-1
$\begingroup$

Which was the first Coulomb's constant value? I didn't found any info in the Internet. I need the first value of K to compare it with my experiment.

Could you please help me?

$\endgroup$
  • 1
    $\begingroup$ Coulomb's constant depends on our choice of units. We can make it 1 or $1/4\pi$, if you wish. When written with this arbitrary choice in mind, Coulombs law can be expressed using the proportionality symbol rather than a choice of a constant. $\endgroup$ – CuriousOne Mar 13 '16 at 19:21
  • $\begingroup$ Oh, I see. I need to compare my results with this form: k=8.9875517873681764×10^9 N·m^2/C^2, so I need the first value in the History because the nature of our procedures. Do you know it? $\endgroup$ – user281697 Mar 13 '16 at 19:33
  • 2
    $\begingroup$ The nature of your procedures? You have an experiment that is precise to 16 digits, yet links back to Coulomb's original work? Wow... ;-) $\endgroup$ – CuriousOne Mar 13 '16 at 19:38
  • $\begingroup$ jaja no! our results were so bad... honestly we didn't control some environmental factors and other errors, so we need the less exact form of K! :( $\endgroup$ – user281697 Mar 13 '16 at 19:48
  • 1
    $\begingroup$ Using the wrong Coulomb constant is not going to save your rear ends. If you mess up an experiment, you admit that and learn from your mistake. Nobody expects you to be closer than one or two digits in a high school or college experiment, anyway. If you can't make one digit... well, that's bad. :-) $\endgroup$ – CuriousOne Mar 13 '16 at 19:54
1
$\begingroup$

I think the exact value for Coulomb's constant was found first mathematically using the equation $k=1/4\pi\epsilon_0$ where $\epsilon_0$ is the permittivity of free space. I believe there is a way to derive the permittivity of free space by using the speed of light.But you can just use the universally accepted value of $\epsilon_0$ which is

$\epsilon_0 = 8.85418782 \times 10^{-12} m^{-3} kg^{-1} s^4 A^2$

Hope that this helps you find a value of k for you to compare to your experimental value.

$\endgroup$
  • $\begingroup$ As you can see from the units of $\epsilon_0$, its numerical value depends on your choice of units for distance, mass, time and charge. You can make it whatever you like by e.g. choosing your unit of charge suitably. In Gaussian units Coulomb's law reads $F=q_1q_2/r^2$. There is nothing to "find". It's a matter of definition. The unit charge in these units is $1 statC=3.33564 \times 10^{−10} C$, which happens to be a rational choice of charge. A $C$ is way large for most applications involving charge (but $1Ampere$ is a useful and rational measure for current, though). $\endgroup$ – CuriousOne Mar 13 '16 at 22:48
  • $\begingroup$ This answer is not responsive. It merely repeats the current accepted value. $\endgroup$ – garyp Mar 14 '16 at 0:05
  • $\begingroup$ I understand @CuriousOne . Just thought, id give the mathematical relationship for k using permittivity of free space, since it might give a good value for comparison with an experimental value, i guess? $\endgroup$ – Armando Ramdass Mar 17 '16 at 0:49

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