Question #9baf6

1 Answer
May 10, 2016

Answer:

#5.7xx10^5ms^-1#, rounded to first decimal place.

Explanation:

When a charged particle having charge #q# is situated in an electric field #vecE#, the force #vecF=qcdotvecE#

Considering only the modulus part #|vecF|=qcdot|vec E|#
We also know from Newton's second law that Force #vecF=mcdot veca#, where #veca# is the acceleration.
again considering the modulus part
#|vecF|=mcdot |veca|#
Equating both we get
#mcdot |veca|=qcdot|vec E|#
or #|veca|=q/mcdot|vec E|#
Kinematic equation for a particle moving with constant acceleration is
#v^2-u^2=2as#,
where #v# is the final velocity, #u# is the initial velocity and #s# is the distance traveled. Inserting given values and solving for #v#

#v^2-0^2=2q/mcdot|vec E|s#
or #v^2=2 xx (3.2xx10^-19)/(6.4xx10^-27)cdot(1.6xx10^5)xx(2xx10^-2)#
or #v=sqrt( 3.2xx10^11)#
or #v=5.7xx10^5ms^-1#, rounded to first decimal place.