Question

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm, and there is a vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 10 eV and with its velocity directed toward the negative plate. How far towards the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.602 x 10-19 J) 0.2 cm 0.1 cm 0.4 cm 0 cm

Answer 1

Answer:

0.4cm

Explanation:

First, you calculate the dis acceleration of the electron generated by the potential difference between the plates:

$F=ma\\\\qE=ma\\\\q\frac{V}{d}=ma\\\\a=\frac{qV}{md}$    (1)

q: charge of the electron = 1.6*10^-19 C

V: potential difference between plates = 100V

m: mass of the electron

d: distance between parallel plates = 1cm = 0.01m

by replacing the values of the parameters you obtain:

$a=\frac{(1.6*10^{-19C})(100V)}{(9.1*10^{-31})(0.01m)}=1.75*10^{15}\frac{m}{s^2}$

you use this value in the following kinematic equation:

$v^2=v_o^2-2ax$    (2)

v: final velocity = 0m/s (until the electron stops)

from the equation (2) you calculate x, the distance traveled by the electron. But before you calculate the initial velocity of the electron by using the following equation:

$K=10eV\\\\\frac{1}{2}mv_o^2=10eV\\\\v_o=\sqrt{\frac{2(10(1.602*10^{-19}J))}{9.1*10^{-31}kg}}\\\\v_o=1.87*10^6\frac{m}{s}$

Next, you replace vo in (2) to obtain x:

$0=v_o^2-2ax\\\\x=\frac{v_o^2}{2a}=\frac{(1.87*10^6m/s)^2}{2(1.75*10^{15}m/s^2)}=0.001m=0.1cm$

Then, the electron travels 0.1cm from the middle of the region between the parallel plates, hence, the distance between the electron and the negatve plate is:

0.5cm-0.1cm=0.4cm