Problem 7 - University Physics with Modern Physics

Question

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{kg}\) is placed on the \(x\) -axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite scparation.) (b) The scond point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) \(5.00 \mathrm{m} ;\) (iii) 50.0 \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

(a) U = 198.74 J; (b) Speeds: 37.42 m/s, 53.01 m/s, and 53.28 m/s.

Step 1: Understanding Electric Potential Energy

The electric potential energy \(U\) for a system of two point charges \(Q\) and \(q\) separated by a distance \(r\) is given by the formula: \[ U = k_e \frac{Qq}{r} \] where \(k_e\) is the Coulomb's constant \(8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2\).

Step 2: Calculating Electric Potential Energy

Substitute the given values: \(Q = 4.60 \times 10^{-6}\, \text{C}\), \(q = 1.20 \times 10^{-6}\, \text{C}\), and \(r = 0.250\, \text{m}\).\[ U = (8.99 \times 10^9) \cdot \frac{(4.60 \times 10^{-6}) \cdot (1.20 \times 10^{-6})}{0.250} \]Calculate to find \(U\) in joules.

Step 3: Substitute and Solve for \(U\)

After substituting the values into the formula, we calculate:\[ U = (8.99 \times 10^9) \cdot \frac{5.52 \times 10^{-12}}{0.250}\]\[ U = 198.74 \text{ J}\]

Step 4: Conservation of Energy Principle

When the charge is released, its potential energy transforms into kinetic energy. The principle of conservation of energy is:\[ U_i + K_i = U_f + K_f \]initially, kinetic energy \(K_i = 0\), and at final point \(K_f = \frac{1}{2}mv^2\). Solving for speed \(v\), " \[ U_i = U_f + \frac{1}{2}mv^2 \] "nwhere \(U_i\) is initial potential energy and \(U_f\) is final potential energy.

Step 5: Calculating Speed at Different Distances

For each of the following distances, we calculate \(U_f\) and solve for \(v\).

Step 6: Sub-step 5a: When Distance is 0.500 m

Using \(r = 0.500\, \text{m}\), \(U_f\) is:\[ U_f = (8.99 \times 10^9) \cdot \frac{5.52 \times 10^{-12}}{0.500} = 99.37 \text{ J}\] " Substitute into energy equation:\[ 198.74 = 99.37 + \frac{1}{2} \cdot 2.80 \times 10^{-4} \cdot v^2\]Solving for \(v\) gives:\[ v = \sqrt{\frac{198.74 - 99.37}{1.40 \times 10^{-4}}} = 37.42 \text{ m/s}\]

Step 7: Sub-step 5b: When Distance is 5.00 m

Using \(r = 5.00\, \text{m}\), \(U_f\) is:\[ U_f = (8.99 \times 10^9) \cdot \frac{5.52 \times 10^{-12}}{5.00} = 1.1 \text{ J}\] " Substitute into energy equation:\[ 198.74 = 1.1 + \frac{1}{2} \cdot 2.80 \times 10^{-4} \cdot v^2\]Solving for \(v\) gives:\[ v = \sqrt{\frac{198.74 - 1.1}{1.40 \times 10^{-4}}} = 53.01 \text{ m/s}\]

Step 8: Sub-step 5c: When Distance is 50.0 m

Using \(r = 50.0\, \text{m}\), \(U_f\) is:\[ U_f = (8.99 \times 10^9) \cdot \frac{5.52 \times 10^{-12}}{50.0} = 0.11 \text{ J}\] " Substitute into energy equation:\[ 198.74 = 0.11 + \frac{1}{2} \cdot 2.80 \times 10^{-4} \cdot v^2\]Solving for \(v\) gives:\[ v = \sqrt{\frac{198.74 - 0.11}{1.40 \times 10^{-4}}} = 53.28 \text{ m/s}\]

Key Concepts

Coulomb's Law

Coulomb's Law provides a vital way to calculate the force between two stationary point charges. This fundamental law indicates that the electric force between these charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. The mathematical expression is:

• \( F = k_e \frac{Qq}{r^2} \)

where \( F \) is the force, \( k_e \) is Coulomb's constant \((8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2)\), \( Q \) and \( q \) are the magnitudes of the charges, and \( r \) is the separation distance between the charges.
This law helps us to understand how powerful the interaction is depending on both the charge amount and the distance apart.
For small distances, the force is much stronger, enforcing the concept of a non-contact force acting through space.

Conservation of Energy

The concept of conservation of energy states that the total energy in an isolated system remains constant. Energy can transform from one form into another—kinetic, potential, thermal, etc.—but cannot be created or destroyed.
In this exercise, potential energy transforms into kinetic energy as the charge moves. The conservation of energy principle is expressed as:

• \( U_i + K_i = U_f + K_f \)

where

• \( U_i \) is initial potential energy,
• \( K_i \) is initial kinetic energy (which is zero as the charge starts from rest),
• \( U_f \) is the final potential energy, and
• \( K_f = \frac{1}{2}mv^2 \) is the final kinetic energy.

This principle guides us to determine how much speed the charge gains as it moves farther from the origin due to the energy shift.

Kinetic Energy

Kinetic energy refers to the energy that a body possesses due to its motion. It plays a crucial role as we calculate how much speed the point charge gains when released.
The formula used to determine kinetic energy is:

• \( K = \frac{1}{2}mv^2 \)

where

• \( K \) is the kinetic energy,
• \( m \) is the mass of the object, and
• \( v \) is its velocity.

In the problem's solution, solving for \( v \) reveals the charge’s final speed at different distances from the origin.
As the charge moves away from the initial point, its potential energy decreases, causing an increase in kinetic energy, hence more speed.

Point Charge Interactions

Point charges are simplified models for charges in physics. We assume them to be located at a single point with no spatial extent. When dealing with interactions:

• They help to model interactions and forces in electrostatics using Coulomb's Law.
• Point charges' influences can vary depending on the distance between the two.

This exercise demonstrates how point charge interactions lead to changes in energy states. The closer the charges, the more potent their interaction, which is why they experience higher forces and greater potential energy initially.
Understanding these interactions is crucial for fields like electromagnetism, electronics, and even quantum mechanics, where particles may behave similarly to point charges due to their small size and specific conditions.