Electric charge, like mass, is a fundamental property of matter. We understand charge because we can observe lightning, see sparking between wires and we can get an electric shock when things like fabrics rub together in the dry air of winter. In more controlled experiments, we can observe that charged objects exert invisible forces on one another.
Charge can exist in one of two states
Benjamin Franklin, an early researcher in electricity and charge, assigned the label positive to the charges that tend to move the most. We now know those to be negatively-charged electrons. Matter may be positively-charged, negatively-charged or neutral (not charged).
Objects that are neutral may be that way because of their inherent nature, like the neutron, but more commonly they are neutral because they contain equal numbers of negatively-charged particles (electrons) and positively charged ones (protons).
Charge is also a conserved quantity. In normal processes (things we would encounter in day-to-day life) charges are neither created from nothing nor destroyed. When the charge of something changes, it's because it loses or gains a negatively-charged electron.
The universe also tends to balance charges; in a given system (a defined piece of the universe), there are generally the same number of positive and negative charges.
For example, in a salt crystal consisting of positively charged sodium ions (Na+) and negatively-charged chloride ions (Cl-), it is highly unlikely that we will have an unpaired charge. Those always tend to pair up evenly.
Charges exert invisible forces on one another in a specific and predictable way.
This ought to make you pause for a second because it is already vastly different than the gravitational force, our other invisible force. Forces between charges can be attractive or repulsive. How many times have you been walking down the street and gotten ejected from Earth by gravity? Gravity has no repulsive component; it is a purely attractive force.
In the following sections we'll figure out the details of how charges interact and the units of charge. As the flow-chart above shows, we can divide our discussion about charge into static electricity phenomena and electric current. Static electricity refers to charge that doesn't move unless "triggered" to do so. A common example you've experienced is getting a shock when you touch something metal, usually in the winter when the air is dry. Accumulated charge on you discharges rapidly into the metal, a "sink" for electrons, producing the shock. Electric current is moving charge.
In theoretical physics, we often refer to "particles" generically. A particle can mean a proton, an electron, an atom, a small atom or a tiny billiard ball. Exactly what a "particle" is will depend on the context of the problem.
The SI unit of charge is the Coulomb (C), [pronounced kool'·ōm] named after French physicist Charles-Augustin de Coulomb, who developed Coulomb's law (below).
A Coulomb is the amount of charge on 6.242 × 1018 electrons. That's not a number you should feel like you need to know. Just remember that our unit of charge is the Coulomb.
The Coulomb is a fundamental unit, from which we derive a number of other useful units in the fields of electricity and magnetism, including the Ampere (A), which is the measure of electric current in C per second.
The charge of a single electron is q = 1.602 × 10-19 C. The symbol for charge is usually the letter q; sometimes the upper case Q is used.
SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units
length | meter | (m) |
mass | Kilogram | (Kg) |
time | second | (s) |
force | Newton | N |
energy | Joule | J |
The unit of charge is the Coulomb, a fundamental SI unit. The charge of one electron is -1.602 × 10-19 C.
The charge of 1 mole 6.022 × 1023 e- is about 96,500 C.
Once in a while, charge is measured in multiples of the electron charge. A charge of -4e would be the same charge as four electrons. Often, we just refer to that as a charge of -4. The "what" (electrons) is implied.
The two fundamental charged particles in our universe are the proton (p+) and the electron (e-). They have charges of
Neutrons have no electric charge.
particle | charge (C) | mass (Kg) |
---|---|---|
proton p+ | $1.602 \times 10^{-19}$ | $1.67 \times 10^{-27}$ |
neutron n0 | $0$ | $1.67 \times 10^{-27}$ |
electron e- | $-1.602 \times 10^{-19}$ | $9.11 \times 10^{-31}$ |
The force between two charges, $q_1$ and $q_2$ is predicted by Coulomb's law:
$$F_c = \frac{k q_1 q_2}{r^2}$$
where $k$ is the Coulomb constant,
$F_c$ stands for "Coulomb force," but sometimes it's written as $F_{es}$ for "electrostatic force." We can view the constant of proportionality, $k$, as being there to get the units right [so that force has units of Newtons, (N)].
The force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance ($r$) between them.
Coulomb's law should remind you of the universal law of gravitation:
$$F_g = \frac{G m_1 m_2}{r^2}$$
Both are called inverse-square laws because of their dependence on the reciprocal of the square of the distance between charges or masses. This is a very important aspect of these equations. If we double the distance between particles, the force between them (gravitational or electrostatic) decreases by a factor of four (22), not two.
Here is a picture of how the force decreases (or "falls off") as a function of separation:
The electrostatic force (written as $F_C$ or $F_{es}$) between two charges, $q_1$ and $q_2$ is given by Coulomb's law:
$$F_c = \frac{k q_1 q_2}{r^2}$$
where $k = 9.0 \times 10^9 \; N·m^2 \cdot C^{-1}$ is the Coulomb constant, the charges have units of Coulombs (C), and r is the separation between them in meters.
Compare the electrostatic force to the gravitational force between the proton and electron of a hydrogen atom in its ground (lowest energy) state, in which the proton-electron separation is about 5.3 × 10-11 m.
The gravitational force is:
$$ \begin{align} \require{cancel} F_g &= \frac{G m_1 m_2}{r^2} \\[5pt] &= \frac{\left(6.674 \times 10^{-11} \frac{m^3}{\cancel{Kg} s^2}\right)(1.67 \times 10^{-27} \cancel{Kg})(9.11 \times 10^{-31} Kg)}{(5.3 \times 10^{-11} m)^2} \\[5pt] &= {\bf 3.61 \times 10^{-47} \; N} \end{align}$$
The electrostatic or Coulomb force is:
$$ \begin{align} F_{es} &= \frac{k q_1 q_2}{r^2} \\[5pt] &= \frac{\left(8.987 \times 10^9 \, \frac{N\cancel{m^2}}{\cancel{C^2}}\right)(+1.6 \times 10^{-19} \cancel{C})(-1.6 \times 10^{-19} \cancel{C})}{(5.3 \times 10^{-11} \cancel{m})^2} \\[5pt] &= {\bf -8.21 \times 10^{-8} \; N} \end{align}$$
$$ \require{cancel} \frac{m^{\cancel{3}} Kg^{\cancel{2}}}{\cancel{Kg} s^2 \cancel{m^2}} = \frac{Kg \cdot m^2}{s^2} = N$$
$$ \require{cancel} \frac{N \cancel{m^2 C^2}}{\cancel{m^2 C^2}} = N$$
The electrostatic force is negative because a proton and a neutron attract each other, but we're really interested in the size (magnitude) of the force here. Now those forces are quite different. To see just how different, let's divide the larger force by the smaller:
$$\frac{8.21 \times 18^{-8} \, N}{3.61 \times 10^{-47} \, N} = \bf{2.3 \times 10^{39}}$$
The electrostatic force is nearly 40 orders of magnitude (40 zeros) stronger than gravity!
That's an immense difference. The electrostatic force is 2,000,000,000,000,000,000,000,000,000,000,000,000,000 times stronger than the gravitational force.
Calculate the electrostatic force between two spheres of radius r = 5 cm, separated by 1.0 m, one with a charge of +1 C and the other with a -1 C charge.
$$ \begin{align} F_{es} &= \frac{k q_1 q_2}{r^2} \\[5pt] &= \frac{(8.987 \times 10^9 N m^2 C^{-2})(+1 C)(-1 C)}{(1.0 \, m)^2} \\[5pt] &= \bf -8.9 \times 10^9 \; N \end{align}$$
First, notice that the radius of the spheres was a bit of a red herring here – not that useful as long as we're treating our spheres as point charges, a common approximation.
9.9 Tera Newtons is a tremendous force. Little could stop those spheres from smashing into one-another.
What this example really shows is that the Coulomb is a pretty large unit of charge. The charge on a mole of electrons (6 × 1023 electrons) is 96,485 C. That number is referred to as the Faraday constant.
The universe, as far as we know, seems to be balanced with respect to charge; there are as many positive charges as negative. A helium atom, for example, contains two protons and two electrons (and two neutrons with no charge), making it neutral overall.
But just because things are neutral doesn't mean we can't move charges around or cause imbalances. Many of the electronic devices we use depend strongly on our ability to accumulate and discharge charge. Here are some examples of how we can create those imbalances.
This is a pretty familiar trick. You rub your hair for a while with an inflated balloon, and a couple of things happen:
What's going on there?
When some materials are rubbed together, the electron clouds are close enough that electrons can be transferred from one kind of atom to another. Electrons usually transfer from substances (atoms) that hold them more weakly to those that hold them more strongly, thus we can usually predict which way they will flow by friction using measures like electron affinity.
Notice that in the diagram above, while the transfer of one electron from one neutral atom to another created a charge imbalance in both (more protons than electrons on the left, more electrons than protons on the right), the charge balance of the universe is unchanged.
The other critical thing to note is that it is electrons that move, not protons. All charge imbalances are created by the movement of electrons, not protons. It's because protons are about 2,000 times more massive than electrons, and they are held to the nucleus by a force stronger than the electrostatic force, the nuclear force.
Here's another example of static charging. If we rub a rubber rod with some fur or wool, the rubber rod will become negatively charged by picking up electrons from the fur. That rod can be used to touch two light-weight foil-covered balls hanging from threads, as shown.
When the rod touches the balls, electrons are transferred in order to reduce the charge imbalance in the rod. Both balls become equally negatively charged, and therefore repel one another, like this:
The same thing would happen if we removed electrons from the balls causing them both to be positively-charged.
When electrostatic charge changes between objects, it is electrons that move, not protons. Electrons are nearly 2,000 times less massive than protons, which are also tightly bound to the heavy nucleus. A positive charge occurs because electrons have moved away.
Charges attract and repel each other through free space. That is, charges don't have to touch to exert a force on one another. We call that kind of charging induction.
Here's an example. In Figure below, we have a spherical object (say a metal ball) on a insulating stand. It is a neutral object with positive and negative charges evenly distributed around the surface of the sphere.
In Fig. we bring a negatively-charged object close to the sphere, but out of contact. The negative charge on the object repels the negative charges in the sphere, causing the sphere to polarize. With the object held near, the sphere develops a positive side (pole) and a negative pole. It is said to be polarized.
Now (Fig. ) imagine that we attach a wire to the sphere on its negative side, where we have an abundance of electrons. We attach that wire to the ground, an infinitely-large "sink" of electrons, essentially something that has a very large capacity to absorb electrons without becoming significantly charged.
To reduce the charge imbalance, electrons will run through the wire into the ground until the repulsive force between all of those excess electrons on the right side of the sphere diminishes. Using an ammeter, we can detect a current, indicating that charges are moving through the wire.
Once the current diminishes, we can open a switch to break the path from the sphere to ground (Fig. ), electrically isolating the sphere, which is now positively charged – not because we added positive charge, but because we removed negatively-charged electrons.
Remember, it's the negative charges that move.
Some materials can "conduct" electric charge. The most familiar are probably metals, like the copper, silver or gold in the wires used in electronics. Conductors and electric current are covered more fully in another section.
We make a great deal of use out of electric current. We use the kinetic energy of flowing electrons to do work, produce heat and light, to charge and discharge elements in electric circuits, and to flip switches in digital circuits.
Calculate the repulsive force between two electrons separated by a distance of 2 Å [1 Angstrom (Å) = 1.0 × 10-10 m]
First, the charge on an electron is $\bf 1.602 \times 10^{-19} C$.
$$ \require{cancel} \begin{align} F_{es} &= \frac{k q_1 q_2}{r^2} \\[5pt] &= \frac{8.99 \times 10^9 \, N \cancel{m^2} \cancel{C^{-2}}(1.602 \times 10^{-19} \cancel{C})^2}{1.0 \times 10^{-10} \, \cancel{m})^2} \\[5pt] &= \bf 2.31 \times 10^{-8} \; N \end{align}$$
The radius of a neon atom (10 protons in the nucleus) is 38 pm (1 pm = 1.0 × 10-12 m). Compare the attractive force between the outermost electron and the nucleus of a neon atom with the outermost electron of a krypton atom (36 protons in the nucleus). The atomic radius of krypton is 88 pm. Assume that the distance between the outermost electron and the nucleus of these atoms is the atomic radius.
Neon
$$ \begin{align} F &= \frac{8.99 \times 10^9 N \cancel{m^2} \cancel{C^{-2}} (10\cdot 1.602 = \times 10^{-19} \cancel{C})(-1.602 \times 10^{-19} \cancel{C})}{(38 \times 10^{-12} \cancel{m})^2} \\[5pt] &= 1.59 \times 10^{-6} \; N \end{align}$$
Krypton
$$ \begin{align} F &= \frac{8.99 \times 10^9 N \cancel{m^2} \cancel{C^{-2}} (36\cdot 1.602 = \times 10^{-19} \cancel{C})(-1.602 \times 10^{-19} \cancel{C})}{(88 \times 10^{-12} \cancel{m})^2} \\[5pt] &= 1.07 \times 10^{-6} \; N \end{align}$$
Even though krypton has a more positive nuclear charge (more protons), the fact that the distance between the nucleus and the outer electron is squared makes the attractive force between them weaker than for the outer electron of the neon atom. Distance matters more.
An particle of unknown charge is situated 2.0 cm from a particle of charge 1 × 10-18 C. Calculate the charge of the unknown particle if the repulsive force between the two particles is 1.0 N.
$$ \begin{align} F &= \frac{k q_1 q_2}{r^2} \\[5pt] \rightarrow \; q_2 &= \frac{F r^2}{k q_1} \\[5pt] &= \frac{(1.0 N)(2.0 \times 10^-2 \, m)^2}{(8.99 \times 10^9 \, Nm^2C^{-2})(1 \times 10^{-18} C)} \\[5pt] &= \bf 44,493 \; C \end{align}$$
Two spheres, each with a mass of 1.0 × 10-5 Kg are situated 0.250 m apart. One sphere has a charge of -1.5 μC (1 μC = 1.0 × 10-6 C), and the other a charge of -3.0 μC. If the spheres are free to move, calculate the initial magnitude of their acceleration as they move apart.
Method 1 Let's assume that one sphere is fixed in place, and that the other can move. First calculuate the force between the spheres:
$$ \begin{align} F &= \frac{8.99 \times 10^9 Nm^2C^{-2})(-1.5 \times 10^{-6}(-3.0 \times 10^{-6} C))}{(0.250)^2} \\[5pt] \text{Now } \; a &= \frac{F}{m} = \frac{0.647 \, N}{1.0 \times 10^{-5} Kg} \\[5pt] &= \bf 64,728 \, m/s^2 \end{align}$$
Method 2 Method 1 is not quite correct, because we expect these particles both to move away from each other. We really should reduce the problem to a one-body problem by using the reduced mass, μ:
$$\mu = \left( \frac{1}{m_1} + \frac{1}{m_2} \right)^{-1} = \left( \frac{2}{1.0 \times 10^{-5} \, Kg} \right) = 5.0 \times 10^{-6} \; Kg$$
$$a = \frac{F}{m} = \frac{0.647 \, N}{5.0 \times 10^{-6} Kg} = \bf 129,400 \, m/s^2$$
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.