Calculating Potential Difference in a Parallel Plate Apparatus

In summary, an electron with a speed of 5x10^6 m/s collides with a negative plate at 1x10^6 m/s in a parallel plate apparatus. The potential difference between the plates can be found by using the Lorentz force equation and integrating the acceleration of the electron. The answer is 68V.
  • #1
Poguer0170
1
0

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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  • #2
Can you show us your calculation, complete with units?
 
  • #3
Perhaps you could show exactly how you got your answer and what it is. Then we can figure out what went wrong.
 
  • Like
Likes jbriggs444
  • #4
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.]
 
  • #5
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
 

What is the concept of charged plates in physics?

Charged plates in physics refer to two parallel plates with opposite charges, creating an electric field between them. This concept is used to understand the behavior of charged particles in an electric field.

How do you calculate the electric field between two charged plates?

The electric field between two charged plates can be calculated by dividing the electric force between the two plates by the magnitude of the charge on one of the plates. This can be represented by the equation E = F/q, where E is the electric field, F is the electric force, and q is the charge on one of the plates.

What is the difference between a positively charged plate and a negatively charged plate?

A positively charged plate has an excess of positive charge, while a negatively charged plate has an excess of negative charge. In an electric field, positively charged particles will be attracted to a negatively charged plate and repelled by a positively charged plate, and vice versa for negatively charged particles.

What is the significance of charged plates in everyday life?

Charged plates have many practical applications in everyday life, including in electronic devices such as capacitors and batteries. They are also used in electrostatic precipitators for air pollution control and in particle accelerators for scientific research.

How does the distance between two charged plates affect the strength of the electric field?

The strength of the electric field between two charged plates is inversely proportional to the distance between them. This means that as the distance decreases, the electric field strength increases, and vice versa. This relationship is represented by the equation E ∝ 1/d, where E is the electric field and d is the distance between the plates.

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