Question:Parallel capacitors and electric beams

AI Thread Summary
An electron beam is directed perpendicular to an electric field between two capacitor plates, with both the beam and the plates experiencing the same potential difference, V. The discussion centers on calculating the deflection angle of the electron beam as it passes through the electric field, emphasizing the need to relate the potential difference to the force acting on the electrons. Participants clarify that the force on the electrons can be expressed using the equation F=qE, where E is the electric field strength, and caution against confusing the charge of the capacitor plates with that of the electrons. The influence of a magnetic field is deemed unnecessary for this scenario, as it does not affect the electron's trajectory. The conversation highlights the importance of understanding the relationship between electric fields and forces in determining the beam's deflection angle.
Joop!!
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An electron beam is send perpendicular onto an electric field between two capacitor plates. The electrons got their kinetic engergi by being accelerated through a potential difference V. The potential difference between the capacitor plates are also V.
The distance between the plates is d, and the length of the plates is a in the direction the electrons move. The electric field is uniform between the plates and 0 outside the plates.The electrons do not hit the plates.

Calculate the angle that the electron beam is deflected by the passage through the space between the plates. How does the angle depend on the potential difference?

This may sound stupid, but i have spend an hour just searching for a formula i could use to solve this problem. I have no idea how to get the angle into a formula.

First i would try to fin the electric field.

So first (Q/V)=e0(A/d) isolate to find Q

and the use F=q(E+VxB)


But I am not really sure if this even makes any sense, can anyone help me get started?


F=q(E+VxB)
 

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HINT: Can you relate the potential different of the plates to the force acting on the electron?
 
Okay.

F=qE and Q=CV

gives

F=CVE so that would give me the force that affects the electron beam between the plates. But it still doesn't make much sense, there must also be a magnetic field that i have to consider.
 
Joop! said:
Okay.

F=qE and Q=CV

gives

F=CVE so that would give me the force that affects the electron beam between the plates. But it still doesn't make much sense, there must also be a magnetic field that i have to consider.
You need to be careful here, Q refers to the charge on the capacitor plates whereas q refers to the charge of the electron. They are two 'unrelated' quantities.

As for the magnetic field, there is no need to consider it here. Yes, the electron produces a magnetic field, but this will not affect the trajectory of the electron since the magnetic field will only affect charges moving relative to the electron.
 
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