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ozgeneral
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Assume a conductor in a rectangle shape for simplicity.

Now, if I only choose one side of this rectangle having length L, and apply external electrical field ∑ only to it(along with the wire), 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∑L, however then I had the following idea, and I started to doubt if I create 2∑L instead.

Here is what bugs me:

(say ∑=I*R∑L=IR, 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∑2∑*L equivalent voltage on the semiconductor by applying only ∑ electrical field.

Do we have ∑ or 2∑L or 2∑L voltage as a result of this experiment? And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

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) :

https://en.wikipedia.org/wiki/Charge_carrier

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? And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

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) :

https://en.wikipedia.org/wiki/Charge_carrier

Assume a conductor in a rectangle shape for simplicity.

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

Here is what bugs me:

(say ∑L=IR, 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∑*L equivalent voltage on the semiconductor by applying only ∑ electrical field.

Do we have ∑L or 2∑L voltage as a result of this experiment? And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

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) :

https://en.wikipedia.org/wiki/Charge_carrier

Source Link
ozgeneral
ozgeneral
  • 563
  • 4
  • 16

Current due to External Electric Field

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? And does this change depending on the material we use for example if I use metal or semiconductor, would this result change?

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) :

https://en.wikipedia.org/wiki/Charge_carrier