Answer to Question #105766 in Electricity and Magnetism for Ali

Question

The gravitational force between two concentrated ("point like") masses is very similc% in its
mathematical stricture to the electrostatic force between two concentrated charges. The •strenotn• of
these two forces is, however, vastly different. To illustrate this. consider the following exanpte.
Somewhere in outer space are two identical spherical dust grains, 50 gm in diameter,
density 2.5 gtcma. They are at a distance d meters apart If the grains were electrically neut•ai. free of
other external forces, and have negligible relative velocity initially, they would eventually collide
Now suppose that both grains are electrically charged, each having n •extra• electrons Fine the
minimum value Of n that would prevent the gravitational collision. Compare this with the approximate
total number of electrons contained in one grain.

Answer 1

First, let us define the units. For instance, the diameter is in "gm". What is it? Gigameter is Gm, g⋅m is gram-meter, is not a unit for diameters. Let the diameter be 50 mm (millimeters), or 0.05 m (this is a SI value and will be used in calculations). The SI unit for density is kg/m³, it can also be g/m³ or g/cm³, so let the density be 2.5 g/cm³, or 2500 kg/m³ (2500 is a SI value and will be used in calculations).

The mass of each grain is

$m=\rho V=\rho\cdot\frac{1}{6}\pi D^3.$

The force of gravitational attraction between two equal grains is

$F_G=G\frac{m^2}{d^2}=G\frac{\rho^2\pi^2 D^6}{36d^2}.$

The Coulomb's force for each grain with n extra electrons with the charge e each is

$F_C=\frac{1}{4\pi\epsilon_0}\frac{q^2}{d^2}=\frac{1}{4\pi\epsilon_0}\frac{(ne)^2}{d^2}.$

These forces must be equal:

$G\frac{\rho^2\pi^2 D^6}{36d^2}=\frac{1}{4\pi\epsilon_0}\frac{(ne)^2}{d^2},\\ \space\\ G\frac{\rho^2\pi^2 D^6}{9}=\frac{1}{\pi\epsilon_0}(ne)^2,\\ \space\\ n=\frac{\rho\pi D^3}{3e}\sqrt{\epsilon_0G\pi}=87999084$

excess electrons.