Calculating Potential Difference For A Bullet To Reach 10 M/s

This article delves into an intriguing physics problem that combines concepts from mechanics and electromagnetism. Specifically, we explore the scenario of a bullet, possessing both mass and charge, accelerating under the influence of an electric potential difference. Our primary goal is to determine the magnitude of this potential difference required for the bullet, starting from rest, to achieve a final speed of 10 m/s. This problem not only reinforces fundamental principles but also highlights the interplay between electric fields and the motion of charged particles. We will explore the physics principles involved, apply relevant equations, and arrive at a solution, offering a comprehensive understanding of the concepts at play. Let's embark on this journey of discovery, unraveling the mysteries of electromagnetism and mechanics as they converge in this fascinating problem.

H2: Problem Statement: A Charged Bullet in Motion

Before diving into the solution, let's clearly define the problem at hand. Imagine a bullet, a tiny projectile with a mass of 2 grams (g), carrying an electric charge of 4 microcoulombs (µC). This bullet is initially at rest. Our challenge is to determine the potential difference it needs to traverse to attain a final velocity of 10 meters per second (m/s). This problem elegantly combines the principles of mechanics, specifically kinematics and energy conservation, with electromagnetism, particularly the relationship between electric potential and electric potential energy. To solve this, we'll need to understand how the electric field does work on the charged bullet, converting electrical potential energy into kinetic energy. The key lies in equating the change in kinetic energy to the work done by the electric field, which is directly related to the potential difference. By carefully applying these concepts and performing the necessary calculations, we can unveil the required potential difference. This problem serves as an excellent illustration of how different branches of physics can intertwine to explain real-world phenomena.

H3: Deciphering the Physics Principles

To effectively tackle this problem, we must first lay the groundwork by understanding the fundamental physics principles at play. The first crucial concept is the relationship between electric potential difference (often denoted as ΔV) and the work done (W) on a charge (q) moving through an electric field. This relationship is beautifully captured by the equation: W = qΔV. This equation tells us that the work done on a charge is directly proportional to both the magnitude of the charge and the potential difference it traverses. The second key principle we'll employ is the work-energy theorem. This theorem states that the net work done on an object is equal to the change in its kinetic energy (KE). Kinetic energy, the energy of motion, is given by the formula KE = (1/2)mv², where m is the mass and v is the velocity of the object. In our scenario, the work done by the electric field on the bullet is entirely converted into the bullet's kinetic energy. Therefore, we can equate the work done (qΔV) to the change in kinetic energy (ΔKE). This connection between electrical work and kinetic energy is the cornerstone of our solution. By understanding and applying these principles, we can bridge the gap between electromagnetism and mechanics, ultimately determining the potential difference required for the bullet to reach its desired speed.

H3: Laying Out the Solution Strategy

Now that we've grasped the underlying physics principles, let's outline a clear strategy for solving the problem. Our primary goal is to find the potential difference, ΔV. To achieve this, we will follow a step-by-step approach that leverages the relationships we discussed earlier. First, we'll calculate the change in kinetic energy (ΔKE) of the bullet. This involves determining the bullet's initial and final kinetic energies using the formula KE = (1/2)mv². Since the bullet starts from rest, its initial kinetic energy is zero. The final kinetic energy can be easily calculated using the given final velocity of 10 m/s and the bullet's mass of 2 g (which we'll need to convert to kilograms). Second, we'll invoke the work-energy theorem, which tells us that the work done on the bullet by the electric field is equal to this change in kinetic energy (W = ΔKE). Third, we'll use the fundamental relationship between work and potential difference (W = qΔV) to connect the work done to the potential difference. Finally, we'll solve this equation for ΔV, using the calculated value of W (which is equal to ΔKE) and the given charge of the bullet (4 µC, which we'll need to convert to Coulombs). By meticulously following these steps, we can systematically unravel the problem and arrive at the correct solution. This structured approach ensures clarity and accuracy in our calculations.

H2: Step-by-Step Solution: Calculating the Potential Difference

Let's now put our solution strategy into action and perform the calculations step-by-step. This will lead us to the final answer for the potential difference.

1. Calculate the change in kinetic energy (ΔKE):

   • First, convert the bullet's mass from grams to kilograms: m = 2 g = 2 × 10⁻³ kg
   • The initial kinetic energy (KE₁) is zero since the bullet starts from rest: KE₁ = 0 J
   • The final kinetic energy (KE₂) is calculated using the formula KE = (1/2)mv²: KE₂ = (1/2) × (2 × 10⁻³ kg) × (10 m/s)² = 0.1 J
   • Therefore, the change in kinetic energy is: ΔKE = KE₂ - KE₁ = 0.1 J - 0 J = 0.1 J

2. Apply the work-energy theorem:

   • The work-energy theorem states that the work done (W) is equal to the change in kinetic energy (ΔKE): W = ΔKE = 0.1 J

3. Use the relationship between work and potential difference:

   • The relationship is given by W = qΔV, where q is the charge and ΔV is the potential difference.
   • First, convert the charge from microcoulombs to Coulombs: q = 4 µC = 4 × 10⁻⁶ C

4. Solve for the potential difference (ΔV):

   • Rearrange the equation W = qΔV to solve for ΔV: ΔV = W / q
   • Substitute the values we have: ΔV = 0.1 J / (4 × 10⁻⁶ C) = 25 × 10³ V
   • Express the result in kilovolts: ΔV = 25 kV

Therefore, the potential difference the bullet must move through to acquire a speed of 10 m/s starting from rest is 25 kV. This detailed step-by-step calculation provides a clear and concise path to the solution, reinforcing the understanding of the underlying physics principles.

H2: Answer and Options Analysis

Based on our calculations, the potential difference required for the bullet to achieve a speed of 10 m/s is 25 kV. Now, let's examine the provided options to identify the correct answer:

• (1) 100 kV
• (2) 75 kV
• (3) 50 kV
• (4) 25 kV

Comparing our calculated result of 25 kV with the given options, we can clearly see that option (4), 25 kV, matches our solution. Therefore, the correct answer is (4) 25 kV. The other options (100 kV, 75 kV, and 50 kV) are significantly higher than the calculated value, indicating that they would impart a much greater speed to the bullet. This analysis reinforces the accuracy of our calculations and provides confidence in the final answer. It also highlights the importance of carefully considering the magnitude of the values involved in physics problems to ensure the solution aligns with physical intuition.

H3: Potential Pitfalls and Considerations

While the solution we've presented provides a clear path to the answer, it's crucial to be aware of potential pitfalls and considerations that can arise when tackling similar problems. One common pitfall is neglecting unit conversions. For instance, failing to convert grams to kilograms or microcoulombs to Coulombs can lead to significant errors in the final result. Another important consideration is the assumption of a uniform electric field. Our solution implicitly assumes that the electric field is uniform, meaning that the force on the bullet is constant throughout its motion. In real-world scenarios, electric fields may not always be uniform, which could complicate the problem. Furthermore, we've neglected any effects of gravity or air resistance. While these effects may be negligible in this specific problem, they could become significant under different circumstances. It's also worth noting that the kinetic energy gained by the bullet is solely due to the work done by the electric field. We've assumed that no other forces are contributing to the bullet's acceleration. By being mindful of these potential pitfalls and considerations, we can approach similar problems with greater awareness and accuracy. This critical thinking is essential for mastering physics concepts and applying them effectively.

H2: Real-World Applications and Significance

This problem, while seemingly theoretical, has significant connections to real-world applications and underscores the importance of understanding the principles of electromagnetism and mechanics. The acceleration of charged particles in electric fields is a fundamental concept that underlies many technologies we use daily. For instance, particle accelerators, used in scientific research and medical treatments, rely on electric fields to accelerate charged particles to incredibly high speeds. The principles we've discussed are directly applicable to understanding how these accelerators work. Similarly, the operation of cathode ray tubes (CRTs), once commonly found in televisions and computer monitors, involves accelerating electrons using electric fields to create images on a screen. Furthermore, the behavior of charged particles in electric fields is crucial in various industrial processes, such as electrostatic painting and separation techniques. Beyond these technological applications, the interaction between charged particles and electric fields plays a vital role in natural phenomena, such as lightning and the behavior of ions in plasmas. Therefore, understanding the principles demonstrated in this problem provides a foundation for comprehending a wide range of phenomena and technologies. It highlights the power of physics to explain the world around us and drive innovation.

H2: Conclusion: Mastering the Interplay of Physics Concepts

In conclusion, we have successfully determined the potential difference required for a bullet with a given mass and charge to reach a specific speed, starting from rest. This problem elegantly illustrates the interplay between fundamental physics concepts, including the work-energy theorem, the relationship between work and potential difference, and the principles of kinematics. By carefully applying these concepts and performing step-by-step calculations, we arrived at the solution of 25 kV. We also analyzed the given options, confirming that our result aligned with one of the choices. Furthermore, we discussed potential pitfalls and considerations that can arise in similar problems, emphasizing the importance of unit conversions, assumptions about the electric field, and the influence of other forces. Finally, we highlighted the real-world applications and significance of this problem, demonstrating its relevance to various technologies and natural phenomena. This exercise not only reinforces our understanding of electromagnetism and mechanics but also showcases the power of physics to explain and predict the behavior of objects in the world around us. Mastering these concepts is essential for anyone pursuing a deeper understanding of physics and its applications.