Taking the case of a hydrogen atom with the relative masses of proton and electron the electrical interaction (module) $F^C$ is much more intense than the gravitational interaction (module) $F^G$ (on a microscopic scale): the gravitation must be seen as a very weak interaction. Therefore the gravitation is seen as a very weak interaction:
$$\frac{F^G}{F^C}\approx 4.4 \cdot 10^{-40}$$
If $1$ and $2$ are two identical spheres of mass $m$, associated to the same electric charge $q$, (see the image in Italian language),
are in equilibrium suspended from two wires of length $\ell$,
Fig. 1
if $$m=\frac{k_0q^2\cos(\alpha)}{4\ell^2 g\sin^3(\alpha)} \tag 1$$
My question are:
- is it a real case if the system of charges is in equilibrium considering that the gravitational force $\mathbf{F}^G$ equilibrate the electrostatic force $\mathbf{F}^C$ (to see the colored vectors in the figure 1.)?
- If $\alpha \to 0$ then $m\to \infty$. Is it possible to consider physically this case, and what can I write?
Proof of the $(1)$ (out of the question for my students of an high school):
If I consider the gravitational force I have found that:
$$m=\frac{k_0q^2}{4\ell^2 g\sin^2(\alpha)} \tag2$$
where $F^G=F^C \iff Gm^2=k_0q^2$ with $g=Gm/(2r)^2$. Hence $g\cdot 4r^2=Gm$ and $m(g\cdot 4r^2)=k_0q^2$; after I will have, being $r=\ell \sin(\alpha)$,
$$m=\frac{k_0q^2}{4gr^2}=\frac{k_0q^2}{4g\ell^2\sin^2(\alpha)}$$
Being $(1)\neq (2)$, I have taken the weight-force $\mathbf{P}$ instead of the gravitational force $\mathbf{F}^G$ like this figure:
Fig. 2
where
\begin{cases} T_y-mg=0 \iff & T_y=P=mg\\ T_x-F^C=0 \iff & T_x=F^C \end{cases}
After $$\tan(\alpha)=\frac{T_x}{T_y}=\frac{F^C}{mg}=\frac{\frac{k_0q^2}{4r^2}}{mg} \iff m=\frac{k_0q^2}{4gr^2\tan(\alpha)}=\frac{k_0q^2\cos(\alpha)}{4\ell^2 g\sin^3(\alpha)}$$ rembering that $r=\ell \sin\alpha$.
Ok, the case is very. Actually, the whole point of the exercise is to find for what mass $m$ of the two spheres so that at the system at equilibrium present the configuration in the picture. So, this case exists if you give a charge $q$ to a sphere (and also $q$ to the other), suspended by wires of length $\ell$, both spheres of mass $$\frac{k_0q^2\cos \alpha}{4\ell^2 g\sin^3\alpha}$$
As for the second question, the idea is that if you don't change $q$, then $F^C$ (the repulsive Coulombian force) is constant. The limit $\alpha\rightarrow0$ means that the two spheres are getting closed, so its either because a third force is applied moving the two spheres closed, or because the gravitational force $P=mg$ is increasing (so as the mass increases, it prevail over the force $F^C$, drawing the spheres together). The reason is that at the equilibrium $$gm\tan\alpha=F^C=c=\text{constant}\rightarrow m=\frac{c}{g\tan \alpha}$$ so as $\alpha\rightarrow0$, then factor $\tan\alpha$ converges to zero, so $gm$ must diverge to $\infty$ so that $gm\tan\alpha$ is the constant value $c$.