11.1 Electric Fields

All right, we're gonna start our next chapter, and it's on electric fields. So today, we're gonna figure out what a field is and how it compares to the gravitational field, which we already know. We're gonna look at some formulas and key things that you need to remember. So, to start with, just like gravity, right? Gravity can act between two objects that are separated. You don't have to be touching it to have a gravitational force. The same thing works with electric fields. We don't have to actually have the charges in contact to have an electric field. They can work over a large distance as well. The idea in the old days of figuring out what happens when things aren't touching was a difficult concept for scientists to figure out. But it does make sense, right? How do we explain a ball that falls? Well, gravity's pulling it downwards. We don't see gravity, we can't really measure it, but we know it's there. The same thing sort of happens when we deal with electric charges. We know two positive charges are gonna separate themselves, but we don't really see it or witness it because they don't touch. A British scientist named Michael Faraday is the one that came up with the idea of a field. He applied it for a lot of electricity and electrostatics. The way he defined it was he said a field is like a sphere of influence. So, he used the word sphere because I think it was sort of dealing with Earth. We knew that the Earth was a sphere, and we knew that gravity was pulling it downwards no matter where you were on Earth. If we're talking about force of gravity, of course, it's gonna be downwards. So, we knew the gravitational field then has to be acting from all different directions. A good example would be if you have a bad smell, like somebody has a fart or something like that. If you're right close to it, the fart is pretty strong, and the further away you get, the less and less of it you smell. Same sort of thing with gravity, right? If you're right close, gravity's strong. If you're further away, it's gonna be not quite as strong. Different types of fields: vectors and scalars. A scalar field would be something where we don't really care about directions. If you're measuring the heat from a fire, you can measure it from any side of the fire, at the top, whatever. You get the heat coming off the fire in every direction, and we just say something like 55 degrees. We wouldn't say north, south, east, west, up, down, left, right, nothing like that. But a vector field does include direction. So, for example, we normally say gravity, the gravitational field strength is 9.81 m/s^2, and it's always downwards towards the center of the Earth. So, you can see here we're talking about the force of gravity, but the actual gravitational field is actually 9.81, which we know is our acceleration of gravity. So, it's important to realize that a field isn't the force. It causes the force. Our gravitational field causes the force of gravity to be whatever your mass times 9.81, but it's not the force itself. The same thing happens with electric fields. We're gonna have some sort of field around electric charge that'll cause the field to go inwards or outwards, depending on what the charge is. But it's the same sort of idea. It's not causing a force or it's not a force, but it does cause a force. Okay, next is how would we detect if there's an electric field, and how strong it would be, and whether it's positive, negative, all those kinds of things. The easiest way to do it would be to have some other known charge nearby. So, if we had a charge that we didn't know, but if we put a positive charge near it, we should be able to figure out what the original one is, right? Because if this one was positive and the original is positive, then the force would be upwards. But if the original was negative, then our force would be towards. It is important to attract or repel to determine the presence and direction of an electric field. However, we must be cautious because the presence of multiple charges can interact and change the field. To address this, scientists have introduced the concept of a test charge, an imaginary charge used as a reference. We typically use a positive test charge, although it could have been negative. We consider the test charge to have no mass and no electric field of its own. It is similar to the idea of a pith ball. We denote the test charge as Q, just like any other charge. If we have a positive source charge and place a positive test charge nearby, we know that the test charge will move away. Similarly, if the source charge is negative, the test charge will attract. Apologies for the confusion in the previous sentences, the diagrams are reversed. The first diagram represents the bottom one, and if the test charge moves towards it, the original charge must be negative. Conversely, if the test charge moves away, the original charge must be positive. Let's consider an example: we have a steel ball with an unknown charge as our source, and we place a test charge nearby. If the test charge moves away to the right, it means that the original charge must have been positive. Therefore, the source charge is positive, and the test charge is also positive. When calculating electric fields, we use the formula E = Force/Q. The force can be calculated using Coulomb's law, which states that the force (F) equals k(q1q2)/r^2. Q represents the charge in coulombs, so the electric field is measured in Newtons per Coulomb. Remember that the formula for electric fields includes a vector symbol above the E, distinguishing it from other formulas that use the same letter. The vector symbol shows that we are specifically referring to electric fields. It can be confusing because there are two E's in different formulas, but understanding the distinction is crucial. Energy formulas do not have the vector symbol, while electric field formulas do. In the formula, Q refers to the charge of the object generating the field. It is important to note that Q is not a special term; it simply represents the charge of the object. Field, not the charge, is what's making the field. So in this case, to calculate the electric field strength of the steel ball in example one, we would have some sort of force that we'd measure using KQ1Q2/R^2 or whatever it is. Our charge would be the charge that's in the electric field, and that's what would give us the electric field strength. Okay, let's do another example. It says we have a 3.7 Coulomb test charge and we have another charge that's -17.94. The distance between them is 2.7 meters and it says there's an attractive force, one's positive and one's negative. So we know this one's gonna get attracted towards it with a force of 3.62 x 10^10. The question is, what is our electric field strength? So, pretty easy question, just plug in Ft/Q. We'd have 3.62 x 10^10 and our Q weight, we have to use the charge that's in the electric field, not what's producing it. So in this case, we even put in the 3.7 coulombs for our answer. Okay, so we don't use the 2.7 distance, we don't use the -17.94 for our source. Those numbers would be what would create the FE, right? If we were to use Kepler's constants KQ1Q2/R^2. Kepler, sorry, Coulomb's law. We could plug all that into the formula and that would give us the 3.62 x 10^10. So in this case, our electric field is just that force divided by the charge that's in the electric field. A couple things to remember: First off, when we're talking about this, you always have to remember that the Q is the test charge in the formula. Sometimes people get that mixed up. And if you ever forget, it's always good to sort of think that electric field works just like a gravitational field. So if you remember back to physics 20, we had force of gravity equals mg. So if we rearrange that to be G equals F/m, it's pretty much the exact same thing as an electric field, as our electric force divided by Q. So instead of a mass and a gravitational field, we now have a charge and electric field. But the logic is exactly the same. So G, our gravity or 9.81, was the gravitational field strength. So now, our E is our electric field strength. That'll be some sort of number, and it's per Coulomb. Usually, mass is our object that's in the gravitational field, right? We measure the mass of the person that's on Earth, not the Earth. So the same thing has to apply for electric fields. We want the charge of the object in the electric field, not what's creating it. So you can see it works very similar to what we did for gravity. Putting all this together, we know that from Coulomb's law, we're gonna have some sort of electric field, and it's going to be different at different locations. But we also know that the direction will be different. So if we have a positive test charge, or serving a positive source charge, its electric field will be upwards because if we put a positive test charge near it, it would get repelled. And if it's negative, then the electric field is inwards. Okay, so we draw arrows going all the way around. You don't need a lot of them, just enough to show that it's going from every direction. And this length of the size of the arrows will indicate how strong the electric field is. For a weak field, I have short arrows. For a strong field, you'd make those arrows a little bit longer. So, a couple... Let's do a couple. More quick examples. So, the first one says that a force of 2.1 Newtons is exerted on a test charge when it's placed in the electric field created by a different charge. It's asking if the force is pushing it west, determine the electric field. So, we have our source charge of +7.5 and our test charge of +9.2 x 10^(-4). Since both charges are positive, we know that the force is pushing the test charge upwards and to the west. We can use the formula E = F/Q, where E is the electric field, F is the force, and Q is the charge. Plugging in our values, we get E = 2.1/9.2 x 10^(-4). Unfortunately, I don't have my calculator with me today, but you can easily compute the answer by dividing those two numbers. Therefore, the answer, whatever it may be, would be directed upwards. I quickly checked, and it should be 2.3 x 10^3 Newtons per Coulomb. But the question is: what is the electric field strength at that point? Since it's positive, we know that our electric field can go upwards from it. However, we need to consider that we are a distance to the west of our charge. Our charge is here, and our point of interest is to the west of it. At this point, our electric field is 2.3 x 10^3 Newtons per Coulomb, and the direction is west. Moving on to example 4, we have the same situation, but with a smaller test charge. Instead of 9.2 x 10^(-4), we now have a charge of 3.7 x 10^(-6). The question remains the same: what's the electric field? Sorry, force acting on it. I already know that at that location, our electric field was 2.3 x 10^3. In this case, we're looking for force. Using the formula F = E * Q, we can calculate that by multiplying the electric field (2.3 x 10^3) by the charge (3.7 x 10^(-6)). This will give us a force of 8.4 x 10^(-3) Newtons. So, the force is 8.4 x 10^(-3) Newtons, and it's going west because our force is upwards. Be careful with the direction based on the location. Now, our last example is the same, but we're using a different charge in the same spot. We solve it in the same way, so our force would still be the same electric field strength. The only thing that's different now is our force is -4.2813 times A by 4.81 x 10^(-2). So, that's gonna give us a new force. But because it's negative, our direction should be east, right? It's gonna get attracted to the source, not pushed away. And if we do that calculation, we get 1.1 times ten to the two Newtons. So, we have everything done right for these three examples. So, pretty simple formula to work with, just be careful of direction, positive negatives, same sort of thing that we were doing before. Our last thing we got to look at today is now how can we combine this with what we already know. So, we know our electric field is force over Q, but if we think back to our last chapter, we also know that electric force is k q1 q2 over R squared. So, we're going to assume in this case, just to make sure we have this straight, q1 is our source charge and q2 will be our test charge. So, we're gonna actually in the formula, we're going to ignore q2, so q1 is our source, we'll just call it Q for a test charge, and that's the same Q that's in our new electric fields formula. So, we can actually substitute these in right. We can call it electric field, we'll put KQ1Q over R squared, and then over Q again. So, you can see that basically our test charge cancels out, so we can get a new formula for electric field which is KQ1 over R squared. And this formula actually works a little bit better because in this formula, we're talking about the source, right? The Q1 in this case is our source that's making electric field, which in a lot of cases is easier. And on your formula sheet, they actually have absolute values around it, which just means we don't worry about whether it's a positive or negative charge, we just solve our answer. And then you use the direction of the electric field to determine which way it would go. Let's compare it to gravity again because it is almost the same formula. On your formula sheet in physics 20, we had G, the gravitational field strength or acceleration of gravity, was big G M over R squared, where M with our mass of the earth or whatever was creating the gravity field. And if we look at this new one, now we have electric field is K, kind of like our big G, Q1 which is our source charge, kind of like the earth, divided by R squared. So, we get basically the same formula now, just dealing with electricity instead of gravity. One more example. So, this one we have an object that has a charge of negative three to the negative six, and the question is, what would the electric field be if you were thirty centimeters away? So, we don't need to worry about a test charge or anything like that anymore. We just have our charge of the object, it's a negative three to the negative six. So, we know that because it's negative, electric field will be inwards. And the question is asking what is it at some point P that's 30 centimeters away from the charge? So, we can use our new formula and put in K 8.99 times ten to the nine times our source charge, this time which is three to the negative six, and divided by thirty centimeters, that would be 30 to the negative two. Squared, plug all that into our formula, so we get an electric field of three times ten to the five Newtons per Coulomb. And for direction, we would say inwards or towards the center because it was negative. Remember our electric field. Goes towards it, so at Point P, any point to P that's 30 centimeters away, our electric field strength would be three to the five, pointing inwards. And one last quick example. So, what happens if we had more than one charge? Just like we just finished doing in the last unit, it works exactly the same way. So, for this example, we have two negatively charged objects. The question is, what's the electric field strength that's exactly halfway between? So, Point P is halfway in between. What we want to do then is use the formula to figure out our electric field for the one on the left. So, we'd have 8.99 times 1.2 divided by our distance, which would be 0.58, so that'll be 0.29 squared. And then we'd also do the same calculation for the other side. So, we'd have 8.99 again, this time our source charge is 3.9, and we're gonna divide it by 0.29 squared again. So, we don't have to worry about directions, just like we did before. That's why we're doing absolute values. We just want to figure out the electric field, and then we will combine them just like we would with vectors. So, in this case, our electric field is negative. That means, Point P is going to be pulled to the left. This one's also negative, so it's going to be pulled to the right. So, we just figure out those two answers and subtract them and combine them like we do with vectors. So, if you get any kind of questions that have angles or whatever, it works exactly the same way. Just figure out each component and combine them using appropriate vectors. So, for this one, we have 1.3 times 10^11 for electric field strength. For the other one, we have 4.2 times 10^11. So, we just need to subtract those two, and that will be our final answer. So, if we subtract those, we get a net answer of 2.9 times 10^11 N/C, and the red one is bigger. So, we would say to the right or towards the red one or towards the negative 3.9, along those lines. And that's it.