Electric Charge and Electric Force

Two protons $1\ \text{fm}$ apart in a nucleus push on each other with about $230\ \text{N}$, roughly the weight of a large dog, acting on particles whose mass is $10^{-27}\ \text{kg}$. That push is the electric force, and its strength is set by one property of the particles, their electric charge, and by how far apart they are.

What charge is

Electric charge is a property of matter, like mass, that determines how strongly an object takes part in electric interactions. It is a scalar: it has a size and a sign but no direction. There are two signs, called positive and negative only by convention (Benjamin Franklin's choice in the 1740s). The SI unit is the coulomb (C).

Ordinary matter gets its charge from three particles:

Here $e = 1.60\times10^{-19}\ \text{C}$ is the elementary charge. In AP Physics it is the smallest amount of charge that exists on its own, so any object's net charge is a whole-number multiple of it: $q = \pm Ne$. (Quarks carry thirds of $e$, but they are never found alone, so the rule holds for every object you will meet.) A bead with $q = -8.0\times10^{-19}\ \text{C}$ has exactly $5$ more electrons than protons; a charge of $2.4\times10^{-19}\ \text{C}$, which is $1.5e$, cannot exist.

A neutral atom has as many electrons as protons. Objects become charged almost always by gaining or losing electrons, because electrons sit on the outside of atoms and the protons are locked in nuclei (topic 8.2 covers how this happens).

A point charge is a model, not a kind of particle: you treat a charged object as if all its charge sat at one point, which is fair when the object's size is small compared with the distances in the problem. Two $2\ \text{mm}$ beads $50\ \text{cm}$ apart are point charges; two $10\ \text{cm}$ metal balls $12\ \text{cm}$ apart are not, because the charge on each crowds toward or away from the other and the "distance between the charges" is no longer one number.

Coulomb's law

In 1785 Charles-Augustin de Coulomb hung a charged pith ball on a thin twisting fiber and measured how hard a second charged ball pushed it at different distances. The force between two point charges turns out to depend on the product of the charges and on the inverse square of the distance between them.

Use absolute values for the magnitude and get the direction from the signs; plugging signed charges into the formula and reading the sign of the result as "attractive or repulsive" works for two charges but becomes a source of errors once there are three.

The inverse square has a geometric reason you will see in topic 8.6: the charge's influence spreads over spheres whose area grows as $r^2$. It makes the law a tool for factor-of-change questions. Doubling one charge doubles the force; doubling the distance divides it by $4$; tripling the distance divides it by $9$.

Several charges. Each pair of charges interacts exactly as if the others were absent. The net force on one charge is the vector sum of the forces from each of the others:

This is the principle of superposition. Because the forces are vectors, you add them by components, never by adding magnitudes, unless every force lies along one line. Superposition holds for any number of charges, but the course framework's boundary statement caps exam force calculations at four or fewer charges, allowing more only when symmetry makes most components cancel. For charge spread continuously over an object, topic 8.4 does the sum as an integral.

Electric force compared with gravity

Charged objects also have mass, so two charged objects attract gravitationally too. The two forces look alike, both inverse-square, but they differ in two ways that shape the whole physical world.

First, gravity only attracts, because mass has one sign. The electric force can attract or repel.

Second, the electric force is enormously stronger. For two protons, with $G = 6.67\times10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$,

The distance cancels, so the ratio is the same at every separation. Between any two particles or small charged objects, gravity is negligible next to the electric force.

So why does gravity, not electricity, hold the Moon in orbit? Because large objects are almost perfectly neutral. Each atom has equal numbers of protons and electrons, so a planet's positive and negative charges nearly cancel, and the electric force between two planets is close to zero. Gravity has no cancellation: every kilogram attracts every other. How close to neutral are they? For Earth ($5.97\times10^{24}\ \text{kg}$) and the Moon ($7.35\times10^{22}\ \text{kg}$) to repel electrically as hard as they attract gravitationally, each would need a charge $q$ with $\frac{kq^2}{r^2} = \frac{GMm}{r^2}$, so

That is about $3.6\times10^{32}$ extra electrons. Earth holds roughly $1.8\times10^{51}$ electrons, so an imbalance of about one electron in $5\times10^{18}$ would cancel lunar gravity entirely. Matter in bulk stays neutral far better than that, because any excess charge attracts the opposite charge from its surroundings.

The electric force is also behind most of the forces in Mechanics. When a book rests on a table, the electron clouds of the surface atoms repel each other; that repulsion, summed over the enormous number of atoms in contact, is the normal force. Friction comes from electric attraction between atoms at contact points, and tension is the electric bonding between neighboring atoms in a rope, stretched slightly. Tracking that many interactions one by one is hopeless, so you model their total as a normal force, friction or tension. These are called contact forces, and they are not new fundamental forces.

Permittivity: how materials respond to charge

Inside an atom, the electron cloud is held to the nucleus but can be shifted. Put a material near a positive charge and in every atom the electrons are pulled slightly toward that charge and the nucleus slightly away. Each atom becomes a tiny separation of $+$ and $-$. This is polarization, and the material is now polarized. The shifted charges make their own field, which points against the one that caused it, so charges embedded in a material push and pull on each other less than they would in vacuum.

Electric permittivity, $\varepsilon$, measures how strongly a material polarizes in response to an electric field. Empty space has nothing to polarize, and its permittivity is the constant $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)$. It appears in Coulomb's law through $k = \frac{1}{4\pi\varepsilon_0}$ and in many relationships later in the course (Gauss's law, capacitance). Air polarizes so little that you treat it as vacuum.

Matter has a permittivity larger than $\varepsilon_0$, and how much larger depends on its composition and structure: how easily its electrons can shift into new arrangements. Water's permittivity is about $80\varepsilon_0$, so the attraction between a sodium ion and a chloride ion dissolved in water is about $80$ times weaker than in vacuum, a large part of why salt dissolves in water. Topic 10.4 uses permittivity in capacitors.

The same idea sorts materials into two classes:

• Conductors are materials in which charge carriers move freely through the whole object. In metals some electrons belong to no single atom and wander the whole sample; in salt water, ions carry the charge.
• Insulators (nonconductors) are materials in which charge carriers can't move through the material. Their electrons only shift slightly within their own atoms or molecules. Glass, rubber, plastic and dry wood are insulators.

Topic 8.2 uses this difference to explain how objects get charged.

Worked examples

Worked exampleExample 2: three charges, two dimensions

Charge $q_1 = +2.0\ \mu\text{C}$ is at the origin. Charge $q_2 = -3.0\ \mu\text{C}$ is at $x = 0.30\ \text{m}$ on the $x$-axis, and $q_3 = +4.0\ \mu\text{C}$ is at $y = 0.40\ \text{m}$ on the $y$-axis. Find the net electric force on $q_1$.

Show solution

Find each force separately, with its direction from the signs, then add components.

Force from $q_2$: opposite signs attract, so it points from $q_1$ toward $q_2$, in the $+x$ direction.

$$F_{2} = \frac{(9.0\times10^{9})(2.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2} = 0.60\ \text{N}.$$

Force from $q_3$: like signs repel, so it points from $q_3$ through $q_1$ and beyond, in the $-y$ direction.

$$F_{3} = \frac{(9.0\times10^{9})(2.0\times10^{-6})(4.0\times10^{-6})}{(0.40)^2} = 0.45\ \text{N}.$$

The components of the net force are $F_x = +0.60\ \text{N}$ and $F_y = -0.45\ \text{N}$, so

$$F_{\text{net}} = \sqrt{(0.60)^2 + (0.45)^2} = 0.75\ \text{N}, \qquad \theta = \tan^{-1}\frac{0.45}{0.60} = 37^\circ$$

below the $+x$ axis. Adding the magnitudes, $1.05\ \text{N}$, would be wrong: the two forces are perpendicular. Notice too that $q_2$ and $q_3$ exert forces on each other, but those forces act on $q_2$ and $q_3$, not on $q_1$, so they play no part here.

Where students lose points

On the exam

Coulomb's law appears both as a calculation (with a calculator) and as a factor-of-change prediction, where the charges may be given symbolically, like $2Q$ and $-3Q$, at distances like $d$ and $2d$. A justification about the direction of a force should name the signs ("opposite charges attract, so the force on $q_2$ is toward $q_1$"). A zero-net-force derivation, like Example 3(c), earns credit for setting the two Coulomb magnitudes equal and for choosing the region correctly with a reason.