Problem 3 - College Physics

Question

How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pair of charges to be \(-0.400 \mathrm{J} ?\) (Take the energy to be zero when the charges are infinitely far apart.)

Step-by-Step Solution

Answer

The point charges must be placed approximately 0.933 meters apart.

Step 1: Identify the given information

We are given a \(-7.20 \mu C\) point charge and a \(+2.30 \mu C\) point charge. The electric potential energy (EPE) of this system is \(-0.400 J\). We also know that this energy is taken as zero when the charges are infinitely far apart.

Step 2: Recall the formula for electric potential energy

The electric potential energy of a pair of point charges is given by the formula: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \(U\) is the electric potential energy, \(k\) is Coulomb's constant \(8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\), \(q_1\) and \(q_2\) are the point charges, and \(r\) is the distance between the charges.

Step 3: Substitute the given values into the formula

Substitute \(U = -0.400 J\), \(q_1 = -7.20 \times 10^{-6} C, q_2 = 2.30 \times 10^{-6} C\), and \(k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\) into the formula to solve for \(r\):\[ -0.400 = \frac{8.99 \times 10^9 \cdot (-7.20 \times 10^{-6}) \cdot 2.30 \times 10^{-6}}{r} \]

Step 4: Solve the equation for distance r

Rearrange the equation to solve for \(r\):\[ r = \frac{8.99 \times 10^9 \cdot (-7.20 \times 10^{-6}) \cdot 2.30 \times 10^{-6}}{-0.400} \]Calculate this to find:\[ r = \frac{8.99 \times 10^9 \cdot 16.56 \times 10^{-12}}{0.400}\]\[ r \approx 0.933 \text{ meters} \]

Step 5: Interpret the result

The calculated distance \(r\) of approximately 0.933 meters is the required separation between the two charges so that their electric potential energy equals \(-0.400 J\).

Key Concepts

Coulomb's law

Coulomb's law is a key principle in understanding electromagnetic forces. It describes the electric force between two stationary, point charges. The law states that the magnitude of the 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 relationship is mathematically expressed as:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:

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

Coulomb's law is foundational for calculating the electric potential energy between charged objects. It illustrates how charges interact and influence each other over distances.

Point charges

Point charges are idealized objects that have a charge but no physical size. In physics, they are modeled as single points in space where the charge is concentrated.
The concept of point charges simplifies the study of electrostatics because:

• It allows for the application of mathematical models like Coulomb's law without the complications of geometry,
• It abstracts real-world objects to focus on the behavior of electric forces and fields.

In the original exercise, we dealt with point charges of \(-7.20 \, \mu \text{C}\) and \(+2.30 \, \mu \text{C}\). These point charges attract each other because they have opposite signs, leading to the calculated electric potential energy. When working with point charges, we often imagine them in a vacuum to exclude environmental effects like air resistance.

Distance between charges

The distance between charges is crucial in calculating interactions such as force and potential energy. As distinctively outlined in Coulomb's law, the distance affects both the magnitude of the electric force and the electric potential energy. The equation for electric potential energy \( U \) between two point charges is expressed as: \( U = \frac{k \cdot q_1 \cdot q_2}{r} \).
Here, as the distance \( r \) increases, the potential energy decreases, demonstrating an inverse relationship. The exercise shows how to solve for \( r \) given the electric potential energy \( U \):

• Negative potential energy means the charges attract,
• The larger the distance, the weaker the attraction, thus resulting in less negative energy.

In our calculation, the distance between the charges—ensuring specific energy criteria—is approximately 0.933 meters. This demonstrates the profound influence of separation on electric interactions.