PHYS102: Introduction to Electromagnetism

Coulomb’s Law

$F=k\frac{\left | q_{1}q_{2} \right |}{r^{2}}$ [equation 18.3]

Coulomb’s law calculates the magnitude of the force $F$ between two point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$. In SI units, the constant $k$ is equal to

$k=8.988\times 10^{9}\frac{N\cdot m^{2}}{C^{2}}\approx 8.99\times 10^{9}\frac{N\cdot m^{2}}{C^{2}}$ [equation 18.4]

The electrostatic force is a vector quantity and is expressed in units of newtons. The force is understood to be along the line joining the two charges. (See Figure 18.19.)

Example 18.1 How Strong is the Coulomb Force Relative to the Gravitational Force?

Compare the electrostatic force between an electron and proton separated by $0.530\times 10^{-10}$ with the gravitational force between them. This distance is their average separation in a hydrogen atom.

Strategy

To compare the two forces, we first compute the electrostatic force using Coulomb’s law, $F=k\frac{\left | q_{1}q_{2} \right |}{r^{2}}$. We then calculate the gravitational force using Newton’s universal law of gravitation. Finally, we take a ratio to see how the forces compare in magnitude.

Solution

Entering the given and known information about the charges and separation of the electron and proton into the expression of Coulomb’s law yields

$F=k\frac{\left | q_{1}q_{2} \right |}{r^{2}}$ [equation 18.5]

$=(8.99\times 10^{9}\: N\cdot m^{2}/C^{2})\times \frac{(1.60\times 10^{-19}C)(1.60\times 10^{-19}C)}{(0.530\times 10^{-10}m)^{2}}$ [equation 18.6]

Thus the Coulomb force is

$F=8.19\times 10^{-8}N$ [equation 18.7]

The charges are opposite in sign, so this is an attractive force. This is a very large force for an electron – it would cause an acceleration of $8.99\times10^{22}\: m/s^{2}$ (verification is left as an end-of-section problem).The gravitational force is given by Newton’s law of gravitation as:

$F_{G}=G\frac{mM}{r^{2}}$ [equation 18.8]

where $G=6.67\times 10^{-11}N\cdot m^{2}/kg^{2}$. Here $m$ and $M$ represent the electron and proton masses, which can be found in the appendices. Entering values for the knowns yields

$F_{G}=(6.67\times 10^{-11}N\cdot m^{2}/kg^{2})\times \frac{(9.11\times 10^{-31}kg)(1.67\times 10^{-27}kg)}{(0.530\times 10^{-10}m)^{2}}=3.61\times 10^{-47}N$ [equation 18.9]

This is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. The ratio of the magnitude of the electrostatic force to gravitational force in this case is, thus,

$\frac{F}{F_{G}}=2.27\times 10^{39}$ [equation 18.10]

Discussion

This is a remarkably large ratio! Note that this will be the ratio of electrostatic force to gravitational force for an electron and a proton at any distance (taking the ratio before entering numerical values shows that the distance cancels). This ratio gives some indication of just how much larger the Coulomb force is than the gravitational force between two of the most common particles in nature.