Calculating Potential Difference in a Parallel Plate Apparatus

AI Thread Summary
An electron is injected into a parallel plate apparatus with an initial speed of 5x10^6 m/s and collides with the negative plate at 1x10^6 m/s. The key equations for this problem are the kinetic energy equation (KE = 1/2 mv^2) and the potential energy equation (PE = qV). The total energy of the electron changes as it moves between the plates, affecting both its kinetic and potential energy. To find the potential difference, one must equate the change in kinetic energy to the electric potential energy. The expected answer for the potential difference is 68V, indicating a need for careful calculation and integration of the forces involved.
Poguer0170
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Homework Statement


an electron with a speed of 5x10^6 m/s is injected into a parallel plate apparatus, in a vacuum, through a hole in the positive plate. The electron collides with the negative plate at 1x10^6 m/s. What is the potential difference between the plates?

Homework Equations


E=1/2mv^2
E=qV

The Attempt at a Solution


I’m not quite sure what to do... i tried finding kinetic energy and setting it equal to the energy in the plates but that gave me a small answer (i checked in the back the answer is 68V)
 
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Can you show us your calculation, complete with units?
 
Perhaps you could show exactly how you got your answer and what it is. Then we can figure out what went wrong.
 
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Hello. Welcome to PF!

Poguer0170 said:

Homework Equations


E=1/2mv^2
E=qV
It will help to use more precise notation. As you know, (1/2)mv2 is the kinetic energy of a particle. So, you can write it as

KE = (1/2)mv2

And, qV is the electric potential energy of a particle with charge q when it is located at a point where the potential is V. So, write

PE = qV

The total energy of the particle at some location is Etot = KE + PE.
(I used the subscript tool on the toolbar to write Etot).

3. The Attempt at a Solution
I’m not quite sure what to do... i tried finding kinetic energy and setting it equal to the energy in the plates but that gave me a small answer (i checked in the back the answer is 68V)

When the electron is at any point in its movement between the plates, it will have a certain amount of KE and a certain amount of PE. Just to see if you are clear on some basics, please answer the following questions (yes or no).

(1) As the electron moves from one point to another, does the KE of the electron change?

(2) As the electron moves from one point to another, does the PE of the electron change?

(3) As the electron moves from one point to another, does Etot of the electron change?

[EDIT: I see that @jbriggs444 and @kuruman replied while I was writing my post. You may ignore my post if you wish.]
 
I suggest the following:
1. Write the Lorentz force: ##m_ea_x=q_eE_x## for electron traveling in positive x direction.
2. Use the fact ## \frac {d^2 x}{dt^2} = v_x \frac{dv_x}{dx}## .
3. Integrate and use the fact that the voltage is the E field times the plate separation..
4. Look up the mass to charge ratio of the electron.

peace,
Fred
 
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