References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.1.

Up to now, our study of electromagnetism has concentrated on static charge distributions (electrostatics) or steady current distributions (giving rise to magnetostatics). Now we start looking in more detail at what happens when currents flow in arbitrary ways.

To begin, we can state the relation between electromagnetic force and current density:

where is the *conductivity* of the material in which the current is flowing. This is an empirical relation, based on observation, rather than a theoretically derived result. Note that here is *not* surface charge density; it is the standard notation for conductivity. Its inverse is , called the *resistivity*, and again should not be confused with volume charge density. Typically, the velocity of the charges is so small that the magnetic force term can be neglected.

As an example, suppose we have two concentric conducting spheres of radii and with , held at a potential difference of . The area between the spheres is filled with a conducting material with conductivity . Our problem is to determine the current that flows between the spheres.

The electric field between the spheres is due entirely to the inner sphere, which we’ll assume has a surface charge density of . Then the field between the spheres is

Ignoring the magnetic term, the current density is

and the total current between the spheres is

where the integral is done over a sphere of radius such that .

We still need to get rid of the explicit reference to the surface charge density , which we can do by imposing the condition that the potential difference between the spheres is . We have

Thus the surface charge density is

and the current is

Note that the current is proportional to the potential difference, which is often true in resistive materials and is known as *Ohm’s law, *which is more usually written as

where is the resistance of the material. In this case

For large , and . This is presumably because there is much more conducting material at a large radius, so it contributes less to the total resistance between the spheres.

A setup used to measure the conductivity of sea water is to place two conducting spheres each of radius a large distance apart (say, at opposite ends of a ship) in the water and pass current between them by holding one sphere at potential zero and the other at potential . (I’m not 100% convinced of this argument, but here goes…) We can view this arrangement as two instances of the concentric sphere setup in this problem. The sphere at potential can be viewed as concentric spheres with the one at at potential and a distant outer sphere at potential . The other sphere (at potential 0) can be viewed as an inner sphere at potential 0 and an outer distant one at potential . The currents within both pair of spheres must be equal (since there is a constant overall current passing from to 0), so if we look at the first sphere, the current is (taking ):

which must also be the current passing into the second sphere. Thus the conductivity of the sea water can be measured as

## Trackbacks

[…] a simple example of Ohm’s law, suppose we have a capacitor with an initial charge of connected in a circuit with a resistor . […]

[…] our definition of conductivity, we’re taking […]

[…] an example of using Ohm’s law to calculate the resistance of a system with variable conductivity. We have a coaxial cylinder of […]

[…] power that can be delivered by varying can be found as follows. The voltage across is (from Ohm’s law), and the power delivered to is , […]

[…] outside vertical edge is outside the field. This motional emf then generates a current according to Ohm’s law: , where is the resistance in the […]

[…] Ohm’s law is derived from the assumption that in a conducting medium . If the currents are steady, then the charge density is independent of time, so , which implies that (if is constant throughout the material) , and we can use the various methods we derived earlier to solve Laplace’s equation and determine the potential. One example of this is given here. […]

[…] another example of calculating the resistance. We have a truncated cone with the radius of the shorter end cap being and of the longer being , […]

[…] from Ohm’s law in the […]

[…] terms of the conductivity of the sea water, the current density due to conduction is and for a capacitor with plates separated by a distance , , […]