Question

an electron is lead with the velocity of 5 into 10 raise to power 6 metre per second in an electric field of 10 Newton per coulomb which had been applied so as to oppose motion what distance will the electron travel and how much time would it take goes before rest

Answer 1

Answer:

Explanation:

if acceleration is constant then

v=u+at(try to derive it using calculus)--------------------------1

where v=final velocity u=initial velocity t=time a=acceleration

according to newton

ma=Eq which imply a=Eq/m

m=mass of electron=9.10938356 × 10^-31-kilograms

a=acceleration

E=electric field=-10N/C

q=charge of electron =1.66 x 10^-19 C

there fore substituting a in equation 1

v=u+(Eq/m)t

here at last electron goes rest means final velocity=v=0

there fore

u/a=t

(u*m)/(E*q)=t

t=(5*10^6*9.10*10^-31)/(10*1.6*10^-19)sec

t=45.5*10^-25/1.6*10^-18 sec

t=28.43*10^-7 sec