The resistivity of a semiconductor can be modified by adding...

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length $L$ and cross-sectional area $A$ lies along the $x$ -axis between $x=0$ and $x=L .$ The material obeys Ohm's law, and its resistivity varies along the rod according to $\rho(x)=$ $\rho_{0} \exp (-x / L) .$ The end of the rod at $x=0$ is at a potential $V_{0}$ greater than the end at $x=L .$ (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude $E(x)$ in the rod as a function of $x .$ (c) Find the electric potential $V(x)$ in the rod as a function of $x .$ (d) Graph the functions $\rho(x), E(x),$ and $V(x)$ for values of $x$ between $x=0$ and $x=L$ .

Key Concepts

Electric Potential and Its Relationship to the Electric Field

The electric potential at any point in a material is obtained by integrating the electric field along a path. Since the field in a non-uniform medium is not constant, the resulting potential function is derived by accounting for the varying contributions from each segment of the material. This results in a potential function that reflects both the geometrical configuration and the spatially varying electrical properties.

Graphical Representation of Physical Quantities

Graphing the spatial dependence of resistivity, electric field, and potential provides a visual understanding of how these quantities change along the length of a material. Such plots are especially useful in cases involving non-uniform materials, as they highlight the regions where variations are most significant and help in interpreting the underlying physical phenomena associated with current flow in variable media.

Electric Field in Non-Uniform Media

The electric field in a conductor is the driving force that pushes charge carriers, and it is related to the local potential gradient via E = -dV/dx. In materials with non-uniform resistivity, the electric field can vary with position even if a constant current flows, due to the variations in the material’s ability to conduct electricity. Understanding this spatial dependence is key to analyzing current flow in inhomogeneous materials.

Integration of Differential Resistance

When dealing with a material whose resistivity changes as a function of position, the overall resistance cannot be obtained by simply multiplying a constant resistivity by geometric factors. Instead, one must consider the differential resistance of small segments, given by dR = ?(x) dx/A, and then integrate over the length of the material. This method accurately captures the influence of spatial variations in material properties on the total resistance.

Ohm's Law in Differential Form

Ohm's law, which relates current density, electric field, and resistivity, is typically expressed as J = E/? for homogeneous materials. In cases where the resistivity varies with position, the relation must be applied locally and then integrated over the region of interest. This differential approach allows for determining quantities such as the overall resistance of a material by considering contributions from infinitesimal elements.

Resistivity Variation

Resistivity in materials is a measure of how strongly the material opposes the flow of electric current. In non-uniform materials, the resistivity can vary spatially, meaning that different parts of the material resist current flow differently. This variation requires considering how local properties affect overall electrical conduction, necessitating integration over the material’s length when calculating quantities like resistance, current, and potential.

Transcript

00:01 We know that the resistance can be equal to integral, ro x, dx over a.

00:06 And the range for such integral is from 0 to l.

00:11 So we know row x, which is the function for the resistivity, is equal to row 0, turns e to the power of negative x over l.

00:18 And we know a here is the area.

00:21 So if we do that our arrangement here, eventually we'll have r is equal to row 0 over a times the integral.

00:30 So now we know the anti -divative.

00:32 For e to the power of x over l, i'm sorry, e to the power of negative x over l, should be negative l times e to the power of a negative x over l.

00:42 And the range is from 0 to l.

00:44 So if you do some arrangement here, eventually we'll have r is equal to row 0 over a times l minus l over e.

00:56 So now we can take the l out.

00:58 We have r is equal to row 0 times l over a times 1 minus to the power of a negative one.

01:07 So we know the current can be equal to potential v0 divided by the resistance.

01:13 As you can tell, this can be also equal to v0 over row 0 l over a, times e, i'm sorry, times 1 minus e to the power of negative 1.

01:24 So if you do some arrangement here, eventually you have the i current is equal to v0a over row 0 times l, times 1 minus e to the power of a negative 1.

01:35 So for part b, we know that the electric fuel function can be equal to current density, multiplied by the resistivity function, which is equal to i over a times row x...

Please give Ace some feedback