A) A proton has aninitial velocity of 2.84 × 107 m/s in the horizontal direction

Question

A) A proton has aninitial velocity of 2.84 ×
107 m/s in the horizontal direction. It enters
a uniform electric field of 13400 N/C directed
vertically.
Ignoring gravitational effects, find the time
it takes the proton to travel 0.123 m horizon-
tally.
Answer in units of ns.


B) What is the vertical displacement of the pro-
ton after the electric field acts on it for that
time?
Answer in units of mm.


C) What is the proton’s speed after being in the
electric field for that time?
Answer in units of km/s. A) A proton has aninitial velocity of 2.84 ×
107 m/s in the horizontal direction. It enters
a uniform electric field of 13400 N/C directed
vertically.
Ignoring gravitational effects, find the time
it takes the proton to travel 0.123 m horizon-
tally.
Answer in units of ns.


B) What is the vertical displacement of the pro-
ton after the electric field acts on it for that
time?
Answer in units of mm.


C) What is the proton’s speed after being in the
electric field for that time?
Answer in units of km/s.

Explanation / Answer

   In this case the owrizonal and verticalmotion of proton can be treated independently.    a.   time take to travelhorizontaldistance   x            t   =   x/ horizontal speed (ux)       t   =   0.123m / 2.84 * 107   m/s              =   4.33* 10-9   s          =   4.33   ns               (1ns   =   10-9   s)    b.   Acceleration ofproton   a   =   force/ mass                                               a     =   e* E / m                                                    =   1.6* 10-19 * 13400 / 1.66 * 10-27                                                    =   1.29* 1012   m/s2          Secondequation of motion is          y   =   uy* t   +   (1/2) * a *t2          uy   =   initialvertical speed   =   0          verticaldisplacement   y   =   0* 4.33 * 10-9   +   0.5 *1.29 * 1012 * (4.33 * 10-9)2                                                 =   1.21* 10-5    m                                                 =   1.21* 10-2   mm    c.   first equation of motionis         vy   =   uy   +   a* t                               finalverticalspeed   vy   =   0   +   1.29* 1012 * 4.33 * 10-9                                                                   =   5.58* 103m/s                                                                   =   5.58   km/s       Final speed of proton willbe rsultant of its horizontal and vertical speed. Hence       v   =   (ux2   +   vy2)             =   {(2.84*107 )2   +   (5.58* 103)2}             =   2.84000005* 107   m/s       Final speed of proton willbe rsultant of its horizontal and vertical speed. Hence       v   =   (ux2   +   vy2)             =   {(2.84*107 )2   +   (5.58* 103)2}             =   2.84000005* 107   m/s

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