Question

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.

Answer 1

To find the reading on the scale, we need to calculate the electrostatic force between the two spheres and then subtract it from the weight of the first sphere. The reading on the scale can be calculated using Coulomb's Law and Newton's second law of motion.

Step-by-step explanation:

The reading on the scale can be calculated using Coulomb's Law and Newton's second law of motion. To find the reading on the scale, we need to calculate the electrostatic force between the two spheres and then subtract it from the weight of the first sphere. The electrostatic force between the two spheres is given by Coulomb's Law:

F = k * (q1 * q2) / r^2

where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the spheres, and r is the distance between the spheres.

Given that q1 = 8 μC, q2 = -3 μC, and r = 0.3 mm, we can calculate the electrostatic force using the formula above. Then, we subtract this force from the weight of the first sphere (which is equal to its mass multiplied by the acceleration due to gravity) to find the reading on the scale.
The weight of the first sphere is given by:

Weight = mass * g

where mass is 5 kg and g is the acceleration due to gravity (approximately equal to 9.8 m/s^2).

By calculating the electrostatic force and subtracting it from the weight of the first sphere, we can determine the reading on the scale.

Answer 2

F_Balance = 46.6 N ,m' = 4,755 kg

Step-by-step 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