A little bit of confusion regarding Coulomb's law

Question

This is a quote from my physics book:

"Coulomb discovered his law without knowing the explicit magnitude of the charge. In fact, it is the other way round: Coulomb’s law can now be employed to furnish a definition for a unit of charge. In the relation, $F = \dfrac{k}{r^2} \big|q_{\ 1} \cdot q_{\ 2}\big|$, $k$ is so far arbitrary. We can choose any positive value of $k$. The choice of $k$ determines the size of the unit of charge".

Can anybody explain why $k$ can be chosen arbitrarily and why the choice of $k$ determines the size of the unit charge?

(If possible, an example will help me to understand this better by choosing the different values of $k$, if $k$ can be chosen arbitrarily).

Answer 1

The equal sign in the equation must be satisfied if the equation is to remain valid. For the case where the magnitude of the unit of charge is unknown, if that magnitude is assumed to be high, the value of "k" must be low enough to keep the equality, and vice versa. Thus, Coulomb was making a mathematical statement, not a physics statement. In addition, once the magnitude of the unit of charge was established by other means, the value of "k" became fixed.

Answer 2

The left hand side is a force with units [kg][m][s]$^{-2}$ so the right hand side must have the same units. In particular, because $r$ has units of [m], it must be that $k q_1q_2$ has units of [kg][m]$^3$[s]$^{-2}$. Obviously $q_1$ and $q_2$ will have the same unit, so this leaves you the freedom to choose the unit of charge and the constant $k$ so the product of their units work out to [kg][m]$^3$[s]$^{-2}$.

For instance, take the units of $q$ to be Coulomb [C]; then $k$ will have units of [kg][m]$^3$[s]$^{-2}$[C]$^{-2}$ and indeed in the SI system $k=\displaystyle\frac{1}{4\pi\varepsilon_0}$ with the units of $\varepsilon_0$ given by [kg]$^{-1}$[m]$^{-3}$[s]$^{2}$[C]$^{2}$ and $\varepsilon_0$ the permittivity of vacuum.

On the other hand, if you choose Gaussian units where $k$ is dimensionless then the unit of charge is the statCoulomb, with [statC]$^2$ having units of [gr][cm]$^3$[s]$^{-2}$, where the units of mass and length as usually taken as the gram and the centimeter for this choice. (Note that, to avoid confusion with the gravitational $g$, I prefer [gr] to [g].)

Answer 3

To add on to what David said, the value of $k$ is constant, but its numerical value is dependent on how you measure charge. For example, you know that the gravitational force is $F=mg$ and $g\approx 10$ in S.I. units (meaning $g=10 \,\frac{\textrm m}{\textrm s^2}$, so you can write $g=10$ if you assume $[m]= \textrm{kg}, [F]= \textrm N$).

However, if you choose to measure your mass in grams, and force in newtons, of course $g$ will not be "10", because

$$10\,\textrm N = 1 \,\textrm{kg}\cdot 10 \,\frac{\textrm m}{\textrm s^2} \Rightarrow 0.01 \,\textrm N = 0.001 \,\textrm{kg}\cdot g \Rightarrow 0.01 \ \textrm N = 1 \,\textrm g \cdot g$$

meaning if you want to give $g$ a numerical value that would make the equation $F=mg$ correct when $[m]= \textrm g, [F]= \textrm N$, you should take $g=0.01$ (of course, $g$ has the same physical units, the value changed due to changing of units).

This is the same thing here: what is the unit of charge measurement?

• Is it the amount of electrons? (Electrons were not discovered yet)
• Is it defined relative to the amount that the force between two charges 1 cm apart will be 1 dyne? (This is called an electro-static unit or e.s.u. for short and when using that definition and measuring distance in cm and force in dyne, $k=1$ by definition)
• Is it defined relative to another amount that is related to different experiments in magnetism?

You decide how you define a "1" in charge units, and that decides what numerical value for $k$ you have.

Answer 4

Assume you live in a time where the concept of "one Coulomb" of charge doesn't exist yet. The concept of charge itself is well known, but there are no numbers connected to it. All you know is that some things carry charge, and they exert forces on other charges. If you take twice as many charge carriers, you get twice the force, so in some sense, the amount of charge influences the force it exerts. But you don't yet have a way to assign a number to a specific amount of charge. But whatever way you choose, that number $q$ should satisfy

$$F\propto\frac{q_1q_2}{r^2}$$

due to experimental evidence. So there is some number $k$ such that

$$F=k\frac{q_1q_2}{r^2}.$$

And this number depends on how we choose to assign a number to an amount of charge. Or we can do it the other way around: The number we choose as $k$ decides how we express charges as numbers! For instance, if the number is $8.85\cdot 10^-12 ~\frac{\mathrm{Nm^2}}{\mathrm C^2}$, then our charges are measured in what we now call Coulombs. But there are other ways to do this. For instance, in the cgs system, we choose $k=1$, and then charge is unitless.

Answer 5

Call the unit of measure of electric charge $C$. $k$ is then the force between two $\textit{unit charges}$ at a distance of $1$ $m$. This way you can measure $k$, by having defined what the unit charge is. We then set $exactly$ the unit charge as $$ 1\;C = \frac{1}{1.602176634\times10 ^{-19}}e $$ where $e$ is (minus) the electric charge of the electron.