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I Electric potential of an electron in a capacitor

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  1. Sep 9, 2018 #1
  2. jcsd
  3. Sep 10, 2018 #2
    In the equations the q is welcome to be positive or negative. When you plug in the negative charge of the electron I think you will see the equations make sense with the indicated direction of z.

    There is no particular reason they had to choose z as up, and had they chosen the other direction the equations could be corrected with a minus sign. However up is a perfectly fine choice and the equations are written correctly for that choice.
     
  4. Sep 10, 2018 #3

    sophiecentaur

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    Direction choice is quite arbitrary as long as you stick with it. Students are often confused by a similar thing in basic problems of acceleration under gravity - when you drop an object, do you choose distance along its downward path to be positive (the way it's actually going) or negative (because up is often positive)? The answer will be correct, whichever you choose.
     
  5. Sep 10, 2018 #4
    It's not the positive or negative sign that doesn't make sense. The complimentary value of the potential is given. If you measure from the bottom, you are giving a different value for the potential energy than when you measure from the top which is the distance potential the electron can accelerate through.
     
  6. Sep 10, 2018 #5

    sophiecentaur

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    Agreed, there will be two values of potential energy, depending which plate you choose to be zero (reference). But the work done shifting the charge from one point to another will be equal to the Potential Difference, which will give you the same answer, whichever your axes are.
    Potential is not defined in terms of "accelerating" an electron - it's the work done in moving the charge from a chosen point. I'm not quite sure where you problem lies but is it something to do with the distance travelled from start position to one plate or the other? You will get two different magnitudes and two different signs for this if not starting from the mid point.
     
  7. Sep 10, 2018 #6
    You're right, I meant to say the energy in joules after moving charge q, not acceleration.
     
  8. Sep 10, 2018 #7

    sophiecentaur

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    OK - so wouldn't you expect two possible results, depending on which side the charge ends up? Is it not just that simple?
     
  9. Sep 10, 2018 #8
    What I didn't understand but now think I do is that the potential energy is negative when you start with the top plate and then goes to zero as you reach the bottom plate with the coordinates chosen.
     
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