Problem 34 - University Physics with Modern Physics

Question

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) 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 \(\mathrm{g}\) 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?

Step-by-Step Solution

Answer

(a) Work required is 3.37 J. (b) No, path along axis is not necessary. (c) It will oscillate with max speed of 67.07 m/s.

Step 1: Calculate Electric Potential at the Center of the Ring

The electric potential \( V \) at the center of a ring with charge \( Q \) and radius \( r \) is given by \( V = \frac{kQ}{r} \), where \( k \) is the electrostatic constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)). The radius of the ring is \( 4.00 \, \text{cm} = 0.04 \, \text{m} \). With \( Q = 5.00 \, \mu\text{C} = 5.00 \times 10^{-6} \, \text{C} \), we find \( V = \frac{(8.99 \times 10^9)(5.00 \times 10^{-6})}{0.04} = 1.12375 \times 10^6 \, \text{V} \).

Step 2: Calculate Work Done to Move Charge to the Center

The work \( W \) done to move a charge \( q \) through a potential difference \( V \) is given by \( W = qV \). Here, the charge \( q = 3.00 \times 10^{-6} \, \text{C} \). Thus, \( W = (3.00 \times 10^{-6})(1.12375 \times 10^6) = 3.37125 \, \text{J} \).

Step 3: Analyze Path Dependency

Since work done only depends on the initial and final potentials, and the potential at any off-axis point will be the same here due to symmetry, the path along the axis or elsewhere through which the charge moves does not affect the work done. Thus, it's not necessary to take a path along the axis.

Step 4: Movement of Displaced Charge

When displaced from the center, the charge ball will experience a net force towards the center due to electric forces without potential difference, causing oscillations. The charge would oscillate within the potential field as potential energy converts to kinetic energy.

Step 5: Calculate Maximum Speed

To find maximum speed, use energy conservation. The initial potential energy at the center is converted into maximum kinetic energy when fully displaced: \( qV = \frac{1}{2}mv^2 \). Solving for \( v \), \( v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{(2)(3.00 \times 10^{-6})(1.12375 \times 10^6)}{0.0015}} \approx 67.07 \, \text{m/s} \).

Key Concepts

Electric Field

The electric field is a fundamental concept in physics that represents the force that a charged particle exerts on another charged particle. Imagine it as an invisible field of influence surrounding a charged object. In the case of our exercise, we have a ring with a positive charge distributed along its circumference. This ring creates an electric field around it.

In general, the strength of the electric field depends on:

• the amount of charge on the object, and
• the distance from the charged object.

The closer you are to the charged object, the stronger the electric field you'll feel.

Now, for a uniform ring, the electric field at the very center is interesting because, due to symmetry, it will cancel out, resulting in zero net electric field force in the center. However, this is not to be confused with electric potential, which does indeed have a specific value at the center of our ring.

Work and Energy

Work and energy are closely intertwined in physics. Work is done when a force acts on an object to move it a certain distance. In our exercise, we're interested in the work done to move a tiny charged ball to the center of a ring.

To calculate this work, we use the relationship between electrical potential energy and the electric field:

• Work (W) is equal to the charge ( q) times the potential difference ( V) it moves through: \( W = qV \).
• Potential energy transforms as the ball moves within the electric field.

This transformation of energy from the electric potential energy to kinetic energy (or vice versa) highlights the conservation of energy principle.

For instance, moving the ball from a distant place to the center of the ring, the work done by us is stored in the potential energy at the center. If the ball were to move freely from the center with no additional forces, it would convert this potential energy into kinetic energy, gaining speed.

Physics Problem Solving

Solving physics problems requires more than just calculations. It involves understanding the principles and how they apply to a specific scenario. In our exercise, several physics principles come into play.

Start by identifying the key concepts involved, like electric fields, work, and energy. Ask yourself:

• What are the initial and final states of the system?
• What quantities remain constant (like total energy)?
• What must I calculate (such as work done, or potential energy)?

Construction a clear plan: Analyze what you know and break the problem down into smaller steps as demonstrated in the solution.

For instance, relate concepts to real-world situations where applicable. Imagine the ramifications of field lines, forces, or potential differences to predict motions, speeds, and energies involved. Following these steps can be very rewarding as the solution comes together piece by piece.