Conductivity and Resistivity

Conductivity is the degree to which a specified material conducts electricity, calculated as the ratio of the current density in the material to the electric field that causes the flow of current. Resistivity is the reciprocal of conductivity, and so the two are interchangeable as long as one tracks the inversions (and, correspondingly, the units). Electrical conductivity tells us how well a material will allow electricity to travel through it, and is similar to thermal conductivity, which tells us the ease with which thermal energy (heat for most purposes) can move through the material.

A conductor is a material which gives very little resistance to the flow of an electric current. Correspondingly, an insulator is a material which is very resistant to the flow of electric current. Semiconductors are materials which display the properties of both conductors and insulators. The most common example of conductive materials are metals, while insulators include materials such as wood or plastics . Semiconductors are generally more rare, but are necessary for the construction of transistors, and as such are present in all computers, with the most common being doped silicon. In addition to these categories, there are superconductors, which have zero resistance under certain conditions (generally extremely low temperatures and/or extremely high pressures), and there also exist a host of other materials with odd behaviors.

The units for conductivity and resistivity are most naturally expressed in terms of the recognizable Ohm ([math]\displaystyle{ \Omega }[/math]). Resistivity has units of [math]\displaystyle{ \Omega \cdot m }[/math], so conductivity has units of [math]\displaystyle{ \frac{1}{\Omega \cdot m} }[/math]. Expressed in terms of SI base units, the unit of resistivity becomes [math]\displaystyle{ \frac{kg\cdot m^3}{s^3\cdot A^2} = \frac{kg\cdot m^3}{s\cdot C^2} }[/math].

Main Idea

Mathematical Method

First, we have the relation between conductivity and resistivity:

[math]\displaystyle{ \sigma = \frac{1}{\rho} }[/math]

By convention, we have [math]\displaystyle{ \sigma }[/math] as the conductivity, and [math]\displaystyle{ \rho }[/math] as the resistivity.

Conductivity and Resistivity are properties of a material, dependent primarily on its chemical composition and structure, but also on its temperature and other environmental factors. As such, for our purposes it will always be either a know value, plugged into an equation, or an unknown value, derived from that equation. Our interest is therefore in those equations. The first is on the idealized relation between resistivity and resistance:

[math]\displaystyle{ R = \frac{\rho L}{A} }[/math]

In this, resistance is [math]\displaystyle{ R }[/math], the length of the wire is [math]\displaystyle{ L }[/math], and [math]\displaystyle{ A }[/math] is its cross sectional area. This assumes that there is a clearly defined length (parallel to the direction of current flow) and cross sectional area (perpendicular to the direction of current flow). Naturally, this is exactly the scenario present in a standard wire. The next two relevant equations are two statements of the same fact, which we call Ohm's law. Precisely, it governs the relationship between the flow of charge and the electric field which produces that flow. The first form states this explicitly:

[math]\displaystyle{ \vec{J} = \sigma \vec{E} }[/math]

Here [math]\displaystyle{ \vec{J} }[/math] is the current density, and [math]\displaystyle{ \vec{E} }[/math] is the electric field as we've seen before. An aside on current density: it is the amount of charge which passes through a given cross sectional area in a given period of time, with SI units [math]\displaystyle{ \frac{A}{m^2} = \frac{C}{m^2 \cdot s} }[/math]. This means that it is a density with area in the denominator, which can be confusing. Given what one has learned so far, this is the operable definition of Ohm's Law. However, another version is more frequently used later on, and so will also be given here:

[math]\displaystyle{ V = I R }[/math]

where [math]\displaystyle{ V }[/math] is the electric potential, [math]\displaystyle{ I }[/math] is the current, and [math]\displaystyle{ R }[/math] is resistance as defined before.

Examples

Simple

An material has a resistivity of [math]\displaystyle{ 200 \; \Omega\cdot m }[/math]. What is its conductivity?

Solution

Since conductivity is the reciprocal of resistivity,

[math]\displaystyle{ \sigma = \frac{1}{\rho} = \frac{1}{200 \; \Omega \cdot m} = 0.005 \frac{1}{\Omega \cdot m} }[/math]

Middling

An electric potential of [math]\displaystyle{ 120 V }[/math] is applied to a circular wire of length [math]\displaystyle{ 2 \cdot 10^4 \; m }[/math] and radius [math]\displaystyle{ 0.001 m }[/math]. The current is equal to [math]\displaystyle{ 1.11 \; A }[/math]. Determine the resistivity, and match it to an elemental metal using an appropriate table (such as )

Solution

We have Ohm's Law

[math]\displaystyle{ V = I R }[/math]

and so plugging in our definition for resistance in terms of resistivity gives that

[math]\displaystyle{ V = \frac{I \rho L}{A} }[/math]

which we rearrange to get

[math]\displaystyle{ \rho = \frac{ V A}{I L} }[/math]

plugging all of the values in gives an answer of [math]\displaystyle{ \rho = 1.7 \cdot 10^{-8} \; \Omega \cdot m }[/math], which is the resistivity of copper.

Difficult

An electric field of [math]\displaystyle{ 1 \cdot 10^{-4} \; N/C }[/math] is pushing charge through a square wire of width [math]\displaystyle{ 0.01 \; m }[/math] made of aluminum (which has conductivity [math]\displaystyle{ \sigma = 3.65 \cdot 10^7 \; \Omega^{-1}\cdot m^{-1} }[/math]) into a metal sphere, which is building up charge. Presuming no other electric fields develop (as shall be seen in later sections, this is not an accurate assumption), determine the charge on the sphere after 10 seconds.

Solution

We have from Ohm's Law that

[math]\displaystyle{ \vec{J} = \sigma \vec{E} }[/math]

Furthermore, we know that current density is charge per second per area, so if we multiply it by time and the cross sectional area, we will find the net charge which has passed through the cross section of the wire, and correspondingly the amount of charge which has built up on the sphere. Thus

[math]\displaystyle{ Q = \vec{J}\cdot A \cdot t = \sigma \cdot \vec{E} \cdot A \cdot t }[/math]

Plugging all of our values in, this lets us compute that

[math]\displaystyle{ Q = ( 3.65 \cdot 10^7 \; \Omega^{-1}\cdot m^{-1})(1 \cdot 10^{-4} \; N/C )(1\cdot 10^{-4} m^2)(10 \; s) = 3.65 C }[/math]