Question

An electron is fired directly toward the center of a large metal plate that has excess negative charge with surface charge density 5.00×10^-6 C/ m². If the initial kinetic energy of the electron is 60 eV and if the electron is to stop (owing to electronstatic repulsion from the plate) just as it reaches the plate, how far from the plate must it be fired?
What is the magnitude of the acceleration?

Answer 1

given

sigma = -5*10^-6 C/m^2

KEi = 60 eV

we know, qe = -1.6*10^-19 C

me = 9.1*10^-31 kg

electric field due to a metal plate, E = sigma/epsilon

= 5*10^-6/(8.854*10^-12)

= 5.65*10^5 N/c

let d is the distance from the plate the electron is fired.

use work-energy theorem,

Workdone on electron = change in kinetic energy

qe*delta_V = KEf - KEi

qe*E*d = 0 - KEi

==> d = KEi/(qE*E)

= 60*1.6*10^-19/(1.6*10^-19*5.65*10^5)

= 1.06*10^-4 m <<<<<<<<<--------------------Answer

a = F/m

= q*E/m

= 1.6*10^-19*5.65*10^5/(9.1*10^-31)

= 9.93*10^16 m/s^2 <<<<<<<<<--------------------Answer