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Electric field question

  • Thread starter Gunman
  • Start date
  • #1
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Homework Statement


I have attached the question.

Homework Equations


a)Potential Energy the electron gains = Kinetic energy it looses
1eV = 1.6 * 10^-19
Total P.E gained by the electron = 100 * 1.6 * 10^-19 = 1.6*10^-17
Therefore the kinetic energy it must possess to reach the plate RS is 1.6 *10^-17 J.
I am wrong but I am not sure where I am incorrect. Any help would be appreciated. =)

b) I am not very sure on how to go about doing this part of the question as I would need the initial velocity, or the acceleration or time taken for it to reach B to find the velocity of electrons that emerge from B?

Thank you.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
20
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You know the initial velocity is your variable, you know the constant force in the y direction, and no force in the x direction. It is akin to a projectile motion problem...
 
  • #3
25
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Yeah, got that part of the question. Thanks. =) But how do I go about doing the first part of the question?
 
  • #4
412
2
The difference between Potential Energy at 'A' and that just beneath the 2nd plate [let's call this point P] is what will give you the Kinetic Energy at 'A'. This is because since the electron has no velocity at P and hence, it's KE is zero and the KE at 'A' is converted to the PE at 'P'. To calculate the PE difference, you can use the formula:

[tex]
U = -W_{ext} = -\int F_{ext}\cdot dr
[/tex]

Also, since the force acts in only one direction, i.e. the direction perpendicular to the plate, all points which are at the same height from the 1st plate will have the same PE. This is because if the electron were moved along a line of similar height, then no work will be done as F and r are perpendicular in this case, and it's dot product will be zero.

Inside a capacitor, at any point, the Electric field is given as:

[tex]
E = \frac{\rho}{\epsilon}
[/tex]

and hence the force on the particle is given by:

[tex]
F = \frac{q\rho}{\epsilon}
[/tex]

Since, the height difference of the two plates is 'd', we have the difference in PE as:

[tex]
U = \frac{q\rho d}{\epsilon}
[/tex]

Equate it with the KE at 'A', and you shall have your answer.
 

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