Electron between two parallel plates

• Mike94
In summary: So vertical velocity is reversed.That's just another way of saying that the angle of launch equals the angle of landing, and the net vertical velocity is zero.In summary, the conversation discusses the equation E = s / E0 and the calculation of s, which is found to be 4.87E-9. The conversation also explores the projection of an electron at an upward angle due to its negative charge, and the calculation of initial speed and force suffered by the electron. The acceleration and time it takes for the electron to exit are also discussed, as well as the speed at which the electron exits the plates. The conversation concludes with a discussion on finding the angle of projection using the symmetry of the problem.
Mike94
Homework Statement
An electron is thrown halfway between two large plates, with an angle Theta and an initial speed v0. The resultant electric field caused by the plates is vertical and its magnitude is noted by E. A neutral and insulating vertical membrane located to the right of the electron has an opening halfway between the two plates.
V0 = 5 * 10^6 m/s
E = 550 N/C
L = 25 cm
Relevant Equations
a) Calculate the value of the parameter s ( sigma )
b) Say if the electron should be projected at an up angle (theta> 0) or down (theta <0) to pass through the opening in the membrane. Justify your answer.
c) Consequently to your answer in b), calculate the two possible angle values in order to that the electron can exit through the opening.

a) E = s / E0 so s is 4.87E-9
b) The electron will be projected at up angle since its charge is negative ( not sure if there's another reason behind it)

c)

Initial speed:

V0 = 5 * 10^6 * cos(theta) + 5 * 10^6 * sin(theta)The force suffered by the electron is:

Fy = q*Ey

Fy = -1.602*10^19 * -550 N/C

Fy = 8.811E-17The acceleration of the electron:

Fy = m*ay

8.811E-17 = 9.11*10^-31 * ay

Ay = 9.6718E13The time it takes for the electron to exit

X= x0 + v0x * t

0.25 = 0 m + 5 * 10^6 * t

t = 5 * 10-E8The speed that the electron quits the plates

Vy = v0y + ay * t

Vy = 0 + 9.6718E13 * 5 * 10-E8

Vy = 4835894.62I’m not sure if I’m on the right path or not.

I think the way I need to find the angle is like:Theta = artcan ( vy / vx )Thank you !

Mike94 said:
a) E = s / E0 so s is 4.87E-9
...
Fy = 8.811E-17
...
Ay = 9.6718E13
Units!
Mike94 said:
b) The electron will be projected at up angle since its charge is negative ( not sure if there's another reason behind it)
That's not what the question asked.
Mike94 said:
c)Initial speed:
V0 = 5 * 10^6 * cos(theta) + 5 * 10^6 * sin(theta)
Without vectors in the equation it makes no sense.
Mike94 said:
The time it takes for the electron to exit
X= x0 + v0x * t
0.25 = 0 m + 5 * 10^6 * t
v0=5 * 10^6m/s. What is v0x?
Mike94 said:
The speed that the electron quits the plates
Can you see a symmetry here?
Mike94 said:
I think the way I need to find the angle is like:
Theta = artcan ( vy / vx )
Yes, or use the symmetry l

haruspex said:
Units!

E = 505 N/C
E0 = 8.854 * 10^-12 C^2 / (Nm^2)

E = s / E0
So my s should be 4.87E-9 uC / m^2

Fy = qE
Fy = -1.602 * 10E-19 * 550 N/C
Fy = 8.881E-17 N

Fy = may
8.811E-17 N = 9.11 * 10E-31 kg * ay
ay = 9.6718E13 m/s^2

x = x0 + v0x * t
0.25 m = 0 m + 5*10^6 * t
t = 5E-8 s

vy = v0y + ay * t
vy = 0 + 9.6718E13 * 5E-8
vy = 4835894.6213 m/2

Theta = arctan(4835894.6213/5E6)
Theta = 44 degrees

haruspex said:
Can you see a symmetry here?

So I guess you mean what I get on the exit, that's my theta at the beginning ?

The other way I see it:

Initial speed:
V0 = 5 * 10^6 * cos(theta) i + 5 * 10^6 * sin(theta) j

x = x0 + vx0t + 1/2axt^2 => t = x-x0 / vx0 => 0.25 m - 0 m/ 5E6*cos(theta)
vy = vy0 + ayt => vy = 5E6*sin(theta) + 9.6718E13 m/s^2 * x
theta = arctan(vy / 5E6*cos(theta))

After working for on it, I see it as a projectile:

R=v0^2*sin(2*theta) / |ay| => 0.25 m = 5E6^2 * sin(2*theta) / 9.6718E13 m/s^2
theta equals to 37.64 degrees or 52.3599 degrees

Last edited:
Mike94 said:
E = 505 N/C
E0 = 8.854 * 10^-12 C^2 / (Nm^2)

E = s / E0
So my s should be 4.87E-9 uC / m^2
That's an extremely small charge. Check exponent/units.
Mike94 said:
x = x0 + v0x * t
0.25 m = 0 m + 5*10^6 * t
You seem to have missed my comment:
haruspex said:
v0=5 * 10^6m/s. What is v0x?
But you have sidestepped that error by using:
Mike94 said:
After working for on it, I see it as a projectile:

R=v0^2*sin(2*theta) / |ay| => 0.25 m = 5E6^2 * sin(2*theta) / 9.6718E13 m/s^2
theta equals to 37.64 degrees or 52.3599 degrees
Quite so... as a projectile it 'lands' at its launch height, so vertical velocity is just reversed.

haruspex said:
That's an extremely small charge. Check exponent/units.

I meant to say that "s" is equals to 4.8697E-9 C

haruspex said:
Quite so... as a projectile it 'lands' at its launch height, so vertical velocity is just reversed.

Not quite sure what you meant by vertical velocity is just reversed. Do you mean that my final angle will be 90 degrees minus the angles from the final vertical component.

Thanks.

Mike94 said:
Not quite sure what you meant by vertical velocity is just reversed. Do you mean that my final angle will be 90 degrees minus the angles from the final vertical component.

Thanks.
I mean that if you throw a ball up with vertical velocity v and horizontal velocity u then when it returns to the height from which it was thrown it will have vertical velocity -v and horizontal velocity u.

Mike94

1. What is the purpose of using two parallel plates in an electron experiment?

The two parallel plates are used to create an electric field, which can be used to manipulate the motion of an electron. By changing the voltage between the plates, the strength of the electric field can be adjusted, allowing for precise control of the electron's movement.

2. How does an electron behave when placed between two parallel plates?

When placed between two parallel plates, an electron will experience a force due to the electric field created by the plates. The direction and magnitude of this force will depend on the charge and orientation of the plates, as well as the charge of the electron.

3. What is the equation for calculating the force on an electron between two parallel plates?

The force on an electron between two parallel plates can be calculated using the equation F = qE, where F is the force, q is the charge of the electron, and E is the strength of the electric field. This equation assumes that the electric field is uniform between the plates.

4. How can the distance between the two parallel plates affect the behavior of an electron?

The distance between the two parallel plates can affect the strength of the electric field, which in turn can affect the force on the electron. As the distance between the plates increases, the electric field becomes weaker, resulting in a smaller force on the electron.

5. What is the significance of studying an electron between two parallel plates?

Studying the behavior of an electron between two parallel plates can provide valuable insights into the principles of electromagnetism and the behavior of charged particles in electric fields. This knowledge has many practical applications, such as in the design of electronic devices and technologies.

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