Respuesta :
To solve this problem it is necessary to apply the concepts related to energy as a function of voltage, load and force, and the definition of Force given by Newton in his second law.
By definition we know that force is equal to
F= ma
Where,
m = mass (at this case of an electron)
a = Acceleration
But we also know that the Energy of an electric object is given by two similar definitions.
[tex]1) E= \frac{F}{q}[/tex]
Where,
F= Force
q = Charge of proton/electron
[tex]2) E = \frac{V}{d}[/tex]
V = Voltage
d = Distance
Equating and rearrange for F,
[tex]\frac{F}{q} = \frac{V}{d}[/tex]
[tex]F = \frac{Vq}{d}[/tex]
The two concepts of force can be related to each other, then
[tex]ma = \frac{Vq}{d}[/tex]
Acceleration would be,
[tex]a = \frac{Vd}{dm}[/tex]
Replacing with our values we have that the acceleration is
[tex]a = \frac{Vq}{dm}[/tex]
[tex]a = \frac{(170)(1.6*10^{-19})}{(2*10^{-2})(9.1*10^{-31})}[/tex]
[tex]a = 1.49*10^{15}m/s^2[/tex]
Now through the cinematic equations of motion we know that,
[tex]V_f^2-V_i^2 = 2ax[/tex]
Where,
[tex]V_f =[/tex] Final velocity
[tex]V_i =[/tex] Initial velocity
a = Acceleration
x = Displacement
Re-arrange to find v_f,
[tex]v_f = \sqrt{v_i^2+2ax}[/tex]
[tex]v_f = \sqrt{2.5*10^5+2*(1.49*10^{15})(0.1*10^{-2})}[/tex]
[tex]v_f = 1.726*10^6 m/s[/tex]
Therefore the electron's speed when it is 0.1 cm from the negative plate is [tex]1.726*10^6 m/s[/tex]
Speed of the electron enters in the gape of two large aluminum plates, when it is 0.1 cm from the negative plate is,
[tex]v_f=1.726\times10^6\rm m/s[/tex]
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
The electric force due to the electric field can be given as,
[tex]F=\dfrac{Vq}{d}[/tex]
Here, (V) is the potential difference, (q) is the charge on the body and (d) is the distance.
The charge on a electron is [tex]1.6\times10^{-19}[/tex] C. As the distance between the two plates is 2 meters and the value of potential difference is 170 V. Thus, put the values in the above formula as,
[tex]F=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]
As the force on a body is the product of mass times acceleration and the mass of the electron is [tex]9.1\times10^{-31}[/tex] kg. Thus, the above equation can be rewrite as,
[tex](9.1\times10^{-31})a=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]
[tex]a=1.49\times10^{15}\rm m/s^2[/tex]
The initial speed of the electron is [tex]2.5\times10^5[/tex] m/s and the acceleration is [tex]1.49\times10^{15}\rm m/s^2[/tex]. Thus from the third equation of motion, the final velocity can be given as,
[tex]v_f=\sqrt{2.5\times10^5+2(1.49\times10^{15})(0.1\times10^{-2})}\\v_f=1.726\times10^6\rm m/s[/tex]
Thus, the speed of the electron, when it is 0.1 cm from the negative plate is,
[tex]v_f=1.726\times10^6\rm m/s[/tex]
Learn more about electric field here;
https://brainly.com/question/14372859
