Problem 60 - University Physics with Modern Physics

Question

(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00\(\mu \mathrm{C}\) and the other a charge of \(-3.50 \mu \mathrm{C},\) with their centers separated by a distance of 0.250 \(\mathrm{m}\) . Assume zero potential energy when the charges are infinitely separated. (b) Suppose that one of the spheres is held in place and the other sphere, which has a mass of \(1.50 \mathrm{g},\) is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere.)

Step-by-Step Solution

Answer

Potential energy: \(-0.252 \; \text{J}\). Escape speed: \(18.3 \; \text{m/s}\).

Step 1: Identify known values and formula for potential energy

First, recognize the given values: charge of the first sphere \( q_1 = 2.00 \times 10^{-6} \; \text{C} \), charge of the second sphere \( q_2 = -3.50 \times 10^{-6} \; \text{C} \), and distance \( r = 0.250 \; \text{m} \). The formula for the electric potential energy \( U \) between two point charges is \( U = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \), where \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_0 = 8.85 \times 10^{-12} \; \text{C}^2/\text{N}\cdot\text{m}^2 \).

Step 2: Calculate potential energy using known values

Substitute the known values into the formula: \[U = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \cdot \frac{(2.00 \times 10^{-6})(-3.50 \times 10^{-6})}{0.250}\]Solving this gives: \[U = -2.52 \times 10^{-1} \; \text{J}\]

Step 3: Understand the condition for escape velocity

To escape, the sphere must reach a velocity of zero at an infinite distance, meaning its initial kinetic energy must equal the magnitude of the potential energy. The kinetic energy \( KE \) formula is \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity.

Step 4: Solve for initial speed using energy conservation

Set the initial kinetic energy equal to the absolute value of the potential energy:\[\frac{1}{2} \times 1.50 \times 10^{-3} \times v^2 = 2.52 \times 10^{-1}\]Solve for \( v \):\[v^2 = \frac{2 \times 2.52 \times 10^{-1}}{1.50 \times 10^{-3}}\]\[v = \sqrt{336.0} \approx 18.3 \; \text{m/s}\]

Step 5: Conclusion

The potential energy of the system is \(-0.252 \; \text{J}\), indicating an attractive force. For the moving sphere to escape, it needs a minimum initial speed of approximately \(18.3 \; \text{m/s}\).

Key Concepts

Potential Energy

In electrostatics, potential energy refers to the energy a charged object possesses due to its position relative to other charged objects. When dealing with point charges, potential energy is the result of the interplay between electrostatic forces.
For two point charges, the potential energy (U) can be calculated using the formula:\[ U = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r} \]where:

• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \( r \) is the distance separating them,
• \( \varepsilon_0 \) is the permittivity of free space.

The sign of the potential energy tells us about the nature of the force. If negative, like in this example, it indicates an attractive force between charges of opposite signs. The potential energy depends on how far apart the charges are and their magnitudes.

Escape Velocity

Escape velocity is the minimum speed needed for an object to "break free" from the gravitational or electrostatic attraction of another object, without any external force contributing after the initial thrust.
In the context of electrostatics, it refers to the speed necessary for a charged particle to leave the electric field of another without being pulled back.
In the given problem, a sphere with mass needs to overcome the attractive force exerted by another charged sphere. Its potential energy must be transformed entirely into kinetic energy at the point of release, while ensuring that the velocity becomes zero at infinite distance:
\[ \frac{1}{2} mv^2 = |U| \]
Here, \( m \) is the mass of the sphere, \( v \) is the escape velocity, and \( U \) is the potential energy. By equating, you can solve for the velocity the sphere requires at initiation.

Point Charges

Point charges are idealized charges situated at a single point in space. This simplification allows physicists and students to easily calculate important interactions without worrying about the complexities introduced by larger objects.
In practice, real objects can often approximate point charges quite well, if the distances involved are much larger than the size of the objects.
The concept is crucial in understanding and calculating forces and potential energy when analyzing electrostatic scenarios. Point charges help to explain the way charged objects interact through their respective electric fields, as well as how these interactions can affect motion and energy exchange, as in this exercise.

Electric Force

The electric force is a fundamental force that occurs between charged objects. It can be attractive or repulsive, depending on the nature of the charges involved.
Opposite charges, such as negative and positive, attract each other. Like charges repel.
The electric force between two point charges can be described by Coulomb's Law:\[ F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \]where:

• \( q_1 \) and \( q_2 \) are the charges,
• \( r \) is the distance between them,
• \( \varepsilon_0 \) is the permittivity of free space.

This law highlights how the force weakens as charges move farther apart, an inverse square relationship. Understanding electric force is key for exploring how charged objects behave and interact over distances in electrostatic fields.