Question

(#966932-48) (Motion of Charge in Field) D2 (Motion Perpendicular to Field) A proton of mass mp-1.67 × 10-27 kg and a charge of gp-1.60 x 10-19 C is moving through vacuum at a constant velocity of 10, 000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of El =2.83e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region

Answer 1

Force acting on a proton which will be given as -

|$\vec{}$$\vec{F}$| = q |$\vec{E}$| $\Leftrightarrow$ m $\vec{a}$ = q |$\vec{E}$|

where, m = mass of a proton = 1.67 x 10^-27 kg

q = charge on a proton = 1.60 x 10^-19 C

|$\vec{E}$| = magnitude of an electric field = 2.83 x 10³ N/C

then, we get

$\vec{a}$ = [(1.60 x 10^-19 C) (2.83 x 10³ N/C)] / (1.67 x 10^-27 kg)

$\vec{a}$ = 2.71 x 10¹¹ m/s²

We know that,  $\vec{v}$ = $\vec{a}$ t $\Leftrightarrow$ t = ($\vec{v}$ / $\vec{a}$)

t = [(10000 m/s) / (2.71 x 10¹¹ m/s²)]

t = 3.7 x 10^-8 s

How far will the proton have been deflected towards the south by the time it exits the region of uniform electric field.

d = v t

d = [(10000 m/s) (3.7 x 10^-8 s)]

d = 3.7 x 10^-4 m