Question

An electron is accelerated by a uniform electric field (1000 V/m) pointing vertically upward. Use energy methods to get the magnitude and direction of its velocity after it moves 0.10 cm form rest. Does the electron gain or lose potential energy?

Answer 1

E=Δφ/Δx
PE->W(work of electric field) -> KE
W=e•Δφ=e•E•Δx
KE= mv²/2
mv²/2= e•E•Δx
v=sqrt{2•e•E•Δx/m}=
=sqrt{2•1.6•10⁻¹⁹•1000•0.001/9.1•10⁻³¹}=
=5.93•10⁵ m/s.