Problem 84 - University Physics with Modern Physics

Question

CP A small sphere with mass 9.00\(\mu g\) and charge \(-4.30 \mu C\) is moving in a circular orbit around a stationary sphere that has charge \(+7.50 \mu \mathrm{C}\) . If the speed of the small sphere is \(5.90 \times 10^{3} \mathrm{m} / \mathrm{s},\) what is the radius of its orbit? Treat the spheres as point charges and ignore gravity.

Step-by-Step Solution

Answer

The radius of the orbit is approximately 3.0 mm.

Step 1: Understand the Problem

We are given a system of two charges, one stationary with charge \( +7.50 \mu C \) and one moving with charge \( -4.30 \mu C \) in a circular orbit. The task is to find the radius of the orbit, knowing the speed of the moving charge is \( 5.90 \times 10^{3} \mathrm{m/s} \).

Step 2: Identify Relevant Principles

The centripetal force needed for circular motion is provided by the electrostatic force between the two charges. We can equate the formula for centripetal force to the Coulomb's law expression for electrostatic force to find the radius.

Step 3: Use Coulomb's Law for Electrostatic Force

Coulomb's Law for the force between two charges is given by:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where \( k = 8.99 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \) is the electrostatic constant, \( q_1 = +7.50 \mu C = +7.50 \times 10^{-6} \, C \) and \( q_2 = -4.30 \mu C = -4.30 \times 10^{-6} \, C \) are the charges, and \( r \) is the radius.

Step 4: Establish Centripetal Force Expression

The centripetal force needed to keep the small sphere in a circular orbit is:\[ F = \frac{m \cdot v^2}{r} \]where \( m = 9.00 \mu g = 9.00 \times 10^{-9} \, kg \) is the mass and \( v = 5.90 \times 10^3 \, m/s \) is the speed of the small sphere.

Step 5: Equate Centripetal and Electrostatic Forces

Set the expressions from Steps 3 and 4 equal to each other:\[ \frac{m \cdot v^2}{r} = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]

Step 6: Solve for Radius

Rearrange the equation from Step 5 to solve for \( r \):\[ r = \frac{k \cdot |q_1 \cdot q_2|}{m \cdot v^2} \]Substitute in the values:\[ r = \frac{8.99 \times 10^9 \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \cdot 7.50 \times 10^{-6} \, C \cdot 4.30 \times 10^{-6} \, C}{9.00 \times 10^{-9} \, kg \cdot (5.90 \times 10^3 \, m/s)^2} \]

Step 7: Calculate the Radius

Calculating the above expression yields:\[ r = \frac{8.99 \times 10^9 \cdot 32.25 \times 10^{-12}}{9.00 \times 10^{-9} \cdot 34.81 \times 10^6} \approx 0.003 \text{ m (or 3.0 mm)} \]

Key Concepts

Electrostatic Force

Electrostatic force is one of the fundamental forces that governs the interactions between charged particles. It is described by Coulomb's Law, which states that the magnitude of the force between two point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them. The formula is given by:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the amounts of the charges, and \( r \) is the distance between the charges.

• Attraction and Repulsion: Charges can attract or repel each other. Opposite charges attract, as seen in the problem with the positive and negative sphere charges.
• Force Direction: The direction of the electrostatic force is along the line joining the charges.

In our exercise, the electrostatic force keeps the smaller sphere in circular motion around the stationary charge. By understanding how this force is calculated using Coulomb's Law, we can infer how such forces impact the motion of charged particles.

Circular Motion

Circular motion refers to the motion of an object along the circumference of a circle or rotation along a circular path. For an object to stay in circular motion, a constant force must act on it, directing it towards the center of its circular path.

• Uniform Circular Motion: When an object moves in a circle at a constant speed.
• Non-uniform Circular Motion: When the speed of the object changes as it moves along its path.

In the given problem, the smaller charged sphere requires a force to keep it moving in a perfect circle around the stationary sphere. This force is the electrostatic force, which acts like a string pulling it inward, preventing it from flying off the path. Understanding circular motion helps in determining the behavior of such systems.

Centripetal Force

Centripetal force is essential in circular motion, providing the necessary force to keep an object moving in a path with a constant radius. This force is always directed towards the center of the circle.

• Definition: Given by \( F = \frac{m \cdot v^2}{r} \), where \( m \) is the object's mass, \( v \) is the velocity, and \( r \) is the radius of the circular path.
• Real-Life Examples: Examples of centripetal forces include gravitational forces in planetary orbits, tension in a string for a spinning object, or the electrostatic attraction in our problem.

In the example exercise, the centripetal force required for the circular motion of the charged sphere is provided by the electrostatic force between the charged spheres. By equating the net inward force to the electrostatic force, we can find the radius of the sphere’s path.

Charge Interaction

Charge interaction involves the forces that arise between charged particles. It is the basis for understanding many phenomena in electromagnetism, including our problem.

• Basic Principle: Like charges repel while opposite charges attract, producing an interaction force between them.
• Applications: Charge interaction is central to technologies like capacitors and electrostatic precipitators.

In the exercise, the interaction between the positive stationary sphere and the negative moving sphere generates an attractive force. This force is responsible for the circular motion of the moving sphere, illustrating how charge interaction governs their dynamics. Recognizing charge interaction helps students understand the significance of forces in electrical systems.