Question

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

Short answer

(a) The work done is 3.36 J. (b) Path is not necessary. (c) Maximum speed is 2.11 m/s.

Step-by-step solution

Calculate Electric Potential at Center

First, we use the formula for electric potential due to a ring at its center, which is derived from the formula for a point charge: \[ V = \frac{kQ}{R} \] where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( Q = 5.00 \, \mu\text{C} = 5.00 \times 10^{-6} \, \text{C} \) is the charge, and \( R = 4.00 \, \text{cm} = 0.0400 \, \text{m} \) is the radius. Plugging in the values,\[ V = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-6}}{0.0400} = 1.12 \times 10^6 \, \text{V} \]

Determine Work Done

The work done to bring a charge \( q \) at this potential point is given by:\[ W = qV \] where \( q = 3.00 \, \mu\text{C} = 3.00 \times 10^{-6} \, \text{C} \).Substituting the values,\[ W = 3.00 \times 10^{-6} \times 1.12 \times 10^6 = 3.36 \, \text{J} \].Thus, the work done to move the charged ball to the center is 3.36 J.

Analyze Path Dependency

(b) Since electric potential is a scalar quantity and only depends on the position (and not the path taken), it is not necessary to take a path along the axis of the ring. Any path leading to the center results in the same amount of work since the potential difference remains constant.

Explore Displacement from Center

(c) If the ball is displaced slightly, it will experience a restoring force due to the electric field of the ring. Since this point is a local equilibrium, it would oscillate back and forth about the center. Maximum speed at center can be calculated using energy conservation, assuming it's released from a small displacement from rest. Maximum kinetic energy is equal to potential energy at displacement:\[ \frac{1}{2} mv^2 = qV \]Where \( m = 1.50 \times 10^{-3} \, \text{kg} \), solving gives:\[ v = \sqrt{\frac{2 \cdot 1.36}{1.5 \times 10^{-3}}} = 2.11 \, \text{m/s} \],Thus, the maximum speed it will reach is 2.11 m/s.

Key concepts

Coulomb's Law

Coulomb's law is a fundamental principle in electromagnetism that describes the force between two charged objects. It states that the electric force (\[ F \]) 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 for this is:

• \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]

where:

• \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the charges.
• \( r \) is the distance between the charges.

Coulomb's law helps explain why the charged ball in the exercise experiences a force when near the ring. The ring's uniformly spread charge creates an electric field that acts as a source of force on the charged ball, bringing it into action as soon as it enters this field, which leads to interesting dynamics such as the work performed on it and its subsequent motion.

Electric Field

An electric field is a region around a charged object where other charges experience a force. It's like an invisible force field.The strength of an electric field (\( E \)) at a point in space can be described by the formula:

• \[ E = \frac{kQ}{r^2} \]

where:

• \( k \) is Coulomb's constant.
• \( Q \) is the charge creating the field.
• \( r \) is the distance from the charge to the point where the field is being calculated.

In the exercise, the charged ball moves within the electric field created by the ring. This field influences the ball by exerting a force on it that can alter its motion. This phenomenon is responsible for the initial work done to bring the ball from far away to the ring's center and its motion when displaced slightly, causing oscillations.

Work-Energy Principle

The work-energy principle connects the concepts of work and energy. It states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it's expressed as:

• \[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]

In the exercise, the work done to bring the charged ball from far away to the center of the ring is calculated using the formula:

• \[ W = qV \]

where:

• \( q \) is the charge of the ball.
• \( V \) is the electric potential at the center of the ring.

This demonstrates how work, in terms of electric potential, results in a transfer of energy to the charged ball, contributing to its kinetic energy upon reaching the center. When the ball is displaced, this stored energy translates into motion, allowing it to oscillate and gain speed.

Oscillations in Electric Fields

Oscillations occur when an object moves back and forth around an equilibrium point. In the context of the exercise, the charged ball generates oscillations in the electric field created by the ring once it's displaced from the center.When an object in such a setup is displaced, it experiences a restoring force from the surrounding electric field, which pulls it back towards the equilibrium position. The ball's maximum speed is reached as it passes through this central point.Using energy conservation principles, the speed can be found by ensuring that the potential energy at maximum displacement converts fully into kinetic energy at the center. Given by:

• \[ \frac{1}{2} mv^2 = qV \]

where:

• \( m \) is the mass of the ball,
• \( v \) is its velocity,
• \) q \( and \( V \) influence how energy is distributed.

This explains how, even with slight displacements, the ball can oscillate with recognizable patterns, demonstrating basic laws of physics within electric fields.