1.6x 10^-19  C    (a)                when electric field applied ,   some external force         acted upon by the field on the proton                   the electric force = F = E q                  but , by the newton's second law ,                     force F = ma                   where m is the mass of the proton = 1.67x 10^-27  kg                                                     a is the acceleration attained in the proton                          on equating ,                            F = Eq = ma                               a = Eq / m   = (15) (1.6x 10^-19  ) / (1.67x 10^-27  )                                  = 14.37x10^8   m/s2       (b)                          by the one standered relation ,                  we have    acceleration = a = V -U /t                                      since ,it starts from rest ,intial velocity is   = 0 m/s                                     thus, V = a t                                            t = V / a = ( 5.0 x 10^4) /(14.37x10^8) = 0.347*10^-4 sec                               After later time , velocity be V'                              by the kinematic realtion                                      V' = V + at                                          = ( 5.0 x 10^4) + (14.37x10^8) (0.347*10^-4 )                                         = 9.98x 10^4 m/s                                                                                                                                             

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