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)

Answer 1

Answer:

0.4 cm

Explanation:

We are given that

Kinetic energy=K.E=qV=10 eV

Distance between plates,d=1 cm=$1\times 10^{-2} m$

1 m=100 cm

Potential difference,V=100 V

$1 eV=1.602\times 10^{-19} J$

We have to find the distance of electron from negative plate.

Mass of electron=$m=9.1\times 10^{-31} kg$

Speed,$v=\sqrt{\frac{2qV}{m}}$

$v=\sqrt{\frac{2\times 10\times 1.602\times 10^{-19}}{9.1\times 10^{-31}}}=1.88\times 10^6 m/s$

Electric field=$E=\frac{V}{d}=\frac{100}{10^{-2}}=10^4 N/C$

Charge on electron,q=$=-1e=-1.6\times 10^{-19} C$

Acceleration,a$=\frac-{eE}{m}=-\frac{1.6\times 10^{-19}\times 10^4}{9.1\times 10^{-31}}=-1.76\times 10^{15} m/s^2$

$v^2-u^2=2as$

Final velocity,v'=0

Using the formula

$-(1.88\times 10^6)^2=2(-1.76\times 10^{15}) s$

$s=\frac{(1.88\times 10^6)^2}{2\times 1.76\times 10^{15}}=1\times 10^{-3} m=0.1 cm$

1 m=100 cm

Mid of separation=$\frac{1}{2}=0.5 cm$

Distance of electron from negative plate=0.5-0.1=0.4 cm