Definition of electric field

January 22, 2001

There was some confusion in class today about how test charges relate to defining the electric field. Here's an attempt to clarify.

First, what use is the electric field? This will hopefully become clearer as the semester goes on. Basically, think of the temperature field T(x,y) defined for the United States. If I put a thermometer in Chicago and T(Chicago) = 20 degrees F, this is useful information. We don't care how T(x,y) was found to be able to use it to determine the temperature at some specific point like Chicago.

Likewise, once you have the electric field defined, E(x,y,z), then you don't care how it was generated to be able to use it. If you put a charge q at a position (x,y,z)=(2 m, 0 m, 0 m) then it feels a force

F=q E(2,0,0)
For example, this might work out to be "12 Newtons in the +x direction" depending on what E and q are. This is useful information, even if we don't know how E was found in the first place.

So what about test charges?

Imagine we have two charges, a +3 C charge at the origin and a +1 C charge at x=2 m. They each exert a repulsive force on the other charge, and this force is given by Coulomb's law to be about 7 x 10^9 Newtons (I'm using 10^9 to mean 10 to the 9th power). So, at x=2 m, that charge feels 7 x 10^9 N of force to the right. The electric field at x=2 m, as caused by the first charge at the origin, is defined as

E = F/q

which is 7 x 10^9 N/C in the +x direction, since in this case I chose to put a 1 C charge there which makes the math easier. This is the answer to the question "what is the electric field E at the point x=2 m due to the charge at the origin?" If we put a different charge at x=2 m, say a charge of 2 micro Coulombs = 2 x 10^(-6) C, the force it feels is
F = Eq

which is (7 x 10^9 N/C)(2 x 10^(-6) C) = 14000 N in the +x direction. If instead we put a -2 micro Coulomb charge there, it would feel a force 14000 N in the -x direction.

Now, notice we have not talked at all about what is happening to the charge at the origin; it feels a force of 7000000000 N at first and then with the smaller charge it only feels a force of 14000 N. The point is, we are not asking about what the electric field is at the origin, or what the electric field is at the origin based on the charges we place at x=2 m. We are only asking about the field at x=2 m, as caused by the charge at the origin, and the answer to this question doesn't depend on what forces the charge at the origin feels.

That's the point of the test charges. They do influence other charges around, but the question is not "what is the influence of the test charge somewhere else?", the question is "what is the force the test charge feels based on everything else?" The answer to the second question is a force, and when this force is divided by the magnitude of the test charge, you get the electric field at the point where you had imagined the test charge to be. The nice thing about this is that in this way the electric field is independent of what test charge you use; that's what makes it such a useful definition, sort of like that temperature field being independent of what test thermometer you use to measure with at any particular point.

A slightly different way to think about it is that the electric field is the force per coulomb that would be felt at a particular location, if a charge was there. The idea of test charges is just one way you could figure out what that force, per coulomb, would be.

If you have more questions and/or still find this confusing, we can talk about it at the help sessions Tuesday and Wednesday night, or you can email me with questions as always.
weeks@physics.emory.edu