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A few questions about electricity and magnetism

  1. Apr 5, 2005 #1
    Hi, I'm new to this site. Anyway, I am studying for the MCAT right now and the physical science review that I have been doing has gone pretty well, I just have a few questions about comparisons that can be made between movement of electrons and movement of solid objects.

    1) What is the difference between V (voltage potential) and Potential Energy. Is voltage the relative amount of attraction that electrons have towards an electronegative pole? Is it allright for me to use stream metaphor. If a waterfall (flow of electrons) can fall from a higher distance than the voltage (potential energy of water at top and kinetic energy at bottom) will be greater.

    2) If this is the case how can one visualize the potential energy that a capacitor can store? PE=.5(Q)V or PE = 1/2(C)V*V. I thought that voltage was the potential energy.

    Thanks for any help contributed.
     
  2. jcsd
  3. Apr 5, 2005 #2

    GCT

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    In most cases it is not a good idea to generalize the concept of electric potential, it is very much mathematically dependent; especially when you're incorporating integrals or when you're attempting to find the electric field as well as other factors that depend on voltage...Unless you're very predisposed or have become very familiar with such relationships.

    However, I'm assuming that the MCAT asks very simple questions about voltage and potential energy, often conceptual. The only things that you'll probably need to know are the general formulas and that yes...voltage is a scalar quantity. If you are perplexed by any specific question, feel free to ask here at PF.

    2)voltage is not potential energy, the voltage represents the eletric potential states around a particular charge system, while you cannot talk about potential energy without referring to a charge which experiences the electric field set up by the original charge system.

    note that the original concept of [itex] \Delta V = \Delta U/q [/itex], thus [itex] \Delta V~(q)= \Delta U[/itex]. In this case however, we're talking about the overal work charging a capacitor-subsequently moving negative charge from one plate and adding it to another until a charge difference of Q exist between the capacitors. The overall work will be half of that in actually moving charge Q between a capacitor between the plates with a net voltage of V. Browse through a physics text to find the integral based derivation.
     
  4. Apr 5, 2005 #3
    [tex] F = qE [/tex]
    And by coulombs law:
    [tex] F = \frac{kq_1q_2}{r^2} [/tex]

    Potential (to do work) is the integral of force with respect to distance, so

    [tex] F = \frac{kq_1q_2}{r^2} [/tex]

    U = [tex] \int \frac{kq}{r^2} dr = \frac{-kq}{r} [/tex]

    This is equal to the potential energy a charge has when between two capacitative plates with constant electric field between.

    A little units analysis:

    [tex] E = \frac{N}{C} = \frac{V}{m} [/tex]

    Multiply all sides by m:

    [tex] E*m = \frac{Nm}{C} = V [/tex]

    Voltage on the right, equals Newton meters per coulomb. Newton meters = Joules which is energy. Thus you can conlude that voltage is potential energy per unit charge.

    Correct me if I'm wrong.
     
    Last edited: Apr 5, 2005
  5. Apr 5, 2005 #4

    GCT

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    [tex] ...= \frac{kq}{r} [/tex]
    should be potential at a point, not work
     
  6. Apr 5, 2005 #5

    GCT

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    As I said, energy does not have a state since it is also dependent on the charge experiencing the electric field of the original charge system.
     
  7. Apr 5, 2005 #6
    Sorry, potential to do work. I'll fix it.
     
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