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Current and Electrical Field

  1. Sep 16, 2015 #1
    Assume a conductor in a rectangle shape for simplicity.

    Now, if I only choose one side of this rectangle, and apply external electrical field ∑ only to it, what EMF would I create on the conductor? I would simply say ∑, however then I had the following idea, and I started to doubt if I create 2∑ instead.

    Here is what bugs me:
    (say ∑=I*R, kirchoffs law: R is the conductors resistance, ∑ is the field we apply)
    Since nucleus of atoms are almost stable, most current will be due to electron movement, accelerating due to the force of the electrical field. Then electrons will create a current I obviously.

    However, there is this topic we covered in semiconductors class in university, that is called hole current. Since electrons move from one atom to other atom, the destination atom is should initially be positively charged to be able to get the atom. When electron completes its movement, destination atom is now neutral, but the source atom is positively charged.

    Although only one electron moved physically, there is also a positively charged 'hole' moved in the opposite direction, which doubles the equivalent current, making it 2I. Then it means we had created 2∑ equivalent voltage on the semiconductor by applying only ∑ electrical field.

    Do we have ∑ or 2∑ voltage as a result of this experiment?

    Here is the wiki page for electron hole:
    https://en.wikipedia.org/wiki/Electron_hole

    In here, its stated that we treat differently to metals and semiconductors, (we ignore holes in metals) which do not answer the question but adds another dimension to it:
    https://en.wikipedia.org/wiki/Charge_carrier
     
    Last edited: Sep 16, 2015
  2. jcsd
  3. Sep 18, 2015 #2

    Simon Bridge

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    V=IR ... here, V would be the potential difference along the conductor. This is Ohm's law, not Kirchoffs.
    The electric field with the electric potential are different things.
    If the same voltage gets a higher current, it means the resistance is lower.

    You state at the start that the test material is a conductor... not a semiconductor.
    Semiconductors don't have to obey Ohm's law.
    Semicondictors are usually modelled with only one type of charge carrier ... also remember that a "hole" is the absence of an electron.

    Note: where there is a minority charge carrier, you do need to account for it when figuring the current... so you'd get more current for the same voltage... which means the resistance is lower than it would be if only one carrier were present.
     
    Last edited: Sep 18, 2015
  4. Sep 19, 2015 #3
    I think you meant conductors here...not semiconductors. Semiconductors generally have two types of charge carriers, electrons and holes. Conductors generally have only electrons contributing to conduction.
     
  5. Sep 21, 2015 #4

    Simon Bridge

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    Dont confuse the model with reality.
     
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