Answer:
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
