Calculating The Electric Potential Of A Charged Sphere A Step-by-Step Guide

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Hey physics enthusiasts! Let's dive into a fascinating problem involving electric potential and charged spheres. We're going to break down the question: "A small conducting sphere has a charge of 3.5 x 10^-8 C. What is the electric potential at 100 cm from it?" into manageable steps. Get ready to flex those physics muscles!

Understanding the Fundamentals of Electric Potential

Before we jump into calculations, let's ensure we're all on the same page regarding electric potential. Electric potential, often denoted as V, represents the amount of electric potential energy a unit charge would possess at a specific location within an electric field. Think of it as the "electrical height" at a given point. Just as objects roll downhill due to gravitational potential difference, positive charges move from regions of high electric potential to low electric potential, and negative charges do the opposite. This movement is driven by the electric force seeking to minimize the potential energy of the system.

It's crucial to distinguish electric potential from electric potential energy. Electric potential energy (U) is the actual energy a charge possesses due to its position in an electric field. The relationship between electric potential and electric potential energy is beautifully simple: U = qV, where q is the charge. This means the electric potential energy is the product of the charge and the electric potential at its location.

For a single point charge, or a spherically symmetric charge distribution (like our conducting sphere), the electric potential at a distance r from the charge is given by the formula: V = kQ/r, where k is Coulomb's constant (approximately 8.99 x 10^9 Nā‹…m2/C2) and Q is the total charge. This formula tells us that electric potential decreases as we move further away from the charge, which makes intuitive sense – the influence of the charge weakens with distance. This inverse relationship between electric potential and distance is a cornerstone of electrostatics.

Think about it this way: the closer you are to a charged object, the stronger the electric force it exerts, and consequently, the higher the electric potential. As you move away, the force diminishes, and so does the electric potential. This understanding of how electric potential varies with distance is key to solving a wide range of problems in electromagnetism. Grasping these core concepts is essential before we tackle the specific problem at hand.

Breaking Down the Problem: A Step-by-Step Approach

Now that we have a solid understanding of the basics, let's dissect our problem. We have a small conducting sphere carrying a charge (Q) of 3.5 x 10^-8 C. We need to find the electric potential (V) at a distance (r) of 100 cm (which is 1 meter) from the sphere. The key here is to recognize that, from a point outside the sphere, a uniformly charged sphere behaves as if all its charge is concentrated at its center. This simplifies our calculation significantly.

The first step is to identify the knowns and unknowns. We know the charge Q (3.5 x 10^-8 C), the distance r (1 meter), and Coulomb's constant k (8.99 x 10^9 Nā‹…m2/C2). We want to find the electric potential V. This clear identification of variables is crucial for any physics problem. It helps us to focus on what information we have and what we need to determine. This organized approach is the hallmark of a successful problem-solver.

Next, we need to select the appropriate formula. As we discussed earlier, the electric potential due to a point charge (or a spherically symmetric charge distribution) is given by V = kQ/r. This formula perfectly fits our scenario, as we're dealing with a charged sphere and want to find the electric potential at a certain distance from it. Choosing the correct formula is often the most critical step in solving a physics problem. Make sure you understand the conditions under which a particular formula is valid.

Now comes the fun part: plugging in the values! We substitute k = 8.99 x 10^9 Nā‹…m2/C2, Q = 3.5 x 10^-8 C, and r = 1 meter into the formula. This gives us V = (8.99 x 10^9 Nā‹…m2/C2) * (3.5 x 10^-8 C) / (1 meter). Carefully substitute each value with its corresponding unit. This not only ensures that your calculation is accurate but also helps you to keep track of the units and perform dimensional analysis to check your answer.

Calculating and Interpreting the Electric Potential

With the values plugged in, it's time to perform the calculation. Multiply 8.99 x 10^9 by 3.5 x 10^-8, and then divide by 1. This yields V ā‰ˆ 314.65 Volts. Make sure you use a calculator for this step to avoid any arithmetic errors. Double-check your calculations to ensure accuracy. A small mistake in the calculation can lead to a significant difference in the final result.

So, what does this result mean? The electric potential at 1 meter from the charged sphere is approximately 314.65 Volts. This means that if we were to place a positive test charge at this location, it would possess 314.65 Joules of electric potential energy per Coulomb of charge. The unit of electric potential is Volts (V), which is equivalent to Joules per Coulomb (J/C). The higher the electric potential, the greater the potential energy a positive charge would have at that point.

Understanding the significance of the value is crucial. It's not just about getting the right number; it's about interpreting what the number represents in the physical world. The electric potential of 314.65 Volts indicates a moderate level of electrical influence at that distance. A charge placed at this point would experience a noticeable force due to the electric field created by the sphere. This interpretation is essential for applying the concept of electric potential to real-world scenarios.

Furthermore, we can consider how this electric potential would change with distance. Since electric potential is inversely proportional to distance (V āˆ 1/r), as we move further away from the sphere, the electric potential will decrease. Conversely, as we move closer, the electric potential will increase. This inverse relationship is a fundamental characteristic of electric potential due to point charges and spherically symmetric charge distributions.

Practical Applications and Further Exploration

Understanding electric potential isn't just an academic exercise; it has numerous practical applications. From the design of electronic circuits to the operation of particle accelerators, electric potential plays a vital role. For example, in circuits, electric potential difference (voltage) drives the flow of current. In particle accelerators, electric potential is used to accelerate charged particles to incredibly high speeds.

Imagine designing a capacitor, a device used to store electrical energy. The electric potential between the capacitor plates directly affects how much energy it can store. Or consider the functioning of an electron gun in a cathode ray tube (remember those old TVs?). The electric potential difference accelerates electrons, creating a beam that illuminates the screen. These are just a couple of examples illustrating the wide-ranging importance of electric potential in technology.

If you're keen to explore further, consider delving into topics like equipotential surfaces, which are surfaces where the electric potential is constant. Visualizing these surfaces can provide a powerful way to understand the electric field. Also, investigate how electric potential relates to electric field (E = -āˆ‡V, where āˆ‡ is the gradient operator). This connection is fundamental in electromagnetism.

You might also want to explore the concept of electric potential energy in systems of multiple charges. How does the total potential energy change when you bring multiple charges together? This leads to fascinating discussions about the stability of systems and the work required to assemble them. The world of electrostatics is rich and rewarding, and the concept of electric potential is a key that unlocks many of its secrets.

So, there you have it! We've successfully calculated the electric potential at a distance from a charged sphere. Remember, the key is to understand the underlying concepts, identify the knowns and unknowns, choose the right formula, and interpret the results. Keep practicing, and you'll become a pro at solving electrostatics problems!

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What is the electric potential at 100 cm from a small conducting sphere that possesses a charge of 3.5 x 10^-8 C?