Item 4


Two 2.0 masses are 1.0 apart on a frictionless table. Each has 1.0 of charge

Part A -

What is the magnitude of the electric force on one of the masses?

Express your answer using two significant figures.


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Part B -

What is the initial acceleration of each mass if they are released and allowed to move?

Express your answer using two significant figures.

To find the magnitude of the electric force on one of the masses, we can use Coulomb's Law.

Coulomb's Law states that the magnitude of the electric force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them.

Given:
- Charge of each mass: 1.0 C
- Distance between the masses: 1.0 m

Using Coulomb's Law:
Electric force = (charge 1 * charge 2) / (distance^2)

Substituting the given values:
Electric force = (1.0 C * 1.0 C) / (1.0 m)^2

Simplifying:
Electric force = 1.0 C^2 / 1.0 m^2

The magnitude of the electric force on one of the masses is 1.0 C^2 / 1.0 m^2.

Moving on to Part B:

If the table is frictionless and the masses are released, they will experience an electric force and accelerate. The electric force will cause acceleration according to Newton's second law, F = ma, where F is the force and a is the acceleration.

In this case, the electric force is the net force acting on each mass, and the mass is 2.0 kg (as specified by "Two 2.0 masses").

Using Newton's second law:
Electric force = mass * acceleration

Since the electric force is given by Coulomb's Law as 1.0 C^2 / 1.0 m^2, and the mass is 2.0 kg, we can rearrange the equation to solve for acceleration:

Acceleration = Electric force / mass

Substituting the given values:
Acceleration = (1.0 C^2 / 1.0 m^2) / 2.0 kg

Simplifying:
Acceleration = 0.5 C^2 / m^2 * kg

The initial acceleration of each mass, when released and allowed to move, is 0.5 C^2 / m^2 * kg.

Part A - To find the magnitude of the electric force on one of the masses, we can use Coulomb's Law. Coulomb's Law states that the electric force between two charges is equal to the product of the charges divided by the square of the distance between them, multiplied by a constant.

Given:
Charge of each mass (q1 and q2) = 1.0 C
Distance between the masses (r) = 1.0 m

Using Coulomb's Law:
Electric force (F) = (k * q1 * q2) / (r^2)

Where k is the electrostatic constant, which has a value of 9.0 x 10^9 N·m^2/C^2.

Calculating the electric force:
F = (9.0 x 10^9 N·m^2/C^2) * (1.0 C) * (1.0 C) / (1.0 m)^2

F = 9.0 x 10^9 N·m^2/C^2

Therefore, the magnitude of the electric force on one of the masses is 9.0 x 10^9 N·m^2/C^2, expressing the answer using two significant figures.

Part B - Since the masses are released and allowed to move, they will experience an initial acceleration due to the electric force acting on them. The electric force and the acceleration are related by Newton's second law, which states that the force on an object is equal to its mass multiplied by its acceleration.

Given:
Mass of each object (m1 and m2) = 2.0 kg

Using the equation F = m*a, we can rearrange it to solve for acceleration:
a = F / m

Calculating the acceleration:
a = (9.0 x 10^9 N·m^2/C^2) / (2.0 kg)

a ≈ 4.5 x 10^9 m/s^2

Therefore, the initial acceleration of each mass, when released and allowed to move, is approximately 4.5 x 10^9 m/s^2, expressing the answer using two significant figures.