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PH 102 Lecture Notes - Lecture 7: Valence And Conduction Bands, Drift Velocity, Electric Field

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silvercicada472
5 Nov 2018
School
University of Alabama
Department
PH
Course
PH 102
Professor
Sergio Fabi

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Document Summary

Specifically, it is the charge per unit time that flows through a given cross-sectional area. (e. g. , the cross section of a wire). The accepted convention for the direction of current flow is the direction of the flow of positive charge. In a metal, however, current consists of electron flow, so that the current direction is opposite to the flow of electrons. In electrostatics the electric field inside a metal is zero, the electric potential is constant, and the charges are at rest. If a metal carries a current, then the electric field is not zero and the potential is not constant. This could happen, for example, by connecting the ends of a wire across the terminals of a battery. In a conductor the conduction electrons travel in zig-zag paths as they collide with atoms. The average distance between collisions is typically tens of nanometers.

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Related Questions

A copper wire of cross-sectional area carries a constant current of 5A. The density of free electrons is electrons per cubic meter. Find the current density and drift velocity of electrons in copper wire. Also, find the electric field and potential difference between two points 100cm apart.

Semi-conductors and the p-n junction, Part F Mastering Chemistry

Part F

The generation currents are not strongly altered by externally applied electric fields. However, applied electric fields can affect the drift current. If the applied field increases the potential of the p-type material, then holes are pushed away toward the n-type side more readily, and electrons are drawn away from the n-type side more readily. What would happen if the field increased the potential of the n-type side, instead?

The figure shows the relative energy levels of the p and n zones, as well as the junction, for an electron. The red curve indicates how the potential difference (set up initially by the recombination currents) increases with a negative bias. Negative bias is the term for an electric field that decreases the potential of the p-type side, thus increasing the energy that an electron needs to enter the p-type side. The black curve indicates how the potential difference decreases with a positive bias (the opposite of a negative bias). To answer the question, think about how these two situations will affect the rate of diffusion of electrons.

Current would flow in the opposite direction.
No current would flow because the increased potential difference would inhibit diffusion.
No current would flow because holes and electrons would combine and thus leave no charge carriers.
Current would continue to flow normally because recombination currents are not affected by external fields.
Current would continue to flow normally because increased recombination currents would cancel the decrease in diffusion currents.
Comments


    The circuit shown above consists of a single battery, whose emf is 1.8 V, and three wires made of the same material, but having different cross-sectional areas. Each thick wire has cross-sectional area 1.1e-6 m2, and is 25 cm long. The thin wire has cross-sectional area 4.8e-8 m2, and is 5.1 cm long. In this metal, the electron mobility is 3e-4 (m/s)/(V/m),and there are 4e+28 mobile electrons/m3.

    a) Which of the following statements about the circuit in the steady-state are true?

    The electron current at location D is the same as the electron current at location A.

    At location B the electric field points toward the top of the page.

    The magnitude of the electric field at locations D and F is the same.

    The magnitude of the electric field at locations F and C is the same.

     

    b) The symbol EF represents the magnitude of the electric field at location F, and the symbol ED represents the magnitude of the electric field at location D. Which of the following equations is a correct energy conservation (loop) equation for this circuit, following a path that starts at the positive end of the battery and goes clockwise?

    0 = +1.8V - EF*0.25m -ED*0.051m -EF*0.25m
    0 = -1.8V+ EF*0.25m +ED*0.051m +EF*0.25m
    0 = -1.8V - EF*0.25m -ED*0.051m -EF*0.25m
    0 = 1.8V +EF*0.25m +ED*0.051m +EF*0.25m