Respuesta :
The speed of the electron before reaching the positive plate is [tex]1.30 \times 10^{7}\ m / s[/tex]
Explanation:
As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density. So,
[tex]\text {Electric field}=\frac{\sigma}{\varepsilon_{0}}[/tex]
And then from the second law of motion, [tex]F=m \times acceleration[/tex]
So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.
[tex]Acceleration =\frac{F}{m}[/tex]
Since force is also calculated as product of charge with electric field in electrostatic force,
[tex]\text {Acceleration}=\frac{q E}{m}=\frac{q \sigma}{m \varepsilon_{0}}[/tex]
So, the charge of electron[tex]q=1.6 \times 10^{-19}\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 \times 10^{-7}\ \mathrm{C} / \mathrm{m}^{2}[/tex]
m is the mass of electron which is equal to [tex]9.11 \times 10^{-31}\ \mathrm{kg}[/tex]
[tex]\varepsilon_{0}=8.85 \times 10^{-12}\ \mathrm{Nm}^{2} \mathrm{C}^{-2}[/tex]
Then,
[tex]\text { Acceleration }=\frac{1.6 \times 2.5 \times 10^{-19} \times 10^{-7}}{9.11 \times 8.85 \times 10^{-31} \times 10^{-12}}=\frac{4 \times 10^{-19-7}}{80.62 \times 10^{-31-12}}[/tex]
[tex]\text { Acceleration }=0.0496 \times 10^{-19-7+31+12}=0.0496 \times 10^{17}\ \mathrm{m} / \mathrm{s}^{2}[/tex]
So the acceleration of the electron in the capacitor will be [tex]4.96 \times 10^{15} m / s^{2}[/tex]
Then, the velocity can be observed from the third equation of motion.
[tex]v^{2}=u^{2}+2 a s[/tex]
As u = 0 and s is the distance of separation between two plates.
[tex]\begin{array}{c}v^{2}=0+\left(2 \times 4.96 \times 10^{15} \times 1.7 * 10^{-2}\right) \\v^{2}=16.864 \times 10^{15-2}=16.864 \times 10^{13}=1.684 \times 10^{14}\end{array}[/tex]
Thus, [tex]v=\sqrt{\left(1.68 \times 10^{14}\right)}=1.30 \times 10^{7}\ m/s[/tex]
So, the speed of the electron before reaching the positive plate is [tex]1.30 \times 10^{7} \mathrm{m} / \mathrm{s}[/tex].
