It's often thought the force between to charges is given by $F = \frac{q\,Q}{4\,\pi\varepsilon_0}\,\frac{1}{r^2}$, because observations lead to this. On the other side I've barely heard about how it's consistent with reality. In fact, I've never heard a professor say it's not valid for tiny distances (force would go up to infinity). Here on stacks I've read Quantum mechanics will take the lead then. But even for normal distances (often set by $1\,\mathrm{m}$) the forces seem to be huge $(\approx 10^6 \,\mathrm{tons})$. I wonder how one can evenly move 2 small charges close to each other without getting an arm ripped off. How are these forces compensated?
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9$\begingroup$ Are you plugging in $1\;\text{C}$ for the charge? That's an unrealistic number. $\endgroup$– HiddenBabelCommented Apr 21, 2021 at 11:42
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$\begingroup$ The extreme situation in nature would be two (or more) protons brought together within a nucleus. $\endgroup$– R.W. BirdCommented Apr 21, 2021 at 13:33
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$\begingroup$ This is 'very often thought' since 1785. $\endgroup$– my2ctsCommented Apr 21, 2021 at 21:03
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1 Answer
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The typical order of magnitude of an electric charge is $10^{-6}$ C or even $10^{-9}$ C in everyday life. For this reason, often, we see that the default unit for electric charges in Lab is a Nano Coulomb (nC). In your example, $10^6$ tons equals to $10^{9}$ kg (or $10^{10}$ N in Earth), and this means you have set the electric charges as $q=Q=1$ C. You assumed that $1$ Coulomb is a small charge, which indeed is not the case; It is a very big charge!
But, in nature, you may observe some phenomena in which an enormous amount of electric charge is transferred. For example, lightning carries an electric current of $30$ kA ($10^3$ Ampere) and transfers $15$ C of electric charge and $10^{9}$ J of energy (see WikiPedia for this example). And, the electric discharge in this phenomenon releases a huge amount of energy.
But, the problem is how to put this huge amount of the electric charge in one place at a distance of one meter from the same amount of charge. It needs a lot of work/force to provide the requirements of this test.
In addition, according to the relation of quantization of electric charge ($q = \pm ne$), the total amount of electrons/protons that you need to have $1$ C is equal to
$$n = \frac{{\Delta q}}{{\left| e \right|}} \approx 6 \times {10^{18}},$$
and, of course, collecting this amount is not easy (where $\left| e \right|$ is the elementary charge of electron).
