How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude 1.00×104n (roughly one ton)? assume that the spheres may be treated as point charges.

  # = 6.6 10¹⁵ electrons for r = 1 m

# = 6.6 10¹³ electrons  for r= 1 cm

Explanation:

The electric force is given by Coulomb's law

       F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m²/C², q are the charges and r the distance between them.

Let's apply this equation to our case. Initially the spheres are not attracted so the net charge in each of them is zero, removing electrons in one of them is a positive net charge of equal value to the negative charge removed. The specific answer of this exercise depends on the distance of the two spheres, by calculation we assume that it is 1 m

The charge of an electron is q₀ = -1.6 10⁻¹⁹ C, the total charge is

            q = #_electron  q₀

           F = k # q₀ # q₀ / r²

          #² = F r² / k q₀²

          # = √ F r² / k q₀²

Let's calculate for r = 1 m

         # = √ [1 10⁴ 1 / 8.99 10⁹ (1.6 10⁻¹⁹)²]

         # = √ [43.45 10³⁰]

         # = 6.6 10¹⁵ electrons for r = 1 m

If the distance is reduced to r = 1 cm = 1 10⁻² m

The number of electrons is reduced to

         # = 6.6 10¹³ electrons

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