Respuesta :
Answer:
a)[tex]v=1.77\times 10^6\ m/s[/tex]
b)[tex]v=3.872\times 10^6\ m/s[/tex]
c)[tex]v=5.5\times 10^6\ m/s[/tex]
d)[tex]v=6.7\times 10^6\ m/s[/tex]
e)[tex]v=7.7\times 10^6\ m/s[/tex]
Explanation:
Given that
d = 2 cm
V = 200 V
[tex]u=4\times 10^5\ m/s[/tex]
We know that
F = E q
F = m a
E = V/d
So
m a = q .V/d b
[tex]a=\dfrac{q.V}{m.d}[/tex] ---------1
The mass of electron
[tex]m=9.1\times 10^{-31}\ kg[/tex]
The charge on electron
[tex]q=1.6\times 10^{-19}\ C[/tex]
Now by putting the all values in equation 1
[tex]a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2[/tex]
[tex]a=1.5\times 10^{15}\ m/s^2[/tex]
We know that
[tex]v^2=u^2+2as[/tex]
a)
s = 0.1 cm
[tex]v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}[/tex]
[tex]v=\sqrt{3.16\times 10^{12}}\ m/s[/tex]
[tex]v=1.77\times 10^6\ m/s[/tex]
b)
s = 0.5 cm
[tex]v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}[/tex]
[tex]v=\sqrt{1.5\times 10^{13}}\ m/s[/tex]
[tex]v=3.872\times 10^6\ m/s[/tex]
c)
s = 1 cm
[tex]v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}[/tex]
[tex]v=\sqrt{3.06\times 10^{13}}\ m/s[/tex]
[tex]v=5.5\times 10^6\ m/s[/tex]
d)
s = 1.5 cm
[tex]v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}[/tex]
[tex]v=\sqrt{4.5\times 10^{13}}\ m/s[/tex]
[tex]v=6.7\times 10^6\ m/s[/tex]
e)
s = 2 cm
[tex]v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}[/tex]
[tex]v=\sqrt{6.06\times 10^{13}}\ m/s[/tex]
[tex]v=7.7\times 10^6\ m/s[/tex]
Answer:
a) d=0.1 cm=0.001 m
[tex]v_{f}=1.9*10^{6}m/s[/tex]
b) d=0.5 cm=0.005 m
[tex]v_{f}=4.2*10^{6}m/s[/tex]
c) d=1.0 cm=0.01 m
[tex]v_{f}=5.9*10^{6}m/s[/tex]
d) d=1.5 cm=0.015 m
[tex]v_{f}=7.3*10^{6}m/s[/tex]
e) d=2 cm=0.02 m
[tex]v_{f}=8.4*10^{6}m/s[/tex]
Explanation:
Let's start with the Coulomb force equation:
[tex]F=qE[/tex]
- F is the force
- q is the electron charge
- E is the electric field
The force is equal to the mass times the acceleration, so we will have:
[tex]ma=qE[/tex]
[tex]a=\frac{qE}{m}[/tex] (1)
Now, the potential difference between the two plates is given by:
[tex]\Delta V=ED[/tex]
[tex]E=\frac{\Delta V}{D}[/tex] (2)
- D is the distance between parallel plates (D = 2.0 cm = 0.02 m).
Let's put (2) in (1)
[tex]a=\frac{q\Delta V}{Dm}[/tex] (3)
By the other side, let's recall the relation between velocities and acceleration of a particle, a kinematic equation:
[tex]v_{f}^{2}=v_{i}^{2}+2ad[/tex] (4)
- vf is the final velocity
- vi is the initial velocity
- a is the acceleration
- d is the distance
If we take square root in both sides, we will get the final velocity equation that depends on the distance.
[tex]v_{f}=\sqrt{v_{i}^{2}+2ad}[/tex] (5)
Let's put (3) in (5)
[tex]v_{f}=\sqrt{v_{i}^{2}+\frac{2dq\Delta V}{Dm}}[/tex]
a) For d=0.1 cm=0.001 m
[tex]v_{f}=\sqrt{v_{i}^{2}+\frac{2dq\Delta V}{Dm}}[/tex]
[tex]v_{f}=\sqrt{(4*10^{5})^{2}+\frac{2*0.001*1.6*10^{-19}*200}{0.02*9.11*10^{-31}}}[/tex]
[tex]v_{f}=1.9*10^{6}m/s[/tex]
b) For d=0.5 cm=0.005 m
[tex]v_{f}=\sqrt{(4*10^{5})^{2}+\frac{2*0.005*1.6*10^{-19}*200}{0.02*9.11*10^{-31}}}[/tex]
[tex]v_{f}=4.2*10^{6}m/s[/tex]
c) For d=1.0 cm=0.01 m
[tex]v_{f}=\sqrt{(4*10^{5})^{2}+\frac{2*0.01*1.6*10^{-19}*200}{0.02*9.11*10^{-31}}}[/tex]
[tex]v_{f}=5.9*10^{6}m/s[/tex]
d) For d=1.5 cm=0.015 m
[tex]v_{f}=\sqrt{(4*10^{5})^{2}+\frac{2*0.015*1.6*10^{-19}*200}{0.02*9.11*10^{-31}}}[/tex]
[tex]v_{f}=7.3*10^{6}m/s[/tex]
e) For d=2 cm=0.02 m
[tex]v_{f}=\sqrt{(4*10^{5})^{2}+\frac{2*0.02*1.6*10^{-19}*200}{0.02*9.11*10^{-31}}}[/tex]
[tex]v_{f}=8.4*10^{6}m/s[/tex]
I hope it helps you! :)
