Understanding Electric Fields and Their Behavior

In this video, we're going to focus on electric fields. Let's begin our discussion with the formula that will help us to calculate the electric field. The electric field is equal to the electric force that is acting on a tiny test charge divided by the magnitude of that. test charge. So it's F over Q. The force is measured in Newtons, the charge Q is measured in Coulombs. So electric field has the units Newtons per Coulomb. Now the electric field is a vector, much like force is a vector. But the good thing about the electric field is, it tells you how the electric force is going to be acting on a positive or negative charge. So let's talk about positive charges first. What happens if we put a positive test charge in an electric field? That test charge is going to feel a force that will accelerate it in the same direction as the electric field. Now what about a negative test charge? What's going to happen if we put it in an electric field? A negative test charge will feel a force that will accelerate it in the opposite direction to the electric field. So make sure you understand that. Positive charges, they accelerate in the same direction as the electric field. The force and electric field vectors will be in the same direction. But for negative charges, they will feel a force that will accelerate in the opposite direction of the electric field. Now, as we said before, an electric field can exert a force on any type of charge, a positive charge or a negative charge. But it turns out that charges can also create their own electric fields. The electric field created by a positive charge extends in all directions away from the positive charge. The electric field created by a negative charge extends in all directions toward the negative charge. So it's going inward towards the negative charge in all directions. Now let's talk about the equation that will help us to calculate the electric field created by a point charge. So let's say we have, let me draw a better circle. Let's say we have this charge, which we'll call big Q. And we want to calculate the electric field at some point A. How can we do that? And let's say A is some distance R. Well, we know that the electric field at point A will be going in this direction. All you need to do is draw an arrow from the positive charge to the point of interest, and that will give you the direction of the electric field. Now, let's say if we place a tiny positive charge at point A. Let me just erase point A. So this tiny positive charge will be considered our test charge. The reason why it has to be tiny, because if it's large enough, it will affect Q. Big Q will be repelled by little q. But if the magnitude of the charge is small, it will be repelled by little q. of little q is very very small it won't affect big Q as much and thus it won't affect the electric field that is created by big Q as much so that's this equation will work if we choose a very tiny positive test charge Now, according to Coulomb's Law, whenever you have two charges next to each other, they will exert a force on each other. Two light charges will feel a force that will repel them. And we can calculate that force using this equation. It's equal to K times Q1, which we'll call big Q, times Q2, which we'll call little q, over R squared, where R is the distance between the two charges. So what we're going to do now is we're going to take this equation and substitute it for f in that equation. But first, I'm going to rewrite this equation as f over 1 times 1 over q. That's the same as f divided by q. So now, let's replace f with what we see here. So this is going to be f, I mean, it's no longer f, but it's going to be k, big Q times little q over r squared. So this whole thing is f, and then we're going to multiply it by 1 over q. And so we could cancel out little q, the test charge. Thus, we get the equation that gives us the electric field, or the magnitude of the electric field, for a point charge, capital Q. So it's KQ over R squared. So if we have our charge, Q, and we wish to calculate the electric field at point A, We could use this formula. K is 9 times 10 to the 9 Newtons times square meters over square coulombs. Q is the magnitude of the charge in coulombs. Now remember, one microcoulomb is 1 times 10 to the minus 6 coulombs. A nanocoulomb is 10 to the negative 9 coulombs, and a millicoulomb is 10 to the minus 3 coulombs. And R is going to be the distance in meters. Now let's say we have a positive charge here. And we have point A, let's say point B, point C, and point D. Determine the electric field created by the charge Q at points A, B, C, and D. So at point A, the electric field, in order to draw it, we're going to draw it starting from the positive charge towards point A. And so the electric field is going north. For point B, we need to draw it towards point B. So the electric field, the direction is west. For point D, it's going to be at an angle. So it appears to be going in the northwest direction. And for point C, just draw it towards point C. So it's going in the... Oh wait, I have to make a correction. That is not northwest. That is northeast. And this is southeast. Now let's do something similar but with a negative charge. So for the sake of practice, go ahead and determine the direction of the electric field at the following points. So point A, point B, point C, and point D. So this time, the electric field is going to point toward the negative charge. So we're going to draw it from the point to the negative charge. So this electric field vector is going west. And then here we're going to draw it from B towards the negative charge. So this is going in the south direction. And then D to the negative charge. Here it's going east, and then from C to the negative charge, this electric field vector is pointed in the northwest direction. So that's how we can determine the direction of any electric field vector that is created by some type of charge, either a positive or negative charge, and we can determine the direction of that vector at any point using examples such as these. But now let's focus on some word problems. Number one, a force of 100 newtons is directed north on a negative 20 microcoulomb point charge. What is the magnitude and direction of the electric field at this point? So feel free to pause the video and work on this example problem. So let's begin by drawing a picture. So let's draw our negative point charge. Now there is an electric force that is acting on this charge, and that electric force is directed north. What is the direction of the electric field? Now if you recall... If we have an electric field pointing east, and if we were to place a positive test charge in that electric field, it will feel a force in the same direction as the electric field. But a negative charge will feel a force that will accelerate it in the opposite direction. So if the electric field is due east, the electric force acting on the negative charge will be west. So if the force is north, what is... the direction of the electric field acting on this charge Q. It has to be in the other direction. It has to be south. So that is the direction of the electric field. It's directed south. Now if we want an angle, we can draw this. This is 0, 90, 180, 270. So the electric field is directed along the negative y-axis. So we can say that it's at an angle of 270 degrees relative to the positive x-axis. And that is in a counterclockwise direction. So those are the ways in which we can describe the direction of the electric field. We could say it's due south, or it's at an angle of 270 degrees. So now that we have the direction of the electric field, let's calculate the magnitude of the electric field. So we could use this formula. We can take the electric force and divide it by the magnitude of the charge. The force acting on it is 100 newtons. The magnitude of the charge, we don't need to worry about the negative sign since we already know the direction. The charge is 20 microcloons, and we can replace micro with 10 to the minus 6. Now let's do some algebra. Let's see if we can get this answer without the use of a calculator. We could divide 100 by 20. How many 20s would make up 100 bucks? That's going to be 520, so 100 divided by 20 is 5. Now the 10 to the minus 6, if we move it to the top, the negative exponent will become a positive exponent. So the negative 6 will change to positive 6. And thus the answer is going to be 5 times 10 to the 6 and the units in newtons per coulomb. So that is the electric field that's acting on this negative charge. Number two, a positive charge of 50 microcoulombs is placed in an electric field of 50,000 newtons per coulomb directed upward. What mass should the charge have to remain suspended in the air? Well, let's talk about how we can create such an electric field. We can create this electric field using a battery and two parallel metal plates. So this is the electrical symbol of a battery. We're going to connect it to these two plates. This is the negative side of a battery and this is the positive side. So this plate is going to acquire a positive charge. And this plate is going to require a negative charge. So if we choose a high enough voltage, and if the distance between the two plates is just right, we can adjust it such that we get an electric field of 50,000 newtons per coulomb. Now we want the electric field to be directed upward. So we want the positively charged plate to be at the bottom, the negatively charged plate to be at the top. Remember, the electric field always extends away from a positive charge and points towards a negative charge. So this is how we can create a uniform electric field using two parallel metal plates. Now let's place our positive charge in the middle between these two plates. If this charge has the right magnitude, mass, it can actually remain suspended in the air. Now let's talk about it. So we have an electric field that is directed upward, and it's acting on this positive charge. What will be the direction of the electric force on this charge? The electric force will be the same direction as the electric field for a positive charge. So the electric force will be directed upward. Now gravity Gravity likes to bring things down, so gravity is going to exert a weight force on this positive charge, bringing it down in the negative y direction. In order for this charge to remain suspended in the air, The electric force that wants to accelerate the charge towards the negatively charged plate, that has to be exactly equal to the gravitational force, or the weight force, that wants to bring the positive charge down towards this plate. So if we can get these two forces to equal each other, then the positive charge will remain suspended in the air. So let's write the forces in the y direction. The sum of the forces in the y direction is equal to the electric force. This is positive. And the weight force is negative because it's going in a negative y direction. And we want the sum of the forces in the y direction to be zero, so that there's no net acceleration, so it remains suspended in the air. Moving this term to the other side, we see that the electric force has to equal the weight force. Now, the electric field is equal to F over Q. Multiplying both sides by Q, we can see that F, the electric force, is the electric field times Q. So let's replace the electric force with E times Q. Now the weight force is simply mg. So now we have everything that we need in order to calculate the mass of this charge. Let's divide both sides by g. So the mass of the charge is going to be equal to the electric field times the magnitude of the charge divided by the gravitational acceleration. The electric field is 50,000 newtons per coulomb. The magnitude of the charge, we're dealing with a 50 microcoulomb charge. So it's going to be 50 times 10 to the negative 6 coulombs. And then we're going to divide that by the gravitational acceleration of 9.8 meters per second squared. So it's 50,000 times 50 times 10 to the minus 6 divided by 9.8, and you should get 0.255 kilograms. So an electric field of 50,000 newtons per coulomb can suspend a positive charge with a mass of 0.255 kilograms, or 255 grams. it can suspend it in the air. If the mass is greater than this number, then the charge will fall down. If it's too light, if the mass is less than this number, then the charge will accelerate towards the negatively charged plate. So the mass has to be at the right number in order for it to remain suspended. Number three, an electron is released from rest in a uniform electric field and accelerates to the east at a rate of of 4 times 10 to the 6 meters per second squared. What is the magnitude and direction of the electric field? So let's draw a picture. So first, let's draw our electron, and then it is accelerating towards the east. Now, according to Newton's second law, the net force is equal to the mass times the acceleration. The net force is in the same direction as the acceleration. So the electric force exerted by this electron due to the electric field is going to be due east as well. Now, if the electric force is east, what is the direction of the electric field? It's going to have to be west. Let me put it here. When dealing with a negative charge, the direction of the force and the electric field, they will be opposite to each other. So now that we have the direction of the electric field, let's focus on getting the magnitude of the electric field. So from the last problem, we saw that the electric force is equal to the electric field times the charge. And using Newton's second law, we can replace the force with the mass times the acceleration. So now we can calculate the electric field if we divide both sides by q. So for this problem, the magnitude of the electric field is going to be the mass times the acceleration, which is the force, divided by the charge. So the mass is... what is the mass of an electron? If you look it up, the mass of an electron is 9.11 times 10 to the negative 31 kilograms. The acceleration given to us in this problem is 4 times 10 to the 6 meters per second squared. And then we're going to divide that by the magnitude of the charge. So what is the charge of an electron? The charge of an electron is negative 1.602 times 10 to the negative 19 coulombs. So these are some numbers that you want to be familiar with. So let's put that here. So let's go ahead and plug in these numbers. By the way, don't worry about the negative sign for Q. It's not going to be relevant here. So the magnitude of the electric field is going to be 2.27 times 10 to the negative 5 Newtons per coulomb. So this is the answer for this problem, and the direction is west. So make sure you write down these numbers. The mass of an electron, as we've considered, is 9.11 times 10 to the negative 31 kilograms. The mass of a proton is 1.67. times 10 to the negative 27 kilograms. The charge of an electron, it's going to be negative 1.602 times 10 to the negative 19 coulombs. The charge of the proton, it's going to have the same magnitude, but the opposite charge. It's going to be positive 1.602 times 10 to the negative 19 coulombs. So those are some numbers that you want to be familiar with when working on problems associated with electric fields if you're dealing with protons and electrons. Number 4. A 40 microcoulomb point charge is placed at the origin. Calculate the magnitude and direction of the electric field created by the point charge at the following locations. So let's draw the point charge first. Now, point P is 5 meters away from the point charge along the x-axis. So this is point P, and this is 5 meters away. What is the electric field at point P? Well, we know the direction. The direction of the electric field is going to be east, if we draw it from the positive charge towards point P. Now to calculate the magnitude of the electric field, it's going to be kq divided by r squared. So k is 9 times 10 to the 9, and then it's newtons times square meters over square coulombs, and I'm running out of space, so I'm not going to put the units here. I'm just going to write 9 times 10 to the 9. q is 40 microcoulombs, so 40 times 10 to the minus 6 coulombs. R is in meters. R is the distance between a charge and a point of interest. So that's 5 meters. So it's going to be 9 times 10 to the 9, times 40, times 10 to the minus 6, over 5 squared. And so the electric field is going to be 14,400 newtons per coulomb. So that's the magnitude of the electric field at point P, and this is the direction. For those of you who want to understand how the units work, here it is. So K is in newtons times square meters over square coulombs. Q is in coulombs, and R is in meters, so we have square meters. As you can see, square meters cancel. Now coulombs squared, we can write that as coulomb times coulomb. So we can cancel. one of the Coulomb units. Thus, we're left with newtons per Coulomb, which is what we have here. So that's it for part A. So it's 14,400 newtons per Coulomb directed east. Now let's move on to part B. So let's calculate the electric field at point S. So, Q is at the origin. Point S is 3 meters east of Q, and then 4 meters north from that point. So, S is at this position here. The electric field can be drawn from Q to S. So the electric field is going in that direction. It's going in the northeast direction. Let's calculate the magnitude of the electric field. So let's use this formula again. It's going to be kq over r squared. So k doesn't change. It's a constant. It's 9 times 10 to the 9. q is still the same. It's 40 times 10 to the minus 6. But r is different. R is no longer the value that we have here. But in actuality, it turns out R is the same. R is the distance between the charge and the point. So we need to use the Pythagorean theorem to calculate the hypotenuse of that right triangle. So this is a, b, and this is c. So c squared is equal to a squared plus b squared. a is 3, b is 4. 3 squared is 9, 4 squared is 16, and then 9 plus 16 is 25. Taking the square root of both sides. we get 5 again. So by coincidence, the electric field is going to have the same magnitude as it did in part A, which was 14,400 N per C. What's going to be different though is the direction of the electric field vector. How can we determine the direction? We know it's somewhat in the northeast direction but not necessarily at a 45 degree angle. What we need to do is calculate theta. So perhaps you remember from trigonometry, Sokotola. If we focus on the Tola part, that tells us tangent, tangent of theta is equal to the opposite side divided by the adjacent side. So tangent of the angle theta is going to be equal to 4 over 3. To calculate theta, we need to take the arc tangent, or the inverse tangent, of 4 over 3. Go ahead and type that in your calculator, and make sure it's in degree mode. So arc tangent 4 over 3 is 53.1 degrees. So that is the angle with respect to the positive x-axis. It's 53.1. So we could say that this is the electric field vector at an angle of 53.1 degrees with respect to the x-axis, or we could say 53.1 degrees north of east. Because here, we're starting from east and we're heading towards the north direction. So it's 53.1 north of east. So that's it for part B. So that's how we can calculate the magnitude of the electric field vector, and also its direction using arc tangent. Number 5. An electron initially at rest is placed in an electric field of 2 times 10 to the 4 newtons per coulomb directed to the west. The distance between the plates is 1 centimeter. What is the acceleration of the electron due to the electric field? So the electric field will emanate away from the positive charge and will point towards the negative charge. So as we can see, the electric field is directed west. Now what effect will it have on the electron? A negatively charged particle will feel a force that will accelerate it in the opposite direction to the electric field. So the electron is going to shoot out of this between these two parallel plates. Let's calculate the acceleration. We know the force acting on a charged particle is equal to the electric field times the charge of that particle. And since this is the only force acting on the electron in the X direction, the net force is going to be the electric force. So we can replace the net force with ma based on Newton's second law. So ma is equal to E times Q. And to solve for A, we're going to divide both sides by m. So for part a, the acceleration is going to be the force, which is E times Q, divided by the mass. So we have an electric field of 2 times 10 to the 4 newtons per coulomb. And the charge of an electron is 1.602 times 10 to the negative 19 coulombs. We're not going to worry about the negative sign. The mass of an electron is 9.11 times 10 to the negative 31 kilograms. So let's go ahead and plug this in. So the acceleration is 3.517 times 10 to the 15 meters per second squared. So that's going to be the acceleration when the electron leaves the negatively charged plate. So now let's move on to part B. What will be the speed of the electron after it leaves the hole? So how can we get that? So now we need to go back to kinematics. The electron is initially at rest, so v initial is zero. We're looking for the final speed, so we'll put a question mark. And we know the distance between the plates. It's approximately, it's one centimeter. So what kinematic formula has acceleration, v initial, v final, and d? It's going to be this one. v final squared is equal to v initial squared plus 2ad. So to solve for the final speed, we simply need to take the square root of both sides. So v final is equal to this. The initial speed is 0. This is going to be 2 times the acceleration, which is 3.517 times 10 to the 15. And the distance between its plates is 1 centimeter. So if we convert 1 centimeter into meters, we need to divide by 100 and it's going to be 0.01 meters. And so we're going to get the square root of 7.034 times 10 to the 13th. And so the final speed is about 8,386,894.5. So we can round that to, let's say, 8.39. And this is times 10 to the 6th. meters per second. So that's how fast the electron is going to be moving when it leaves the hole. Number six, a 200 microcoulomb charge is placed at the origin and a negative 300 microcoulomb charge is placed one meter to the right of it. What is the magnitude and direction of the electric field midway between the two charges? And then for the second part, 30 centimeters to the right of the negative charge. So let's start with the first part of the problem. Let's begin by drawing a picture. So this is going to be the first charge, we'll call it Q1, and the second charge, Q2. So these two charges are separated by a distance of 1 meter. And we want to determine the magnitude and the direction of the electric field midway between the two charges. So that's going to be at this point. How can we do that? Well, we need to determine the direction of each electric field at that point. Q1 is going to create an electric field called E1, which will be directed east. Now Q2 will create an electric field, E2, which starts from the point but points towards the negative charge, and that's going to be directed east as well. So remember, the electric field created by a positive charge extends away from the positive charge, but the electric field... field created by a negative charge points towards a negative charge. So at the same E1 and E2, they're in the same direction. So the net electric field is going to be E1 plus E2. Along the x-axis, or the horizontal axis, both of these are positive because they're going in the positive x direction. E1 is k times q1 over r1 squared. E2 is k times q2 over r2 squared. Now, what's r1 and r2? So r1 is the distance between q1 and the point of interest. r2 is the distance between q2 and the point of interest. So R1 and R2, they're both half of point of one meter, which means that R1 and R2, they're both equal to 0.5. So because R1 and R2 are the same, we can simply call it R. So let's replace R1 with R. And let's replace R2 with R as well. So now we could simplify this equation by factoring out the GCF, the greatest common factor, which is going to be k over r squared. And then we're left with q1 plus q2. So this is the formula that we could use to calculate the net electric field for this particular part in the problem. Now let's go ahead and plug in the numbers. So it's going to be k, which is 9 times 10 to the 9, over r squared. r is 0.5. and then times Q1. Q1 is 200 times 10 to the negative 6. Now for Q2, we're going to use a positive value, not negative 300 times 10 to the negative 6, because we already know the direction of E2. It's going to the right, and it's going to have a positive value because it's heading in the eastward direction. So whenever you're calculating the magnitude for electric field or electric force, you don't need to include the negative charge. You can just find the direction based on where the arrow is going. So let's replace Q2 with 300 times 10 to the negative 6 coulombs. So 200 plus 300, this becomes 500. So it's 9 times 10 to the 9 times 500 times 10 to negative 6 divided by 0.5 squared. Thus, the net electric field is 18 million, which is 1.8 times 10 to the 7 newtons per coulomb. So that's the answer for part A. Now let's move on to part B. So let's redraw the picture for that. So here we have our positive charge and the negative charge. So we got Q1, Q2, and they're separated by a distance of 1 meter. But 30 centimeters to the right, or 0.3 meters, we're going to have our new point of interest. And let's call this point B. So we want to determine the net electric field at point B. So E1, the electric field created by Q1, if we draw it from Q1 to point B, we can see that it's going east. Now, if we draw an electric field from point B to Q2, because it's a negative charge, it needs to go towards a negative charge, it's going to the left. Now, which of these two electric fields is greater, E1 or E2? What would you say? Notice that point B is closer to the negative charge than it is to the positive charge. So E2 is going to have a greater effect than, I mean Q2 is going to have a greater effect on point B than Q1 because it's closer. So if you go back to the equation for electric field, there's two things that the electric field depends on. The magnitude of the charge. and the inverse square of the distance. But the distance is squared, so the distance has a greater impact than the charge. But also, Q2 has a bigger charge than Q1. So those are two factors that favor Q2 over Q1. Point B is closer to Q2, and Q2 has a greater charge magnitude than Q1. So therefore, we can conclude that E2 is going to be bigger than E1. Now, the net electric field is going to be positive E1, because it's going along the positive x-axis, and then plus negative E2, because that's moving towards the west, or the negative x-axis. Now, Now if E1 is bigger than E2, the net electric field will be positive. If E2 is greater than E1, it's going to be negative. And we've confirmed that E2 is going to be bigger. So we should get a negative value. If we get a positive value for the net electric field, we did something wrong. So let's go ahead and do the math. So the net electric field is going to be kq1 over r1 squared plus kq2 over r2 squared. So this time r1 and r2 will be different. So we can't simplify this process by factoring. So let's plug in the numbers. This is k, q1 is 200. Oh, this should be a negative sign, by the way. based on what we have here. So Q1 is 200 times 10 to negative 6. Now R1, R1 is the distance between Q1 and point B. So R1 is going to be the sum of 1 and 0.3. Thus R1 is 1.3 meters squared. and then minus now because we've considered the direction of e2 we've assigned a negative value we don't need to plug in this negative value for q2 we've already taken that into an account so it's going to be minus k and then q2 we're going to use the positive value of q2 300 times 10 to the minus 6. And R2, that's the distance between Q2 and point B, that's 0.3 meters. So let's calculate E1 first. Let's focus on this fraction. 9 times 10 to the 9th times 200 times 10 to negative 6 divided by 1.3 squared. That's 1.065 times 10 to the 7th Newton's percolum. Now, focusing on E2, that's going to be 9 times 10 to the 9th times 300 times 10 to the negative 6 divided by 0.3 squared. So this is... 3 times 10 to the 7 Newtons per Coulomb. So we can see that this number is bigger than that number. So the net electric field is going to be negative. 1.065 minus 3. Wait, something is wrong. Let me double check my work. This should be times 10 to the 6, not 10 to the 7. That's 1,065,088. So that's 1.065 times 10 to the 6. E2 is 30 million, which is 3 times 10 to the 7. So now, if we subtract those two numbers, we get this answer. Negative 2.89 times, so it's 28,935,000, so it's negative 2.89 times 10 to the 7 newtons per coulomb. So we can see why it's negative. E2 is significantly larger than E1. And so that's it for this problem. Number seven, the electric field at point X, two meters to the right of a certain positive charge, is 100 newtons per coulomb. What will be the magnitude of the new electric field if the magnitude of the positive charge doubles in value? So let's draw a picture first. So here is our positive charge, and let's say this is point X, and the distance between these two is two meters. Now at that point the electric field is 100 N per C when the magnitude of the charge is, we'll call it Q. But what happens if we double the magnitude of the charge? So the electric field is KQ over R squared. If you double Q, the electric field is going to double. A quick and simple way to get the answer is to plug in one for everything that doesn't change. Q doubles, so we're going to plug in two. R remains the same, so the electric field is going to double. It's going to go from 100 to 200. Now what about part B? Let's say the distance between the charge and point X doubles. So let's say point X is now over here, and the magnitude is Q. What will be the new electric field? So this time, Q doesn't change. K is the same, so we're going to replace it with a 1. Everything that doesn't change, replace it with a 1. Now the distance doubles. 2 squared is 4, so the electric field is going to be 1 fourth of its original value. 1 fourth of 100, or 100 divided by 4, that's 25. So what you need to take from this is that the electric field is weaker at a greater distance away from the charge. The closer you move towards the point of charge, the closer you move towards the point of charge. charge, the greater the electric field will be. So as the distance increases, the electric field decreases. But as the distance from the point charge decreases, the electric field increases. So there's two ways in which you can increase the electric field. You can increase the magnitude of the charge, which will cause the electric field to go up. Or you could reduce the distance between the point of interest and the charge. And that will also increase the electric field. Now let's move on to part C. The distance between the charge and point X reduces by a factor of 3. So what's going to be the magnitude of the new electric field in that case? So we're bringing it a lot closer to Q. So here's the new position of X. So here the distance doubled to 4 meters. But now it's going to be reduced by factor 3, so it's 2 thirds of a meter. So let's use this formula again. For part C, k and q doesn't change. r is now 1 third of its original value. 1 times 1 is 1, 1 squared is 1, 3 squared is 9. Now we need to multiply the top and the bottom by 9. 1 times 9 is 9, 1 ninth times 9, the 9th cancel, we get 1. So the electric field is going to increase by a factor of 9. 100 times 9 is 900. So as you can see, as we get closer to the point charge, the magnitude of the electric field greatly increases. Now what happens if we triple the magnitude of the charge, but at the same time reduce the distance to 1 fourth of its original value? So I won't draw a new picture for this. Let's just get the answer. So k doesn't change, q triples, and the distance is reduced to 1 fourth of its value. So we have 1 times 3, which is 3, 1 squared is 1, 4 squared is 16. So now I'm going to multiply the top and the bottom by 16. So it's going to be 3 times 16, which is 48. So the electric field will increase by a magnitude of 48. So the original electric field was 100. If we multiply that by 48, the new electric field will be 4800 newtons per coulomb. So this right here is the answer to part D. So the reason why it's so high is because we've increased the charge, which causes E to go up, and at the same time we reduce the distance, which greatly increased the value of E. So that's it for this problem. Number 8. Two identical point charges with a magnitude of 100 microclumes are separated by a distance of 1 meter, as shown below. Part A. At what point will the net electric field be equal to 0? Will it be to the left? Will it be between Q1 and Q2? Or to the right? So let's identify three points of interest. The first point will be somewhere to the left, which we'll call point A. The second point will be somewhere in the middle between Q1 and Q2, likely the midpoint. And then C will be to the right. Now, to draw the electric field vector created by Q1, we need to draw a line from Q1 to point A. This is going to be E1. And for E2, we're going to draw it from Q2 to point A. So at point A, both electric fields are moving to the left. So there's going to be a net electric field at point A. It's not going to be zero. But here's a question for you. Which electric field will be greater, E1 or E2? Now remember, these charges are identical, so the magnitude of the charge is the same. The only thing that's different is the distance. Q1 is closer to point A than Q2. So E1 is going to be bigger than E2. So if this is E1, E2 is going to be a smaller vector. Nevertheless, the net electric field at point A will be directed west. Now what about at point C? To draw E1, we're going to draw a line from Q1 to point C. And E2 will be from Q2 to point C. Now which one is bigger? Q2 is closer to point C than Q1, so E2 is going to be bigger at point C. So for E1, we're going to draw a small vector, and for E2, we're going to draw a bigger vector. Nevertheless, the net electric field at point C will be directed east. Now what about at point B? E1 is going to be directed away from Q1, but towards point B. E2 will be directed away from Q2, but towards point B. And Q1 and Q2 have the same charge, and at the midpoint, at point B, they will be equally distant from point B. So if the charges are the same and the distances are the same, the magnitude of E1 and E2 will be the same, but they're opposite in direction, which means that E1 and E2 cancels. So at point B, the net electric field is zero. So let's assume that Q1 is the origin. It's at position 0. Point B will be at 0.5 meters. So at 0.5 meters relative to the first charge, the net electric field will be 0. E1 and E2 will cancel completely. So that's the answer for part A. At point B, or at the midpoint between Q1 and Q2, the net electric field will be 0. Now what about part B? If the charge on Q2 doubles to 200 microcouplems, where along the x-axis relative to the first charge will the net electric field be equal to zero? And the distance between these two is still the same. Do you think the net electric field will be equal to zero between Q1 and B or between B and Q2? Now, Q2 is bigger than Q1. So at point B, where they're equidistant, E1 is going to be a smaller vector than E2. E2 is going to be twice as large. In order to make these vectors equal, we need to increase E1 and decrease E2. If we can't change Q1 and Q2, the only thing we could change is location. We need to move closer to Q1. As we move closer to Q1, E1 is going to get bigger, E2 will get smaller, and at some point, they're going to equal each other. So, we're going to place point P between Q1 and B. Somewhere between Q1 and B, the net electric field will be equal to zero, and we need to find that point. So that point is going to be R1. R2. is the distance between Q2 and point P. Now let's call R1 X. If R1 is equal to X, what's R2? Notice that the total distance is 1. So R2 is going to be 1 minus X. If you add X and 1 minus X, you're going to get 1. So at this point, All we need to do to get the answer for part B is get the value of x, because x represents the distance relative to the first charge where the net electric field will be equal to zero. So how can we calculate x? The net electric field at point B, remember at point B we have E1, which is going towards the right, and E2, I mean at point, not B, but point P. At point P, we have E1 going to the right, and E2 is going to be going to the left. So because E1 is going to the right, it's positive. E2 is going to the left, it's negative. Now, we want, point P is defined as the point where the net electric field is 0. So E1 minus E2 will be 0. If we add E2 to both sides, we'll get that E1 is equal to E2. They have to be the same in magnitude, but opposite in direction. E1 is KQ1 over R1 squared. E2 is KQ2 over R2 squared. Now let's divide both sides by K, so we can cancel that term. And let's replace r1 with x. So we have q1 over x squared. And let's replace r2 with 1 minus x. Don't forget to square it. Now let's cross multiply. So here we're going to have Q2 times x squared and then this is going to be Q1 times 1 minus x squared. Now let's replace Q2 with its value and let's keep the unit microclooms. So we have 200 microclooms times x squared. Q1 is 100 microclooms. And what we're going to do at this point is we're going to divide both sides by 100 microcoulombs. So the unit microcoulombs will cancel on the left side. On the left, we have 200 over 100, which is 2. So we get 2x squared is equal to 1 minus x squared. Now we don't need to FOIL 1 minus x squared. What we could do is take the square root of both sides. So we're going to have the square root of 2, and the square root of x squared is just x. The square root of 1 minus x squared is just 1 minus x. Now the square root of 2 is 1.414. So now we need to do some algebra. Let's add 1x to both sides. Let me write that better. So there's a coefficient of 1 here. As we add 1x to both sides, My handwriting is just not working today. I don't know why. 1.41x plus 1x is 2.41x. And we can bring down the 1 on the right side. So we have 2.41x, I mean 2.414x is equal to 1. So to get x by itself, we need to divide both sides by 2.414. So x is going to be 1 divided by 2.414, and so we get 0.414. So at 0.414 meters to the right of Q1, which is point P, at that point the net electric field will be equal to zero. So that's how we can calculate the exact location along the x-axis where the net electric field will be equal to zero if Q1 and Q2 have different magnitudes of charge.

So it's F over Q. The force is measured in Newtons, the charge Q is measured in Coulombs. So electric field has the units Newtons per Coulomb.

Now the electric field is a vector, much like force is a vector. But the good thing about the electric field is, it tells you how the electric force is going to be acting on a positive or negative charge. So let's talk about positive charges first. What happens if we put a positive test charge in an electric field? That test charge is going to feel a force that will accelerate it in the same direction as the electric field.

Now what about a negative test charge? What's going to happen if we put it in an electric field? A negative test charge will feel a force that will accelerate it in the opposite direction to the electric field.

So make sure you understand that. Positive charges, they accelerate in the same direction as the electric field. The force and electric field vectors will be in the same direction. But for negative charges, they will feel a force that will accelerate in the opposite direction of the electric field. Now, as we said before, an electric field can exert a force on any type of charge, a positive charge or a negative charge.

But it turns out that charges can also create their own electric fields. The electric field created by a positive charge extends in all directions away from the positive charge. The electric field created by a negative charge extends in all directions toward the negative charge. So it's going inward towards the negative charge in all directions. Now let's talk about the equation that will help us to calculate the electric field created by a point charge.

So let's say we have, let me draw a better circle. Let's say we have this charge, which we'll call big Q. And we want to calculate the electric field at some point A.

How can we do that? And let's say A is some distance R. Well, we know that the electric field at point A will be going in this direction. All you need to do is draw an arrow from the positive charge to the point of interest, and that will give you the direction of the electric field. Now, let's say if we place a tiny positive charge at point A.

Let me just erase point A. So this tiny positive charge will be considered our test charge. The reason why it has to be tiny, because if it's large enough, it will affect Q.

Big Q will be repelled by little q. But if the magnitude of the charge is small, it will be repelled by little q. of little q is very very small it won't affect big Q as much and thus it won't affect the electric field that is created by big Q as much so that's this equation will work if we choose a very tiny positive test charge Now, according to Coulomb's Law, whenever you have two charges next to each other, they will exert a force on each other.

Two light charges will feel a force that will repel them. And we can calculate that force using this equation. It's equal to K times Q1, which we'll call big Q, times Q2, which we'll call little q, over R squared, where R is the distance between the two charges.

So what we're going to do now is we're going to take this equation and substitute it for f in that equation. But first, I'm going to rewrite this equation as f over 1 times 1 over q. That's the same as f divided by q.

So now, let's replace f with what we see here. So this is going to be f, I mean, it's no longer f, but it's going to be k, big Q times little q over r squared. So this whole thing is f, and then we're going to multiply it by 1 over q. And so we could cancel out little q, the test charge.

Thus, we get the equation that gives us the electric field, or the magnitude of the electric field, for a point charge, capital Q. So it's KQ over R squared. So if we have our charge, Q, and we wish to calculate the electric field at point A, We could use this formula. K is 9 times 10 to the 9 Newtons times square meters over square coulombs. Q is the magnitude of the charge in coulombs.

Now remember, one microcoulomb is 1 times 10 to the minus 6 coulombs. A nanocoulomb is 10 to the negative 9 coulombs, and a millicoulomb is 10 to the minus 3 coulombs. And R is going to be the distance in meters. Now let's say we have a positive charge here. And we have point A, let's say point B, point C, and point D.

Determine the electric field created by the charge Q at points A, B, C, and D. So at point A, the electric field, in order to draw it, we're going to draw it starting from the positive charge towards point A. And so the electric field is going north. For point B, we need to draw it towards point B.

So the electric field, the direction is west. For point D, it's going to be at an angle. So it appears to be going in the northwest direction. And for point C, just draw it towards point C.

So it's going in the... Oh wait, I have to make a correction. That is not northwest.

That is northeast. And this is southeast. Now let's do something similar but with a negative charge. So for the sake of practice, go ahead and determine the direction of the electric field at the following points. So point A, point B, point C, and point D.

So this time, the electric field is going to point toward the negative charge. So we're going to draw it from the point to the negative charge. So this electric field vector is going west. And then here we're going to draw it from B towards the negative charge. So this is going in the south direction.

And then D to the negative charge. Here it's going east, and then from C to the negative charge, this electric field vector is pointed in the northwest direction. So that's how we can determine the direction of any electric field vector that is created by some type of charge, either a positive or negative charge, and we can determine the direction of that vector at any point using examples such as these.

But now let's focus on some word problems. Number one, a force of 100 newtons is directed north on a negative 20 microcoulomb point charge. What is the magnitude and direction of the electric field at this point?

So feel free to pause the video and work on this example problem. So let's begin by drawing a picture. So let's draw our negative point charge.

Now there is an electric force that is acting on this charge, and that electric force is directed north. What is the direction of the electric field? Now if you recall...

If we have an electric field pointing east, and if we were to place a positive test charge in that electric field, it will feel a force in the same direction as the electric field. But a negative charge will feel a force that will accelerate it in the opposite direction. So if the electric field is due east, the electric force acting on the negative charge will be west. So if the force is north, what is...

the direction of the electric field acting on this charge Q. It has to be in the other direction. It has to be south.

So that is the direction of the electric field. It's directed south. Now if we want an angle, we can draw this.

This is 0, 90, 180, 270. So the electric field is directed along the negative y-axis. So we can say that it's at an angle of 270 degrees relative to the positive x-axis. And that is in a counterclockwise direction.

So those are the ways in which we can describe the direction of the electric field. We could say it's due south, or it's at an angle of 270 degrees. So now that we have the direction of the electric field, let's calculate the magnitude of the electric field.

So we could use this formula. We can take the electric force and divide it by the magnitude of the charge. The force acting on it is 100 newtons.

The magnitude of the charge, we don't need to worry about the negative sign since we already know the direction. The charge is 20 microcloons, and we can replace micro with 10 to the minus 6. Now let's do some algebra. Let's see if we can get this answer without the use of a calculator.

We could divide 100 by 20. How many 20s would make up 100 bucks? That's going to be 520, so 100 divided by 20 is 5. Now the 10 to the minus 6, if we move it to the top, the negative exponent will become a positive exponent. So the negative 6 will change to positive 6. And thus the answer is going to be 5 times 10 to the 6 and the units in newtons per coulomb.

So that is the electric field that's acting on this negative charge. Number two, a positive charge of 50 microcoulombs is placed in an electric field of 50,000 newtons per coulomb directed upward. What mass should the charge have to remain suspended in the air? Well, let's talk about how we can create such an electric field. We can create this electric field using a battery and two parallel metal plates.

So this is the electrical symbol of a battery. We're going to connect it to these two plates. This is the negative side of a battery and this is the positive side.

So this plate is going to acquire a positive charge. And this plate is going to require a negative charge. So if we choose a high enough voltage, and if the distance between the two plates is just right, we can adjust it such that we get an electric field of 50,000 newtons per coulomb. Now we want the electric field to be directed upward. So we want the positively charged plate to be at the bottom, the negatively charged plate to be at the top.

Remember, the electric field always extends away from a positive charge and points towards a negative charge. So this is how we can create a uniform electric field using two parallel metal plates. Now let's place our positive charge in the middle between these two plates.

If this charge has the right magnitude, mass, it can actually remain suspended in the air. Now let's talk about it. So we have an electric field that is directed upward, and it's acting on this positive charge. What will be the direction of the electric force on this charge? The electric force will be the same direction as the electric field for a positive charge.

So the electric force will be directed upward. Now gravity Gravity likes to bring things down, so gravity is going to exert a weight force on this positive charge, bringing it down in the negative y direction. In order for this charge to remain suspended in the air, The electric force that wants to accelerate the charge towards the negatively charged plate, that has to be exactly equal to the gravitational force, or the weight force, that wants to bring the positive charge down towards this plate.

So if we can get these two forces to equal each other, then the positive charge will remain suspended in the air. So let's write the forces in the y direction. The sum of the forces in the y direction is equal to the electric force.

This is positive. And the weight force is negative because it's going in a negative y direction. And we want the sum of the forces in the y direction to be zero, so that there's no net acceleration, so it remains suspended in the air.

Moving this term to the other side, we see that the electric force has to equal the weight force. Now, the electric field is equal to F over Q. Multiplying both sides by Q, we can see that F, the electric force, is the electric field times Q.

So let's replace the electric force with E times Q. Now the weight force is simply mg. So now we have everything that we need in order to calculate the mass of this charge. Let's divide both sides by g.

So the mass of the charge is going to be equal to the electric field times the magnitude of the charge divided by the gravitational acceleration. The electric field is 50,000 newtons per coulomb. The magnitude of the charge, we're dealing with a 50 microcoulomb charge. So it's going to be 50 times 10 to the negative 6 coulombs. And then we're going to divide that by the gravitational acceleration of 9.8 meters per second squared.

So it's 50,000 times 50 times 10 to the minus 6 divided by 9.8, and you should get 0.255 kilograms. So an electric field of 50,000 newtons per coulomb can suspend a positive charge with a mass of 0.255 kilograms, or 255 grams. it can suspend it in the air.

If the mass is greater than this number, then the charge will fall down. If it's too light, if the mass is less than this number, then the charge will accelerate towards the negatively charged plate. So the mass has to be at the right number in order for it to remain suspended. Number three, an electron is released from rest in a uniform electric field and accelerates to the east at a rate of of 4 times 10 to the 6 meters per second squared. What is the magnitude and direction of the electric field?

So let's draw a picture. So first, let's draw our electron, and then it is accelerating towards the east. Now, according to Newton's second law, the net force is equal to the mass times the acceleration.

The net force is in the same direction as the acceleration. So the electric force exerted by this electron due to the electric field is going to be due east as well. Now, if the electric force is east, what is the direction of the electric field? It's going to have to be west. Let me put it here.

When dealing with a negative charge, the direction of the force and the electric field, they will be opposite to each other. So now that we have the direction of the electric field, let's focus on getting the magnitude of the electric field. So from the last problem, we saw that the electric force is equal to the electric field times the charge.

And using Newton's second law, we can replace the force with the mass times the acceleration. So now we can calculate the electric field if we divide both sides by q. So for this problem, the magnitude of the electric field is going to be the mass times the acceleration, which is the force, divided by the charge. So the mass is... what is the mass of an electron?

If you look it up, the mass of an electron is 9.11 times 10 to the negative 31 kilograms. The acceleration given to us in this problem is 4 times 10 to the 6 meters per second squared. And then we're going to divide that by the magnitude of the charge. So what is the charge of an electron?

The charge of an electron is negative 1.602 times 10 to the negative 19 coulombs. So these are some numbers that you want to be familiar with. So let's put that here.

So let's go ahead and plug in these numbers. By the way, don't worry about the negative sign for Q. It's not going to be relevant here.

So the magnitude of the electric field is going to be 2.27 times 10 to the negative 5 Newtons per coulomb. So this is the answer for this problem, and the direction is west. So make sure you write down these numbers. The mass of an electron, as we've considered, is 9.11 times 10 to the negative 31 kilograms. The mass of a proton is 1.67.

times 10 to the negative 27 kilograms. The charge of an electron, it's going to be negative 1.602 times 10 to the negative 19 coulombs. The charge of the proton, it's going to have the same magnitude, but the opposite charge. It's going to be positive 1.602 times 10 to the negative 19 coulombs.

So those are some numbers that you want to be familiar with when working on problems associated with electric fields if you're dealing with protons and electrons. Number 4. A 40 microcoulomb point charge is placed at the origin. Calculate the magnitude and direction of the electric field created by the point charge at the following locations.

So let's draw the point charge first. Now, point P is 5 meters away from the point charge along the x-axis. So this is point P, and this is 5 meters away.

What is the electric field at point P? Well, we know the direction. The direction of the electric field is going to be east, if we draw it from the positive charge towards point P. Now to calculate the magnitude of the electric field, it's going to be kq divided by r squared. So k is 9 times 10 to the 9, and then it's newtons times square meters over square coulombs, and I'm running out of space, so I'm not going to put the units here.

I'm just going to write 9 times 10 to the 9. q is 40 microcoulombs, so 40 times 10 to the minus 6 coulombs. R is in meters. R is the distance between a charge and a point of interest.

So that's 5 meters. So it's going to be 9 times 10 to the 9, times 40, times 10 to the minus 6, over 5 squared. And so the electric field is going to be 14,400 newtons per coulomb.

So that's the magnitude of the electric field at point P, and this is the direction. For those of you who want to understand how the units work, here it is. So K is in newtons times square meters over square coulombs. Q is in coulombs, and R is in meters, so we have square meters. As you can see, square meters cancel.

Now coulombs squared, we can write that as coulomb times coulomb. So we can cancel. one of the Coulomb units.

Thus, we're left with newtons per Coulomb, which is what we have here. So that's it for part A. So it's 14,400 newtons per Coulomb directed east.

Now let's move on to part B. So let's calculate the electric field at point S. So, Q is at the origin. Point S is 3 meters east of Q, and then 4 meters north from that point.

So, S is at this position here. The electric field can be drawn from Q to S. So the electric field is going in that direction.

It's going in the northeast direction. Let's calculate the magnitude of the electric field. So let's use this formula again. It's going to be kq over r squared.

So k doesn't change. It's a constant. It's 9 times 10 to the 9. q is still the same. It's 40 times 10 to the minus 6. But r is different.

R is no longer the value that we have here. But in actuality, it turns out R is the same. R is the distance between the charge and the point. So we need to use the Pythagorean theorem to calculate the hypotenuse of that right triangle. So this is a, b, and this is c.

So c squared is equal to a squared plus b squared. a is 3, b is 4. 3 squared is 9, 4 squared is 16, and then 9 plus 16 is 25. Taking the square root of both sides. we get 5 again.

So by coincidence, the electric field is going to have the same magnitude as it did in part A, which was 14,400 N per C. What's going to be different though is the direction of the electric field vector. How can we determine the direction? We know it's somewhat in the northeast direction but not necessarily at a 45 degree angle.

What we need to do is calculate theta. So perhaps you remember from trigonometry, Sokotola. If we focus on the Tola part, that tells us tangent, tangent of theta is equal to the opposite side divided by the adjacent side. So tangent of the angle theta is going to be equal to 4 over 3. To calculate theta, we need to take the arc tangent, or the inverse tangent, of 4 over 3. Go ahead and type that in your calculator, and make sure it's in degree mode.

So arc tangent 4 over 3 is 53.1 degrees. So that is the angle with respect to the positive x-axis. It's 53.1.

So we could say that this is the electric field vector at an angle of 53.1 degrees with respect to the x-axis, or we could say 53.1 degrees north of east. Because here, we're starting from east and we're heading towards the north direction. So it's 53.1 north of east. So that's it for part B.

So that's how we can calculate the magnitude of the electric field vector, and also its direction using arc tangent. Number 5. An electron initially at rest is placed in an electric field of 2 times 10 to the 4 newtons per coulomb directed to the west. The distance between the plates is 1 centimeter.

What is the acceleration of the electron due to the electric field? So the electric field will emanate away from the positive charge and will point towards the negative charge. So as we can see, the electric field is directed west. Now what effect will it have on the electron? A negatively charged particle will feel a force that will accelerate it in the opposite direction to the electric field.

So the electron is going to shoot out of this between these two parallel plates. Let's calculate the acceleration. We know the force acting on a charged particle is equal to the electric field times the charge of that particle.

And since this is the only force acting on the electron in the X direction, the net force is going to be the electric force. So we can replace the net force with ma based on Newton's second law. So ma is equal to E times Q. And to solve for A, we're going to divide both sides by m.

So for part a, the acceleration is going to be the force, which is E times Q, divided by the mass. So we have an electric field of 2 times 10 to the 4 newtons per coulomb. And the charge of an electron is 1.602 times 10 to the negative 19 coulombs. We're not going to worry about the negative sign. The mass of an electron is 9.11 times 10 to the negative 31 kilograms.

So let's go ahead and plug this in. So the acceleration is 3.517 times 10 to the 15 meters per second squared. So that's going to be the acceleration when the electron leaves the negatively charged plate. So now let's move on to part B. What will be the speed of the electron after it leaves the hole?

So how can we get that? So now we need to go back to kinematics. The electron is initially at rest, so v initial is zero.

We're looking for the final speed, so we'll put a question mark. And we know the distance between the plates. It's approximately, it's one centimeter.

So what kinematic formula has acceleration, v initial, v final, and d? It's going to be this one. v final squared is equal to v initial squared plus 2ad.

So to solve for the final speed, we simply need to take the square root of both sides. So v final is equal to this. The initial speed is 0. This is going to be 2 times the acceleration, which is 3.517 times 10 to the 15. And the distance between its plates is 1 centimeter.

So if we convert 1 centimeter into meters, we need to divide by 100 and it's going to be 0.01 meters. And so we're going to get the square root of 7.034 times 10 to the 13th. And so the final speed is about 8,386,894.5. So we can round that to, let's say, 8.39.

And this is times 10 to the 6th. meters per second. So that's how fast the electron is going to be moving when it leaves the hole. Number six, a 200 microcoulomb charge is placed at the origin and a negative 300 microcoulomb charge is placed one meter to the right of it.

What is the magnitude and direction of the electric field midway between the two charges? And then for the second part, 30 centimeters to the right of the negative charge. So let's start with the first part of the problem.

Let's begin by drawing a picture. So this is going to be the first charge, we'll call it Q1, and the second charge, Q2. So these two charges are separated by a distance of 1 meter.

And we want to determine the magnitude and the direction of the electric field midway between the two charges. So that's going to be at this point. How can we do that?

Well, we need to determine the direction of each electric field at that point. Q1 is going to create an electric field called E1, which will be directed east. Now Q2 will create an electric field, E2, which starts from the point but points towards the negative charge, and that's going to be directed east as well.

So remember, the electric field created by a positive charge extends away from the positive charge, but the electric field... field created by a negative charge points towards a negative charge. So at the same E1 and E2, they're in the same direction.

So the net electric field is going to be E1 plus E2. Along the x-axis, or the horizontal axis, both of these are positive because they're going in the positive x direction. E1 is k times q1 over r1 squared. E2 is k times q2 over r2 squared. Now, what's r1 and r2?

So r1 is the distance between q1 and the point of interest. r2 is the distance between q2 and the point of interest. So R1 and R2, they're both half of point of one meter, which means that R1 and R2, they're both equal to 0.5. So because R1 and R2 are the same, we can simply call it R.

So let's replace R1 with R. And let's replace R2 with R as well. So now we could simplify this equation by factoring out the GCF, the greatest common factor, which is going to be k over r squared.

And then we're left with q1 plus q2. So this is the formula that we could use to calculate the net electric field for this particular part in the problem. Now let's go ahead and plug in the numbers.

So it's going to be k, which is 9 times 10 to the 9, over r squared. r is 0.5. and then times Q1. Q1 is 200 times 10 to the negative 6. Now for Q2, we're going to use a positive value, not negative 300 times 10 to the negative 6, because we already know the direction of E2.

It's going to the right, and it's going to have a positive value because it's heading in the eastward direction. So whenever you're calculating the magnitude for electric field or electric force, you don't need to include the negative charge. You can just find the direction based on where the arrow is going. So let's replace Q2 with 300 times 10 to the negative 6 coulombs.

So 200 plus 300, this becomes 500. So it's 9 times 10 to the 9 times 500 times 10 to negative 6 divided by 0.5 squared. Thus, the net electric field is 18 million, which is 1.8 times 10 to the 7 newtons per coulomb. So that's the answer for part A. Now let's move on to part B. So let's redraw the picture for that.

So here we have our positive charge and the negative charge. So we got Q1, Q2, and they're separated by a distance of 1 meter. But 30 centimeters to the right, or 0.3 meters, we're going to have our new point of interest. And let's call this point B.

So we want to determine the net electric field at point B. So E1, the electric field created by Q1, if we draw it from Q1 to point B, we can see that it's going east. Now, if we draw an electric field from point B to Q2, because it's a negative charge, it needs to go towards a negative charge, it's going to the left.

Now, which of these two electric fields is greater, E1 or E2? What would you say? Notice that point B is closer to the negative charge than it is to the positive charge.

So E2 is going to have a greater effect than, I mean Q2 is going to have a greater effect on point B than Q1 because it's closer. So if you go back to the equation for electric field, there's two things that the electric field depends on. The magnitude of the charge.

and the inverse square of the distance. But the distance is squared, so the distance has a greater impact than the charge. But also, Q2 has a bigger charge than Q1. So those are two factors that favor Q2 over Q1.

Point B is closer to Q2, and Q2 has a greater charge magnitude than Q1. So therefore, we can conclude that E2 is going to be bigger than E1. Now, the net electric field is going to be positive E1, because it's going along the positive x-axis, and then plus negative E2, because that's moving towards the west, or the negative x-axis.

Now, Now if E1 is bigger than E2, the net electric field will be positive. If E2 is greater than E1, it's going to be negative. And we've confirmed that E2 is going to be bigger. So we should get a negative value.

If we get a positive value for the net electric field, we did something wrong. So let's go ahead and do the math. So the net electric field is going to be kq1 over r1 squared plus kq2 over r2 squared.

So this time r1 and r2 will be different. So we can't simplify this process by factoring. So let's plug in the numbers.

This is k, q1 is 200. Oh, this should be a negative sign, by the way. based on what we have here. So Q1 is 200 times 10 to negative 6. Now R1, R1 is the distance between Q1 and point B.

So R1 is going to be the sum of 1 and 0.3. Thus R1 is 1.3 meters squared. and then minus now because we've considered the direction of e2 we've assigned a negative value we don't need to plug in this negative value for q2 we've already taken that into an account so it's going to be minus k and then q2 we're going to use the positive value of q2 300 times 10 to the minus 6. And R2, that's the distance between Q2 and point B, that's 0.3 meters. So let's calculate E1 first.

Let's focus on this fraction. 9 times 10 to the 9th times 200 times 10 to negative 6 divided by 1.3 squared. That's 1.065 times 10 to the 7th Newton's percolum.

Now, focusing on E2, that's going to be 9 times 10 to the 9th times 300 times 10 to the negative 6 divided by 0.3 squared. So this is... 3 times 10 to the 7 Newtons per Coulomb. So we can see that this number is bigger than that number.

So the net electric field is going to be negative. 1.065 minus 3. Wait, something is wrong. Let me double check my work.

This should be times 10 to the 6, not 10 to the 7. That's 1,065,088. So that's 1.065 times 10 to the 6. E2 is 30 million, which is 3 times 10 to the 7. So now, if we subtract those two numbers, we get this answer. Negative 2.89 times, so it's 28,935,000, so it's negative 2.89 times 10 to the 7 newtons per coulomb. So we can see why it's negative.

E2 is significantly larger than E1. And so that's it for this problem. Number seven, the electric field at point X, two meters to the right of a certain positive charge, is 100 newtons per coulomb. What will be the magnitude of the new electric field if the magnitude of the positive charge doubles in value? So let's draw a picture first.

So here is our positive charge, and let's say this is point X, and the distance between these two is two meters. Now at that point the electric field is 100 N per C when the magnitude of the charge is, we'll call it Q. But what happens if we double the magnitude of the charge?

So the electric field is KQ over R squared. If you double Q, the electric field is going to double. A quick and simple way to get the answer is to plug in one for everything that doesn't change. Q doubles, so we're going to plug in two. R remains the same, so the electric field is going to double.

It's going to go from 100 to 200. Now what about part B? Let's say the distance between the charge and point X doubles. So let's say point X is now over here, and the magnitude is Q.

What will be the new electric field? So this time, Q doesn't change. K is the same, so we're going to replace it with a 1. Everything that doesn't change, replace it with a 1. Now the distance doubles.

2 squared is 4, so the electric field is going to be 1 fourth of its original value. 1 fourth of 100, or 100 divided by 4, that's 25. So what you need to take from this is that the electric field is weaker at a greater distance away from the charge. The closer you move towards the point of charge, the closer you move towards the point of charge.

charge, the greater the electric field will be. So as the distance increases, the electric field decreases. But as the distance from the point charge decreases, the electric field increases. So there's two ways in which you can increase the electric field. You can increase the magnitude of the charge, which will cause the electric field to go up.

Or you could reduce the distance between the point of interest and the charge. And that will also increase the electric field. Now let's move on to part C.

The distance between the charge and point X reduces by a factor of 3. So what's going to be the magnitude of the new electric field in that case? So we're bringing it a lot closer to Q. So here's the new position of X.

So here the distance doubled to 4 meters. But now it's going to be reduced by factor 3, so it's 2 thirds of a meter. So let's use this formula again.

For part C, k and q doesn't change. r is now 1 third of its original value. 1 times 1 is 1, 1 squared is 1, 3 squared is 9. Now we need to multiply the top and the bottom by 9. 1 times 9 is 9, 1 ninth times 9, the 9th cancel, we get 1. So the electric field is going to increase by a factor of 9. 100 times 9 is 900. So as you can see, as we get closer to the point charge, the magnitude of the electric field greatly increases. Now what happens if we triple the magnitude of the charge, but at the same time reduce the distance to 1 fourth of its original value? So I won't draw a new picture for this.

Let's just get the answer. So k doesn't change, q triples, and the distance is reduced to 1 fourth of its value. So we have 1 times 3, which is 3, 1 squared is 1, 4 squared is 16. So now I'm going to multiply the top and the bottom by 16. So it's going to be 3 times 16, which is 48. So the electric field will increase by a magnitude of 48. So the original electric field was 100. If we multiply that by 48, the new electric field will be 4800 newtons per coulomb. So this right here is the answer to part D. So the reason why it's so high is because we've increased the charge, which causes E to go up, and at the same time we reduce the distance, which greatly increased the value of E.

So that's it for this problem. Number 8. Two identical point charges with a magnitude of 100 microclumes are separated by a distance of 1 meter, as shown below. Part A. At what point will the net electric field be equal to 0?

Will it be to the left? Will it be between Q1 and Q2? Or to the right?

So let's identify three points of interest. The first point will be somewhere to the left, which we'll call point A. The second point will be somewhere in the middle between Q1 and Q2, likely the midpoint.

And then C will be to the right. Now, to draw the electric field vector created by Q1, we need to draw a line from Q1 to point A. This is going to be E1.

And for E2, we're going to draw it from Q2 to point A. So at point A, both electric fields are moving to the left. So there's going to be a net electric field at point A. It's not going to be zero. But here's a question for you.

Which electric field will be greater, E1 or E2? Now remember, these charges are identical, so the magnitude of the charge is the same. The only thing that's different is the distance.

Q1 is closer to point A than Q2. So E1 is going to be bigger than E2. So if this is E1, E2 is going to be a smaller vector. Nevertheless, the net electric field at point A will be directed west.

Now what about at point C? To draw E1, we're going to draw a line from Q1 to point C. And E2 will be from Q2 to point C.

Now which one is bigger? Q2 is closer to point C than Q1, so E2 is going to be bigger at point C. So for E1, we're going to draw a small vector, and for E2, we're going to draw a bigger vector.

Nevertheless, the net electric field at point C will be directed east. Now what about at point B? E1 is going to be directed away from Q1, but towards point B.

E2 will be directed away from Q2, but towards point B. And Q1 and Q2 have the same charge, and at the midpoint, at point B, they will be equally distant from point B. So if the charges are the same and the distances are the same, the magnitude of E1 and E2 will be the same, but they're opposite in direction, which means that E1 and E2 cancels.

So at point B, the net electric field is zero. So let's assume that Q1 is the origin. It's at position 0. Point B will be at 0.5 meters.

So at 0.5 meters relative to the first charge, the net electric field will be 0. E1 and E2 will cancel completely. So that's the answer for part A. At point B, or at the midpoint between Q1 and Q2, the net electric field will be 0. Now what about part B?

If the charge on Q2 doubles to 200 microcouplems, where along the x-axis relative to the first charge will the net electric field be equal to zero? And the distance between these two is still the same. Do you think the net electric field will be equal to zero between Q1 and B or between B and Q2?

Now, Q2 is bigger than Q1. So at point B, where they're equidistant, E1 is going to be a smaller vector than E2. E2 is going to be twice as large.

In order to make these vectors equal, we need to increase E1 and decrease E2. If we can't change Q1 and Q2, the only thing we could change is location. We need to move closer to Q1. As we move closer to Q1, E1 is going to get bigger, E2 will get smaller, and at some point, they're going to equal each other.

So, we're going to place point P between Q1 and B. Somewhere between Q1 and B, the net electric field will be equal to zero, and we need to find that point. So that point is going to be R1. R2. is the distance between Q2 and point P.

Now let's call R1 X. If R1 is equal to X, what's R2? Notice that the total distance is 1. So R2 is going to be 1 minus X. If you add X and 1 minus X, you're going to get 1. So at this point, All we need to do to get the answer for part B is get the value of x, because x represents the distance relative to the first charge where the net electric field will be equal to zero.

So how can we calculate x? The net electric field at point B, remember at point B we have E1, which is going towards the right, and E2, I mean at point, not B, but point P. At point P, we have E1 going to the right, and E2 is going to be going to the left.

So because E1 is going to the right, it's positive. E2 is going to the left, it's negative. Now, we want, point P is defined as the point where the net electric field is 0. So E1 minus E2 will be 0. If we add E2 to both sides, we'll get that E1 is equal to E2. They have to be the same in magnitude, but opposite in direction. E1 is KQ1 over R1 squared.

E2 is KQ2 over R2 squared. Now let's divide both sides by K, so we can cancel that term. And let's replace r1 with x. So we have q1 over x squared. And let's replace r2 with 1 minus x.

Don't forget to square it. Now let's cross multiply. So here we're going to have Q2 times x squared and then this is going to be Q1 times 1 minus x squared. Now let's replace Q2 with its value and let's keep the unit microclooms. So we have 200 microclooms times x squared.

Q1 is 100 microclooms. And what we're going to do at this point is we're going to divide both sides by 100 microcoulombs. So the unit microcoulombs will cancel on the left side.

On the left, we have 200 over 100, which is 2. So we get 2x squared is equal to 1 minus x squared. Now we don't need to FOIL 1 minus x squared. What we could do is take the square root of both sides. So we're going to have the square root of 2, and the square root of x squared is just x.

The square root of 1 minus x squared is just 1 minus x. Now the square root of 2 is 1.414. So now we need to do some algebra.

Let's add 1x to both sides. Let me write that better. So there's a coefficient of 1 here. As we add 1x to both sides, My handwriting is just not working today. I don't know why.

1.41x plus 1x is 2.41x. And we can bring down the 1 on the right side. So we have 2.41x, I mean 2.414x is equal to 1. So to get x by itself, we need to divide both sides by 2.414.

So x is going to be 1 divided by 2.414, and so we get 0.414. So at 0.414 meters to the right of Q1, which is point P, at that point the net electric field will be equal to zero. So that's how we can calculate the exact location along the x-axis where the net electric field will be equal to zero if Q1 and Q2 have different magnitudes of charge.

Transcript for:Understanding Electric Fields and Their Behavior

Transcript for:
Understanding Electric Fields and Their Behavior