Hey guys! Let's dive into a super interesting physics problem today that deals with the flow of electrons in an electrical device. It's a fundamental concept in understanding how electricity works, and we'll break it down step by step so it's crystal clear. So, the problem we are going to solve together is: An electric device delivers a current of $15.0 A$ for 30 seconds. How many electrons flow through it?
Breaking Down the Basics of Electric Current
Let's start with the basics. Electric current is essentially the flow of electric charge through a conductor. Think of it like water flowing through a pipe; the current is the amount of water passing a certain point in a given time. In the case of electricity, the charge carriers are usually electrons, which are tiny negatively charged particles. The standard unit for measuring current is the ampere (A), named after the French physicist André-Marie Ampère. One ampere is defined as the flow of one coulomb of charge per second (1 A = 1 C/s). This means that if a device has a current of 15.0 A, it's like saying 15.0 coulombs of charge are flowing through it every single second. Now, a coulomb (C) is a unit of electric charge. But how many electrons make up one coulomb? That's where the elementary charge comes in. The elementary charge, often denoted as e, is the magnitude of the electric charge carried by a single proton or electron. Its value is approximately $1.602 \times 10^-19}$ coulombs. This is a tiny number, which means it takes a lot of electrons to make up one coulomb. In fact, about 6.242 × 10¹⁸ electrons are needed to make up one coulomb. So, when we talk about a current of 15.0 A, we're talking about an immense number of electrons flowing every second! To really grasp this concept, imagine a crowded concert venue. Each person trying to get through a door is like an electron, and the number of people passing through the door per second is like the current. The more people (electrons) that pass through the door (conductor) each second, the higher the current. Now, let's consider the time factor. Our problem states that the current of 15.0 A flows for 30 seconds. This means we have a constant stream of electrons flowing for this duration. To find the total number of electrons that flowed during this time, we need to calculate the total charge that passed through the device and then determine how many electrons that charge represents. This is where the formula relating current, charge, and time comes into play. We know that current (I) is the amount of charge (Q) flowing per unit time (t), which can be expressed ast}$. This is a fundamental relationship in electrical circuits and is key to solving our problem. Understanding this relationship is crucial because it allows us to connect the macroscopic measurement of current (something we can easily measure with an ammeter) to the microscopic world of electron flow. It's like having a bridge that connects the observable electrical phenomena to the underlying movement of charged particles. In summary, before we jump into the calculations, let's recap the key concepts{t}$. With these basics in mind, we're well-equipped to tackle the problem and find out how many electrons flowed through the electrical device. So, let's get to the math and see how it all works out!
Solving for Total Charge and Number of Electrons
Alright, guys, now that we've got a solid understanding of the basics, let's get our hands dirty with the calculations! Remember, the problem asks us to find the total number of electrons that flow through an electric device when it delivers a current of 15.0 A for 30 seconds. We've already established that current (I) is the rate of flow of charge (Q) over time (t), and we have the formula: $I = \fracQ}{t}$. The first step in solving this problem is to find the total charge (Q) that flowed through the device. We know the current (I = 15.0 A) and the time (t = 30 s). We can rearrange the formula to solve for Q \times 30 \text s}$. Calculating this, we get$. So, the total charge that flowed through the device is 450 coulombs. That's a significant amount of charge! But remember, we're not just interested in the charge; we want to know the number of electrons. To find this, we need to use the value of the elementary charge (e), which is approximately $1.602 \times 10^-19}$ coulombs per electron. We know that the total charge (Q) is equal to the number of electrons (n) multiplied by the elementary charge (e)e}$. Now, let's plug in the values}1.602 \times 10^{-19} \text{ C/electron}}$. Calculating this, we get \text{ electrons}$. Wow! That's a massive number of electrons! It's approximately 2.81 sextillion electrons. To put it in perspective, that's more than the number of stars in the observable universe! This huge number highlights just how many electrons are involved in even a relatively small electric current. When we talk about a current of 15.0 A, we're talking about an almost unimaginable number of electrons zipping through the device every second. This calculation really brings home the scale of electrical phenomena at the microscopic level. So, to recap, we first calculated the total charge that flowed through the device using the formula $Q = I \times t$. Then, we used the value of the elementary charge to determine the number of electrons that made up that charge, using the formula $n = \frac{Q}{e}$. The result, approximately 2.81 × 10²¹ electrons, is a testament to the sheer number of charge carriers involved in electric current. This problem not only reinforces our understanding of the relationship between current, charge, and time but also gives us a sense of the scale of electron flow in electrical devices. Next, we'll discuss why this understanding is so important in practical applications and how it relates to other concepts in electricity. So, keep those thinking caps on, and let's dive deeper into the world of electrons!
Practical Implications and Further Exploration
Okay, so we've crunched the numbers and found that a whopping 2.81 × 10²¹ electrons flowed through the device. That's awesome! But you might be thinking,