Two large conducting parallel plates are located at x 18 pl

Two large conducting parallel plates are located at x = -1.8 (plate A) and x = 1.8 (plate B). A uniform field of 1200 V/m, in the positive x-direction, is produced by charges on the plate. The center plane at x = 0.0 m is equipotential surface on which V = 0. An electron is projected from x = 0.0 m, with an initial kinetic energy K = 500 ev of 2.04 x 10^7 in the positive x-direction. Determine the kinetic energy of the electron when it reaches plate A.

Solution

The Electric field between then two conducting parallel plates is perpendicular to plates.

the initial kinetic energy=500eV=8_*10^-17J

Thus we can find the initial velocity,

K.E_initial=mv²/2

v_initial²=(2_*8_*10^-17)/m=1.75643092_*10¹⁴m/s

We can find the force by the given electric field.

F_e=qE

Thus we know acceleration

a=F_net/m=qE/m

using the kinematical eqn,

v_final²=v_initial²+2ax =v_initial²+2qEx/m = 1.75643092*10¹⁴ + (2_*1.6_*10^-19_*1200_*1.8)/9.1_*10^-31 = 7.587751876_*10¹⁴ m/s

v_final²= 9.344211576_*10¹⁴m/s

K.E_final= mv_final²/2 = 4.256_*10^-16J