Question

22 spring2019 HW01.pdf Physics 122 Spring 2019 Homework #01 page 9 of Problem 8: S0 Dd σο In the cathode ray tube of an old-fashioned television, a beam of electrons is sent between two parallel charged plates. By controlling the charge on the plates, the spot where the beam hits a phosphorescent screen can be swept up and down. (A second pair of plates sweeps the spot side to side) Consider two infinite parallel horizontal charged sheets separated from each other by a vertical distance D. The top sheet has surface charge density +0o while the bottom sheet has surface charge density-o Suppose an electron of mass m and charge q is initially somewhere between the plates, traveling in some direction with speed so. A time later, the electron is observed to be at a position d lower. At this later time, find the speed s of the electron. Note: here assume that the electron does not hit either plate at any time, and do not neglect the effects of gravity on the electron. Hint: See if you can solve this problem using conservation of energy

Answer 1

Charge on electron $q=-e$

Work done by all the forces = change in kinetic energy

Work done by gravitational force $W_g=mgd$

Work done by electrostatic force $W_e=qEd=\frac{q\sigma_0 d}{\epsilon_0}$

Change in kinetic energy $\Delta KE=\frac{1}{2}m(s_f^2-s_0^2)$

$mgd+\frac{q\sigma_0 d}{\epsilon_0}=\frac{1}{2}m(s_f^2-s_0^2)$

$s_f^2=2gd+\frac{2q\sigma_0 d}{\epsilon_0 m}+s_0^2$

$s_f=\sqrt{2gd+\frac{2q\sigma_0 d}{\epsilon_0 m}+s_0^2}$

Speed of electron is $s_f=\sqrt{2gd+\frac{2q\sigma_0 d}{\epsilon_0 m}+s_0^2}$