Problem 58 - 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

(a) The potential energy is \(-0.25172 \mathrm{J}\). (b) The minimum speed needed is \(18.32 \, \mathrm{m/s}\).

Step 1: Calculate Potential Energy

The potential energy \(U\) between two point charges can be calculated using the formula: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2) \), \(q_1\) and \(q_2\) are the charges \((2.00 \, \mu\mathrm{C} \text{ and } -3.50 \, \mu\mathrm{C})\), and \(r\) is the separation distance \((0.250 \, \mathrm{m})\). Substituting the values, we have: \[ U = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-6} \times (-3.50 \times 10^{-6})}{0.250} \]. Calculating this, \[ U = -0.25172 \, \mathrm{J} \].

Step 2: Determine Minimum Initial Speed for Escape

The kinetic energy needed to overcome the potential energy and allow the sphere to escape is equal to the magnitude of the potential energy calculated. Therefore, \( K = 0.25172 \) J. The kinetic energy \(K\) is given by \( K = \frac{1}{2} mv^2 \), where \(m\) is the mass \(1.50 \, \mathrm{g} = 0.0015 \, \mathrm{kg}\). Solving for \(v\), we have: \[ \frac{1}{2} \times 0.0015 \times v^2 = 0.25172 \]. Simplifying: \( v^2 = \frac{0.25172 \times 2}{0.0015} = 335.6267 \). Thus, \(v = \sqrt{335.6267} \approx 18.32 \, \mathrm{m/s} \).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that describes the force between two charged objects. This law is used to determine the potential energy between two point charges in your physics problems. It tells us how the electrostatic force varies with the magnitude of the charges and the distance between them. The formula for the force is given by:- \[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \] This equation might remind you of gravity’s inverse-square law. However, in electrostatics, charges can repel or attract, depending on their types:* **Like charges** (both positive or both negative): **repel** each other. * **Opposite charges** (one positive, one negative): **attract** each other. Understanding Coulomb's Law is critical, because it not only allows us to calculate forces but the associated potential energy as well. Potential energy, represented as \( U \), can be linked to the charges and the distance:- \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] Notice the similarity to the force formula? Instead of \( r^2 \), potential energy inversely depends on \( r \), which changes how we compute energy versus force. Let's recap:- **Key constant**: \( k \) (Coulomb's constant) - \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\).- **Interaction**: Determine whether charges attract or repel.Grasping these principles gives a sound foundation for understanding complex electrostatic interactions!

Point Charges

Point charges are idealized charges that help us simplify the complex interactions of electrostatic forces in physics. Imagine them as tiny spots or particles carrying a charge in an otherwise empty space. They are invaluable in calculations involving electrostatic fields and forces, enabling us to apply formulas easily.**Why are point charges important?**They serve as:- Fundamental elements, making complex problems more manageable.- Simplified models for real-world objects like electrons or ions.When working with point charges:- Assume the charge is concentrated at a single point regardless of the object's actual size.- Treat all interactions as happening at a distance \( r \) between them.In the original exercise, we deal with two point charges:- One charge of \(2.00 \mu \mathrm{C}\), and another of \(-3.50 \mu \mathrm{C}\).These charges seem to occupy no space, but their interactions capture the fundamental principles of electrostatics. The simplicity of point charges allows for easy interaction modeling over a range of distances, making it possible to apply concepts like Coulomb’s Law accurately.By understanding point charges, you demystify many electrostatic phenomena, providing a clearer pathway to practical problem-solving.

Kinetic Energy

Kinetic energy is a key concept when dealing with moving objects or particles. In physics, it represents the energy an object possesses due to its motion and is expressed as:- \[ K = \frac{1}{2}mv^2 \]Where:* \( m \) is the mass of the object.* \( v \) is the velocity at which the object is moving.In the context of the given exercise, when the sphere is shot away, it requires kinetic energy sufficient to escape the attraction of the other charged sphere. This needs to be equal to the magnitude of the potential energy calculated earlier:- **Kinetic energy needed**: \( 0.25172 \) J.To find the minimum speed \( v \) required for the sphere to escape, use the kinetic energy formula. Rearrange it to solve for velocity:- \[ v = \sqrt{\frac{2K}{m}} \]This calculation ensured that our sphere has just enough energy to escape, reaching a speed of zero at an infinite distance, known as escape velocity. Remember, achieving escape velocity guarantees that the moving object breaks free from the other’s electrostatic grip, reflecting how kinetic and potential energy interplay in dynamic systems.