Question

Each of the electrons in a particle beam has a kinetic energy of 1.36 ✕ 10⁻¹⁷ J. The mass of an electron is 9.11 ✕ 10⁻³¹ kg.

(a) What is the magnitude of the uniform electric field (pointing in the direction of the electrons' movement) that will stop these electrons in a distance of 11.0 cm?
  N/C

(b) How long will it take to stop the electrons?
  ns

(c) After the electrons stop, the electric field will continue to act on them, causing the electrons to accelerate in a direction (perpendicular to the field, towards the field or, opposite to the field) with a magnitude of acceleration of ...
  m/s²

Answer 1

a) When an electron moves distance Ar in the direction of an electric field, the work done on it is, W = F (Ar) coso = E(Ar) cos180° =-eE(Ar) From work energy theorem we have, Whet = KE, - KE -eE(Ar) = 0) – KE ΚΕ. e(x) 1.36x10-17 J (1.60x10-19C)(0.11m) = 772.72 N/C b) The magnitude of the retarding force acting on the electron is F = e E and Newton's second -F -EE law gives the acceleration as a= = E. Thus the time required to bring the electron mm to rest is,

-V-V 0-2(KE)/m -eEm V2m(KE) eE 29.11x10-5 kg)(1.36x10-17J) (1.60x10-C)(772.72 N/C) = 40.26x10-ºs -40.26ns The magnitude of acceleration is, lal=eE 71 (1.60x10-1°C)(772.72N/C) 9.11x10-kg =1.357 10 m/s