course index

Ch. 16: Electric Forces and Fields

The first evidence for electric forces was the observation that, after rubbing with wool, silk, or hair, some objects are able to attract other objects, like a comb attracting pieces of paper, or a balloon sticking to a wall. Three thousand years ago, a thousand years ago, and up until just a few hundred years ago, this behavior must have seemed odd, almost magical. We now understand that it is simply explained as the transfer of some electrons between the two rubbed objects, resulting in a net charge on the comb or balloon which produces an electric force on other objects.

16.1 Electric Charge

Electric charge comes in two types, positive and negative. The nuclei of atoms are positively charged, as are protons, and cations. Electrons and anions are negatively charged.

There are a few basic rules for charges:

The last rule means that we can only have charges that equal an integral number of electrons or protons. We can never have a fraction of an electron charge. The electron charge is very small, and when dealing with larger quantities of charge, we usually neglect the discreteness of charge and treat it as a continuously varying quantity -- the same as we do with mass.

The SI unit of charge is the coulomb, abbreviated C. The charge of a single electron is -1.60219×10-19C, or put another way, -1.0C of charge contains 6.24×1018 electrons. While much less than a mole of electrons (6.02×1023), this is nevertheless a substantial number. The same holds true for protons if the sign of charge is made positive.

16.2 Electric Conductors and Insulators

It is found that most materials fall into one of two classes distinguished by how freely charge is able to move in the material. In conductors charges can move about freely. Most metals are good conductors. On the other hand, charges in insulators are basically fixed, not free to move. Glass, rubber, and most plastics are insulators.

While some conductors are better conductors than others, and some insulators are better insulators than others, dividing the world into these two broad categories is a good starting point. Differentiating between degrees of conductors and insulators is just a refinement of thee broad categories.

A small number of materials fall into a third category between conductors and insulators. These are known as semiconductors, the most widely known example being silicon.

16.3 Coulomb's Law

After people realised that electric charges exist (a major contribution was made by Benjamin Franklin), experiments were undertaken to understand the force exerted by one charge on another. The most detailed experiments were done by Charles Coulomb. He observed that the electric force is:

These properties are summarized in an equation known as Coulomb's law:

F = k |q1| |q2| / r²

where k = 9×109Nm²/C² is the Coulomb constant, q1 and q2 are the two charges in coulombs, and r is the distance between them in meters.

Notice the similarities between the electrical force and the gravitational force.

The electrical force is a force just like all those encountered in Physics 2130 (spring force, normal force, gravitational force). As such, the electrical force obeys Newton's third law (for every action there's an equal and opposite reaction). Therefore, for a pair of charges, 1 and 2, F12 = -F21.

Example P15.4

An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electrical force acting on the alpha particle when it is 2.0×10-14m from the gold nucleus?

We'll use Coulomb's law to calculate the electrical force. F = kq1q2/r² = (9×109Nm²/C²)(2.0e)(79e)/ (2.0×10-14m)² = (360×1019N/c²)(1.6×10-19C)2 = 9.1×10-17N.

Example P15.8

A 2.2×10-9C charge is on the x axis at x = -1.5m and a 5.4×10-9C charge is on the x axis at x = 2.0m. Find the net force exerted on a 3.5×10-9C charge located at the origin.

Start by making a sketch of the problem. Draw the x axis and the locations of the 3 charges. Show the forces exerted on the charge at the origin; there are 2 of them, and since all the charges are positive, all the forces are repulsive. One force points left and the other points right.

F1 = kq1Q/r²1 = (9.0×109Nm²/C²)(2.2×10-9C)(3.5×10-9C)/(1.5m)² = 31×10-9N = 3.1×10-8N. F1 points to the right (+x direction).

F2 = kq2Q/r²2 = (9.0×109Nm²/C²)(5.4×10-9C)(3.5×10-9C)/(2.0m)² = 43×10-9N = 4.3×10-8N. F2 points to the left (-x direction).

The net force is Ftot = F1 - F2 = -1.2×10-8N.

Copyright © Robert Harr 2013