Two spherical objects are separated by a distance that is 1.94 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 7.61 x 10-21 N. How many electrons did it take to produce the charge on one of the objects

   n = 1  10¹⁰  electron

Explanation:

For this exercise we will use Coulomb's law

          F = k q₁ q₂ / r²

where it indicates that the value of the force is 7.61 10⁻²¹ N, the distance between the objects is r = 1.94 10⁻³ m and the charge on the objects has the same charge q₁ = q₂ = q,

           F = k q² / r²

            q =

let's calculate

          q =

          q = √(3.182 10⁻¹⁸)

           q = 1.783 10⁻⁹ C

the charge of an electron is q₀ = 1.6 10⁻¹⁹ C, therefore

         q = n q₀

         n = q / q₀

         n = 1,78 10⁻⁹ / 1.6 10⁻¹⁰

         n = 1.06   10¹⁰ electron

the number of electrons must be integer

        n = 1  10¹⁰  electron

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.