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Net Electron Acceleration and Potential Energy

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    I was watching a Khan Academy video (here) on electric potential energy (not Voltage.) Here was the problem:

    There is an infinitely extending positively charged plane (uniform.) The charged plane has a positive electric field with 5 N/C (Newtons per Coulomb.)

    Electric Potential Energy is the amount of energy it would take to move an electron x distance in y direction.

    F = Eq (where E is the electric field, q is the amount of charge, and f is the force.)

    Here's what I don't understand.

    Sal says that the amount of force it would take to move the electron is E*q. In this case 10 N, (2C * 5N/C.) And that the amount of work to move 2 C of charge against this electric field would be 10N*3M = 30NM = 30 Joules (J). Yet how could the electrons possibly accelerate towards the 3M if the electric field pushing the electron away and the electric field pushing the electron toward the electric field are the same. (They are both 5 N/C.) Shouldn't the electron just say where it is? And if that is the case, since W = Fd and d is 0, the work is 0. And if the work is zero then the potential energy would also be zero.

    What am I misunderstanding?

    2. Relevant equations
    W = Fd
    F = Eq

    3. The attempt at a solution

    I've searched my book, Practical Electronics for Inventors as well as The Physics Classroom and Wikipedia for answers. So far I have had no luck.

    Thank you so much for your help
     
  2. jcsd
  3. Nov 29, 2014 #2

    Orodruin

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    Think of an electron that already has a non zero velocity. The work necessary to lift the electron with maintained velocity against the electric field will be what?
     
  4. Nov 29, 2014 #3

    gneill

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    The electron need not accelerate. It is sufficient that it does not change its velocity if it is already moving in the correct direction :)

    So if the electron is moving in the appropriate direction and the net force acting on it is zero, then it will continue to move without acceleration to its destination. The force that counters the electric force to achieve that zero balance is the one that you evaluate for the work it does.

    If the electron begins at rest then the argument goes that in the limit where the time it takes to move the charge is irrelevant (can approach infinite time) and the speed of the electron as it proceeds can be negligible, then the applied force can be indistinguishable from a force that exactly balances the electric force, so the work done is the same as in the above case. Or, if you prefer, a force negligibly greater than the electric force is applied.

    The same sort of argument is made when evaluating the work done in moving a mass in a gravitational field. When all is said and done you evaluate the PE at the starting and ending positions and the difference is the work required.
     
  5. Nov 29, 2014 #4
    The work necessary to move an electron against the electric field, assuming it already had a velocity in the same direction, as the force you're applying to it, and an opposite direction to the electric field would be 30 J.
    By "lift" do you mean move?

    So are you saying that an electron only has potential energy if it is already moving in the direction opposite of natural forces?

    For example if I throw up a ball it gains potential energy as it goes up. Then it loses it as it goes down because it is closer to the zero reference point. So an electron has zero potential energy unless it is already moving in the opposite direction of natural forces?

    Tell me if I'm understanding this correctly. Thank you so much for the help.
     
  6. Nov 29, 2014 #5

    gneill

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    No, you are definitely not understanding this correctly :)

    Potential energy is strictly a function of location in a field (location being with respect to some reference location). The speed is irrelevant to potential energy. The concept of assigning an initial speed to the electron (or a mass in a gravitational field) for purposes of demonstrating the work involved in moving the object from one place to another is to avoid the complications of dealing with accelerating the object from rest, then decelerating back to rest at its destination. It can be done, but it just makes the intervening maths more complex without changing the result.
     
  7. Nov 29, 2014 #6
    Wait, or is it that the counter force prevents the electron from moving more than it would have otherwise and so W = Fd is evaluated for the distance the electron was stopped from traveling?

    What I don't understand I think is:
    Is my definition of work wrong in that electrical work does not involve distance?
    Or is my definition of potential energy wrong?
     
  8. Nov 29, 2014 #7
    So my definition of potential energy is wrong then. Thanks so much.

    What does it mean though? Is potential energy just how much force is exerted between two objects or particles, one stationary and one possibly in motion connected by a field?

    On a Hyperphysics page I read, it states that that: U = Ke(Q1*Q2)/(r)
    This is reminiscent of a form of coulomb's law, which the only difference is that it is over r2, not r, and it is used for finding the force between two separated charges.
     
  9. Nov 29, 2014 #8

    gneill

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    The counter force prevents acceleration. It ensures a net zero force acting on the electron so that you don't have to deal with accelerations.

    Work ALWAYS involves distance. Something is moved from A to B. When the object involved is interacting with a field then the difference in the potential of the field at the start and destination locations is tells you the work involved in moving the object from A to B (assuming a conservative field).
     
  10. Nov 29, 2014 #9

    gneill

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    You should look up the definition of potential energy. It involves the work required to transport an object from some reference location (often infinitely far removed) to a given location in space.

    When we deal with gravitational potential energy near the Earth's surface we assume that the field is uniform and choose a reference height with respect to the Earth's surface as a convenient zero reference, but in a larger context where the gravitational force is not assumed to be a constant but instead follows Newtons law of gravitation, the zero reference is taken to be at infinite distance.

    The Hyperphysics information is correct. If you assume a stationary charge Q1 and calculate the work required to bring charge Q2 from an infinite distance to a distance r from Q1, then that is the expression you will derive from Coulomb's force law.

    Energy has the unit Joules. The same units apply to all forms of energy whether potential or kinetic. The Joules is what is known as a "composite unit", as it is defined in terms of other units. Force x distance yields joules.
     
  11. Nov 29, 2014 #10
    Okay thanks for the help. I think I get it now.
     
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