5.3 Coulomb's law  (Page 2/11)

These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, the permittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant:

The force on the electron in hydrogen

A hydrogen atom consists of a single proton and a single electron. The proton has a charge of $\text{+}e$ and the electron has $\text{−}e$ . In the “ground state” of the atom, the electron orbits the proton at most probable distance of $5.29\times {10}^{\mathrm{-11}}\text{m}$ ( [link] ). Calculate the electric force on the electron due to the proton.

Strategy

For the purposes of this example, we are treating the electron and proton as two point particles, each with an electric charge, and we are told the distance between them; we are asked to calculate the force on the electron. We thus use Coulomb’s law.

Solution

Our two charges and the distance between them are,

$\begin{array}{ccc}\hfill q_1& =\hfill & \text{+}e=\text{+}1.602\times {10}^{\mathrm{-19}}\text{C}\hfill \\ \hfill q_2& =\hfill & \text{−}e=\mathrm{-1.602}\times {10}^{\mathrm{-19}}\text{C}\hfill \\ \hfill r& =\hfill & 5.29\times {10}^{\mathrm{-11}}\text{m}.\hfill \end{array}$

The magnitude of the force on the electron is

$F=\frac{1}{4\pi {ϵ}_0}\frac{{\left|e\right|}^2}{r^2}=\frac{1}{4\pi \left(8.85\times {10}^{\mathrm{-12}}\frac{{\text{C}}^2}{\text{N}·{\text{m}}^2}\right)}\frac{{\left(1.602\times {10}^{\mathrm{-19}}\text{C}\right)}^2}{{\left(5.29\times {10}^{\mathrm{-11}}\text{m}\right)}^2}=8.25\times {10}^{\mathrm{-8}}\text{N}.$

As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on the electron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressed as

$\overrightarrow{\text{F}}=\left(8.25\times {10}^{\mathrm{-8}}\text{N}\right)\widehat{\text{r}}.$

Significance

This is a three-dimensional system, so the electron (and therefore the force on it) can be anywhere in an imaginary spherical shell around the proton. In this “classical” model of the hydrogen atom, the electrostatic force on the electron points in the inward centripetal direction, thus maintaining the electron’s orbit. But note that the quantum mechanical model of hydrogen (discussed in Quantum Mechanics ) is utterly different.

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Multiple source charges

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given N charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges.

Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual test charges. Thus, we can calculate the net force on the test charge Q by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition    , and is one of the more important features of the electric force. In mathematical form, this becomes