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.
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 s 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?
1 answer
1 answer