Question

an electron with a speed of 5.00 x 10^8 cm/s enters an electric field of magnitude 1.00 x 10^3 N/C, traveling along field lines in the direction that retards its motion.

a) How far will the electron travel in the field before stopping momentarily
b) how much time will have elapsed
c) in the region with the E field is only 8.00 mm long, what fraction of the electron's initial KE will be lost in that region?

Answer 1

F=ma
F=eE
ma = eE

a=eE/m= 1.6•10⁻¹⁹•10³/9.1•10⁻³¹= =1.76•10¹⁴ m/s²
s=v₀²/2a = v₀²m/2eE=
=(5•10⁶)²•9.1•10⁻³¹/2•1.6•10⁻¹⁹•10³= =0.071 m

v=v₀-at,
v=0,
t= v₀/a=5•10⁶/1.76•10¹⁴=2.84•10⁻ ⁸ s.

s₁=(v₀²-v₁²)/2a,
v₁² = v₀²-2s₁a.

KE₁/KE₀=(m v₁²/2)/ (m v₀²/2) =
= (v₁²/ v₀)² =( v₀²-2s₁a)/ v₀²=
=1-(2s₁a/ v₀²)= 1- (2•8•10⁻³•1.76•10¹⁴/25•10¹⁶)=
= 1-1.13•10⁻⁵=0.99999.