Problem 12 - College Physics

Question

It takes +6.0 J of work to move two charges from a large distance apart to \(1.0 \mathrm{~cm}\) from one another. If the charges have the same magnitude, (a) how large is each charge, and (b) what can you tell about their signs?

Step-by-Step Solution

Answer

(a) Each charge is approximately \(2.58 \times 10^{-6} \mathrm{C}\); (b) both charges have the same sign.

Step 1: Understanding Coulomb's Law

The work done in bringing two charges from a large distance to a distance of \( r = 0.01 \, \text{m} \) is given by the change in electric potential energy, which can be derived from Coulomb's law. Coulomb’s law states that the magnitude of the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is \( F = \frac{k |q_1 q_2|}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). The electric potential energy at distance \( r \) is \( U = \frac{k q_1 q_2}{r} \).

Step 2: Calculate Electric Potential Energy Change

We are given that the work done \( W = 6.0 \, \text{J} \). The work done is equal to the final potential energy when the charges are brought close from infinity: \( W = \frac{k q_1 q_2}{r} \). Since \( q_1 = q_2 = q \), we have \( W = \frac{k q^2}{r} \).

Step 3: Solve for Charge Magnitude

Plug in the given values and solve for \( q \):\[ 6.0 = \frac{(8.99 \times 10^9) q^2}{0.01} \]Rearranging gives:\[ q^2 = \frac{6.0 \times 0.01}{8.99 \times 10^9} \]Calculate \( q \):\[ q = \sqrt{\frac{6.0 \times 0.01}{8.99 \times 10^9}} \approx 2.58 \times 10^{-6} \, \text{C} \]

Step 4: Determine Charge Signs

Since the work done is positive, the electric potential energy increases, indicating that work needs to be done against the repulsive force. This means the charges have the same sign; either both are positive or both are negative.

Key Concepts

Electric Potential Energy

Electric potential energy is a crucial concept in the study of electrostatics. It helps us understand how energy is stored between charged particles. When two charges, say like a pair of charged spheres, are moved closer together or farther apart, this kind of energy changes.

Electric potential energy arises because charged particles either attract or repel one another. If you push two like charges closer together, you do work against the repulsive force, increasing their potential energy. Conversely, if you allow them to separate, their potential energy decreases as it is converted into kinetic energy.

In mathematical terms, the electric potential energy (U) between two point charges can be expressed as:

• \( U = \frac{k q_1 q_2}{r} \)
• where \( U \) is the potential energy in joules, \( k \) is the Coulomb constant \((8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.

This equation shows that potential energy is inversely proportional to the distance \( r \), meaning as charges get closer, potential energy increases.

Charge Magnitude

Understanding the magnitude of a charge is essential to predicting how charges will interact. Charge magnitude refers to the amount of electric charge an object carries. This quantity is typically measured in coulombs (C). In our exercise, we deal with charges of the same magnitude, which simplifies calculations.

The magnitude determines the strength of the charge's electric field and, subsequently, the electric force it can exert on another charge. The charge's magnitude does not provide information on whether it is positive or negative, just on the quantity of the charge.

In the given problem, we derived the magnitude by using the formula for electric potential energy. Equipped with the electric potential energy and distance, knowing that both charges have the same magnitude, we solved for a single charge \( q \):

• Given: \( W = \frac{k q^2}{r} \)
• Solve for \( q \) gives: \( q = \sqrt{\frac{W \, r}{k}} \approx 2.58 \times 10^{-6} \, \text{C} \)

This calculation determines that each charge has a magnitude of approximately \( 2.58 \times 10^{-6} \, \text{C} \).

Electric Force

Electric force is the interaction force that occurs between two charged objects. This force can either be attractive or repulsive depending on the nature of the charges involved. According to Coulomb's Law, the electric force \( F \) between two point charges is:

• \( F = \frac{k |q_1 q_2|}{r^2} \)

Here, \( k \) is the Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation distance. This force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This means the closer and larger the charges, the stronger the force.

In our exercise, the electric force plays a role in understanding why work is needed to bring two like charges together. They repel each other, necessitating work to decrease the separation distance. Hence, doing positive work against the repulsive force indicates the like sign of the charges.