We know that Coulomb discovered experimentally that the electrical force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance, all this multiplied by Coulomb's constant $k_e$. But how was the value of $k_e$ discovered? Why is $k_e$ equal to $\frac 1 {4\pi\epsilon_0}$?

  • $\begingroup$ Should this be moved to hsm SE? $\endgroup$
    – FGSUZ
    Mar 29, 2020 at 0:55

1 Answer 1


Coulomb used a torsion balance to measure the force between two known charges at a known separation. Knowing the charges, the separation, and the force allows $k_e$ to be computed.

Here are the details, from Wikipedia:

The torsion balance consists of a bar suspended from its middle by a thin fiber. The fiber acts as a very weak torsion spring. If an unknown force is applied at right angles to the ends of the bar, the bar will rotate, twisting the fiber, until it reaches an equilibrium where the twisting force or torque of the fiber balances the applied force. Then the magnitude of the force is proportional to the angle of the bar. The sensitivity of the instrument comes from the weak spring constant of the fiber, so a very weak force causes a large rotation of the bar.

In Coulomb's experiment, the torsion balance was an insulating rod with a metal-coated ball attached to one end, suspended by a silk thread. The ball was charged with a known charge of static electricity, and a second charged ball of the same polarity was brought near it. The two charged balls repelled one another, twisting the fiber through a certain angle, which could be read from a scale on the instrument. By knowing how much force it took to twist the fiber through a given angle, Coulomb was able to calculate the force between the balls. Determining the force for different charges and different separations between the balls, he showed that it followed an inverse-square proportionality law, now known as Coulomb's law.

To measure the unknown force, the spring constant of the torsion fiber must first be known. This is difficult to measure directly because of the smallness of the force. Cavendish accomplished this by a method widely used since: measuring the resonant vibration period of the balance. If the free balance is twisted and released, it will oscillate slowly clockwise and counterclockwise as a harmonic oscillator, at a frequency that depends on the moment of inertia of the beam and the elasticity of the fiber. Since the inertia of the beam can be found from its mass, the spring constant can be calculated.

Note that the force measured by the torsion balance had to be calibrated using a mechanical technique that did not involve electrical charges.


After this answer appeared, the OP added a new question:

Why is $k_e$ equal to $\frac 1 {4\pi\epsilon_0}$?

This is simply an alternate way to write the same constant in Coulomb’s Law. There is no physical significance to the equality; it is true by definition. Sticking in a $4\pi$ here makes it disappear in Maxwell’s equations for electrostatics.

By the way, in Gaussian units $k_e$ is particularly simple: it is exactly $1$ and dimensionless! Gaussian units accomplish this by defining that system’s unit of charge, the statcoulomb, to be such that when two point charges of one statcoulomb are separated by one centimeter, the electrostatic force between them is one dyne.

  • $\begingroup$ Thanks a lot for your answer. I will now look up something about Gaussian units and see if I can understand better $\endgroup$
    – Tom Avery
    Mar 28, 2020 at 18:49
  • $\begingroup$ You didn't actually answer the question. You described (or rather, pasted a Wikipedia description of) Coulomb's experiment, but Coulomb never calculated the k constant in his original experiments because he never knew the exact charges of the materials he used - only the proportionalities. It's true that k can be found if you know the magnitudes of the charges, but people had no way of measuring exact charges in 1785. That's what Tom Avery is asking about; when and how the value of k was first determined, historically. $\endgroup$
    – K.T.B.
    Jul 7, 2022 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.