Coulomb’s law

∙Charles-Augustin de Coulomb studied charge, and discovered an inverse square law for the electric force F between two point charges q1 and q₂ separated by distance r :

∙k is called Coulomb’s constant. Beware. There is another constant that is designated k called Boltzmann’s constant used in thermodynamics.
∙There is an alternate form of Coulomb’s law:

PRACTICE:

Show that the numeric value of $\frac{1}{4\pi\epsilon_0}$ equals the numeric value of $k$.

SOLUTION:

We know that $k = 8.99 \times 10^9$
We know that $\epsilon_0 = 8.85 \times 10^{-12}$
Then $\frac{1}{4\pi\epsilon_0} = \frac{1}{4\pi(8.85 \times 10^{-12})} = 8.99 \times 10^9 = k$.

FYI

Either $F = \frac{kq_1q_2}{r^2}$ or $F = \frac{q_1q_2}{4\pi\epsilon_0r^2}$ can be used. It is your choice. The first is easiest, though.
In Topic D.1 we discovered that the gravitational force also follows an inverse square law.

Coulomb’s law – extended distribution

∙We can use integral calculus to prove that a spherically symmetric shell of charge Q acts as if all of its charge is located at its center.
∙Thus Coulomb’s law works not only for point charges, which have no radii, but for any spherical distribution of charge at any radius.

∙Be very clear that r is the distance between the centers of the charges.

EXAMPLE:

A conducting sphere of radius 0.10 m holds an electric charge of Q = +125 μC. A charge q = -5.0 μC is located 0.30 m from the surface of Q. Find the electric force between the two charges.

SOLUTION:

Use $F = \frac{kq_1q_2}{r^2}$, where r is the distance between the centers of the charges.
Then r = 0.10 + 0.30 = 0.40 m. Thus
$F = \frac{kq_1q_2}{r^2} = \frac{(8.99 \times 10^9)(125 \times 10^{-6})(-5.0 \times 10^{-6})}{0.40^2}$
= -35 N.

Electric field – definition

∙Suppose a charge q is located a distance r from a another charge Q.
∙We define the electric field strength E as the force per unit charge acting on q due to the presence of Q.

∙The units are Newtons per Coulomb (N C^-1).

EXAMPLE:

Let q be a small charge located a distance r from a larger charge Q. Find the electric field strength due to Q at a distance r from the center of Q.

SOLUTION:

Use $E = \frac{F}{q}$ and $F = \frac{kq_1q_2}{r^2}$

From $E = \frac{F}{q}$, we have $E_q = F$.
And from Coulomb’s law we can write:
$E_q = F = \frac{kq_1q_2}{r^2} = \frac{kQq}{r^2}$
$E = \frac{kQ}{r^2}$

Thus the electric field strength is given by:

Electric field – the field model

∙In 1905 Albert Einstein published his Special Theory of Relativity, which placed an upper bound on the speed anything in the universe could reach.
∙According to relativity, nothing can travel faster than the speed of light c = 3.00×108 ms-1.
∙Thus the “Coulomb force signal” cannot propagate faster than the speed of light.
∙Since the electric force was thought to be an action-at-a-distance force, a significant problem arose with the advent of relativity which required a paradigm shift from action-at-a-distance to the new idea of field.

∙In the action-at-a-distance model, if the Coulomb force signal cannot propagate faster than the speed of light, electrons trapped in orbits about nuclei cannot instantaneously feel the Coulomb force, and thus cannot instantaneously adjust their motion on time to remain in a circular orbit.
∙Since electrons in orbit around nuclei do not spiral away, a new model called the field model was developed which theorized that the space surrounding a nucleus is distorted by its charge in such a way that the electron “knows” how to act.

∙In the field model the charge Q distorts its surrounding space.
∙Then the electrons know which way to curve in their orbits, not by knowing where Q is, but by knowing the local curvature of their immediate environment.

Electric monopoles and dipoles

∙Because there are two types of electric charge, electric fields can have field lines pointing inward AND outward.
∙A single charge is called a monopole.

Electric monopoles and dipoles

∙If two opposite electric monopoles are near enough to each other their field lines interact as shown here: