A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having a charge of −− 3 μCμC is positioned directly above the first sphere. The distance between the two spheres is 0.3 mm . Part A What is the reading on the scale? Express your answer in newtons to three significant figures.

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

         F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

be subject to a force andchange its motion