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Parallel Plates Question.

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Two oppositely charged parallel plates are seperated by 0.20m and have an electric potentiial difference of 1.2*10^3V across them. Locations I and II are in the region between the plates. Distance between top and bottom plate is 0.20m.
    [itex]\stackrel{}{postive plate\leftrightarrow}[/itex]
    | Region 1 : .16m in the y axis, down to region two.
    |
    |
    |
    | Region one ends here, Region II is .06m from where region one ends to the point II
    [itex]\leftrightarrow[/itex]

    I and II create a right angle triangle where 0.16m is the y from I to the right angle, and from the right angle to II is .06m. The voltage is shown to be across the whole two plates.

    2. Relevant equations



    3. The attempt at a solution

    I really dont have one, i tried V=ΔE/q, but it was just too simple, worth a shot i suppose. I tried finding the difference in voltage didnt work.
     
  2. jcsd
  3. Jan 26, 2013 #2

    Simon Bridge

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    1. I don't see a question: what are you supposed to find?
    2. I don't understand the geometry...

    You have a field between two plates, parallel to the x-z plane, with a given PD between them
    ... from that you can find the electric field between the plates and things like the function V(y).


    You have a couple of regeons marked out inside the plates ...
    from the "top" plate going down - regeon I extends from the top plate down 0.16m
    regeon II starts there, and ends 0.06m further down ... at "point II" - which must be below the bottom plate since point II is 0.22m down from the top plate and the plates are 0.2m apart.

    Then you say that "I and II create a right angle triangle" which will be gard since you have defined them as regeons (though II could refer to a point instead) and not line-segments.

    Perhaps you should scan the diagram?
     
  4. Jan 26, 2013 #3
    The Question is how much work is required to move an electron from location I to II, sorry the II is .06m to the right of where I ends, creating the right trangle.

    I will take a picture, and link/upload the pick in a second.

    Sorry.
     
  5. Jan 26, 2013 #4
  6. Jan 26, 2013 #5
    3011us7.jpg
     
  7. Jan 26, 2013 #6

    Simon Bridge

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    Great - now it's clear!

    Consider the point where there is a right-angle in the triangle ... call it point III.

    Note: the +Q plate is at the top - so the electric field is pointing downwards - so it is a bit like gravity except that electrons are negatively charged. So which direction does the force point? Is the electric field conservative or non-conservative? What does this mean about the relationship between the path electron takes between the endpoints and work needed to follow that path?

    What is the relationship between potential energy and work?

    What is the potential energy of an electron at points I, II, and III?
    Now you can figure out the work.
     
  8. Jan 26, 2013 #7
    Well force is opposite of the field right, electric field is the constant in parallel plates from what i remember. W= change in energy. But how do i start with the distances and such, or are they red herring information? Do i find the hypotenuse?
     
  9. Jan 26, 2013 #8

    Simon Bridge

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    F=qE ... so it is in the same direction as the field for positive charges and opposite for negatve charges.

    The distances are not a red herring.
    Try answering the other questions...

    What is the relationship between the potential and the y position?
    Between potential and potential energy?
    Between potential energy and work?

    You've done this with a uniform gravitational field before.
     
  10. Jan 27, 2013 #9
    Well wouldnt an electron want to move up toward the positive plate?

    1. I dont really know, between I and III the electric potential is smaller seeing the shorter distance, but electric field remains constant throughout the parallel plates right.
    Well if an electron at I wants to move to the negative plate, it has high potential energy right, or is the opposite? Because work is needed to move an electron
     
  11. Jan 29, 2013 #10

    Simon Bridge

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    The linchpin is getting the potential function V(y).
    Treat the bottom plate as V(y=0)=0, then the top plate is V(y=Y)=?
    How does the voltage vary between the plates?
    (Hint: the electric field is related to the slope of the potential function - how?)

    From there you can get the potential energy of the electron at each point - the change in potential energy is the work.
     
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