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Electrical Potential Difference between parallel plates

  1. Jun 21, 2009 #1
    Alright before I ask the question I'm going to be pretty blunt, I'm not very good at physics at all. I have a great interest and desire to learn physics but it just isn't very easy for me so explanations are going to have to be pretty detailed and even then I still might not get them.

    1. The problem statement, all variables and given/known data
    The electrical potential differences between two parallel plates is 3.50x10^3 V. If a single electron increases its electric potential by moving a distance of 0.100 m, the electron had to move toward the:
    A. positive plate through a potential difference of 1.40x10^3 V
    B. positive plate through a potential difference of 3.50x10^3 V
    C. negative plate through a potential difference of 1.40x10^3 V
    D. negative plate through a potential difference of 3.50x10^3 V


    2. Relevant equations
    W=F(d)


    3. The attempt at a solution
    I have no idea where to start as I don't know how to relate the potential difference to a single charge.
     
  2. jcsd
  3. Jun 21, 2009 #2
    Can you see whether it moved toward the positive or negative plate?
     
  4. Jun 21, 2009 #3
    I would assume that because it was a positive charge that it would move towards the negative plate since opposites attract.
    Edit: Oops obviously I was feeling dyslexic I meant because it was a negative charge it would go to the positive plate.
     
    Last edited: Jun 21, 2009
  5. Jun 21, 2009 #4
    If the negative charge moved toward the positive plate, it would experience a decrease in potential energy, as it would be going "with the flow". However, the text says that the charge experiences an increase in potential energy.
    Think of a ball in a gravitational field near the Earth's surface. If you let the ball go, it falls with the field, losing potential energy and gaining kinetic energy. when you raise the ball up, you are doing work on the ball to increase its potential energy. It now has more "potential" to do work (by falling with the field again).
    In this problem, the fact that the particle experiences an increase in electric potential means something is pushing the particle "against the flow", specifically against the direction of the acceleration it should experience due to the electric field of the capacitor.
    Please check out this page for more examples.
     
    Last edited: Jun 21, 2009
  6. Jun 21, 2009 #5
    So then the negative charge is migrating to the negative plate? Gaining electrical potential energy?
     
  7. Jun 21, 2009 #6
    So I think I got an answer. I did 0.1/0.25 and got .4 and did 3.5x10^3 * 0.4 = 1.4x10^3. If it is moving towards the negative plate (gaining potential energy) then the answer is C (I think).
     
  8. Jun 21, 2009 #7

    ideasrule

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    Homework Helper

    "So then the negative charge is migrating to the negative plate? Gaining electrical potential energy?"

    Yes, that's right. However, to answer the rest of the question, the distance between the plates must be known. Does the question give this?
     
  9. Jun 21, 2009 #8
    My bad I forgot to write part of the question. The distance between the two plates was 0.25 m.
     
  10. Jun 21, 2009 #9
    Right. It's being pushed "the wrong way", resisting the pull of the electric field and thus gaining electric potential energy.
    Now, potential energy is only meaningful in terms of change in potential energy; ie., a particle doesn't have a canonical amount of potential energy in a field, it only has a difference in potential energy compared to another point it could possibly occupy in the field. That is why potential energy is defined as a difference [itex]E = -\delta U[/itex] or a differential E = -dU. In electrostatics, since the field does not change in time, we shall take the potential energy of the electron to only be a function of its position in the electric field.
    Practically, if the plates are large enough, this problem is one-dimensional, as motion parallel to the plate does not change the potential energy of the electron (the equipotentials are parallel to the plates of the capacitor and everywhere normal to the electric field lines. The electron changes potential by punching through these equipotential surfaces). Let's call the negative plate U(n) = 0, because once the electron hits that plate, it's not really going to decrease its potential energy anymore. Then the positive plate inherits a potential energy of U(p) = 3.50x10^3 V by the problem statement.
    Now, we know that in a parallel plate capacitor, to a point particle, the field is a constant vector of magnitude E0 pointing from the positive plate to the negative plate, which means the potential energy is simply U(x) = E0x where x is the distance from the negative plate.
    However, without the distance between the plates or at least one more piece of information, I don't think you can work out the change in potential the electron passed through to travel 0.1 m.
     
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