Coulomb's Law (Electric Force)

Concept: Coulomb's Law

12m
Video Transcript

Hey guys. In this video we're going to be talking about Coulomb's law which tells us the electric force between two charges. Because we're talking about electricity Coulomb's law is obviously going to be very important. So, make sure that you watch this video as many times as you need and do all the practice problems so that you understand this well before moving on, okay? Let's get to it as, we know the electric force can be either attractive or repulsive, this is a direct consequence of the fact that charges can be both positive or negative. Alright, mass which can only be positive only produces an attractive gravitational force, electric charges because they can be positive or negative can produce an attractive electric force or a repulsive electric force, okay? What Coulomb's law tells us is that if we would have two charges which we can call q1 and q2 separated by some distance r then the magnitude of the force that separates them, sorry, the magnitude of the force between them is going to be k q1 q2 over r squared where k which is known as Coulomb's constant is 8.99 ten times 10 to the ninth. Now, there are units associated with k, Newton's meter squared per Coulomb squared but these units are not, sorry, these units are not important to remember, okay? We don't have to remember those, the reason is, is that as long as you're using SI units you'll get SI units as a result. So, we know that as long as we use SI units our force is going to be in Newtons, okay? Now, this equation k is incredibly important you must remember this equation, sorry, this constant. Alright, this is one of those fundamental constants that you are always going to have to remember, okay? The professors most likely not going to give you this in an exam problem. So, please commit this to memory, it's very important. Alright, that's half of what the force is, force is a vector. So, we need to know both the magnitude and the direction.

Coulomb's law tells us the magnitude. Now, we need to talk about the direction, okay? The force is always going to point along the line connecting the charges, what do I mean by that? Well, if we look at these two charges right here, let's say on the top charge the force is attractive, well it's going to point from the top charge towards the bottom charge along this line between them, this is an attractive force, okay? Always on the line between them, what if the force between the two was repulsive? Well then on the top charge it would point away from the bottom charge but still on the line connecting them, okay? So, those are the two halves of what a force is, the magnitude and the direction. Now, remember like the charges do what? They repel, opposite charges do what? They attract, this is an absolutely key universal fact that you must remember, okay? Now, let's do an example, what is the ratio of the electric force to the gravitational force in a hydrogen atom, okay? So, all we need to do is we need to figure out, what is the electric force between the proton and the electron in hydrogen and then what is the gravitational force between the proton and the electron, okay? A bunch of constants will be given to you if this were, let's say in a homework problem or an exam problem, the first is the distance between the two, okay? That's going to be a 0.5 times 10 to the negative 10 meters, next the mass of the proton 1.67 times 10 to the negative 27 kilograms, finally, the mass of the electron which is 9.11 times 10 to the negative 31 kilograms, it will be up to you to remember the charge on the proton and the electron, okay? Which is what? the elementary charge, okay? So, let's start with finding out, what the gravitational force is, this is going to be given by g, m1, m2 over r squared where g the gravitational constant is 6.67 times 10 to the negative 11 with some units associated with it, mass of the proton, mass of the electron and the distance between them squared, okay? Plugging all this into your calculator getting an answer, it's going to be 4.06 times 10 to the negative 47 Newtons. Now this is an incredibly small number, this is a 0 decimal point and forty six zeros before the 4, okay? It's almost hard to describe how small this number is. Alright, let's see if the electric force is any bigger, the electric force, remember, Coulomb's law is k, q1, q2 over r squared so that is Coulomb's constant which you must commit to memory 8.99 times 10 to the 9 with its units, the elementary charge which is the charge on a proton 1.6 times 10 to the negative 19, elementary charges also the charge on an electron 1.6 times 10 to the negative 19 divided by that distance squared, okay? Plugging this into your calculator getting an answer, this is about 9.21 times 10 to the negative 8 Newton's which is considerably larger, how much larger, let's find the ratio which is what the problem is asking for anyway, the electric force divided by the gravitational force is going to be 9.1 times 10 to the negative 8 divided by 4.06 times 10 to the negative 47 which is about 2.25 times 10 to the 39, okay? This number is a massive number how big is it? It's the two with 39 zeroes behind it, okay? This number is also almost indescribable large it's really, really big you're going to be hard-pressed to find any sort of analogy that you can make that tells you how big 10 to the 39 is, okay? The thing to take away from this problem is just that the electric force is much, much, much stronger than gravity, gravity is a pretty course, okay?

Now, our last problem in this video is going to be, if two identical charges are connected by a 5 centimeter wire with 10 Newton's of tension, what is the magnitude of the charges? Well, let's think about what's going to happen here, we take two like charges they're going to want to repel what, if we were going to connect them by a stream okay? Well, then that string is going to snap with some tension, right? As these two charges try to repel they can't because of the tension stream. So, there's going to be some electric force, let's say on the right charge, that's pushing it away from the left charge but the tension in the string is holding it back. Now, because the charge is not moving, we know by Newton's second law that that tension and that electric force have to be equal, since the tension is 10 Newtons that means the electric force is also 10 newtons okay? Now, we don't know what this charge is. So, we can't simply plug in Coulomb's law and get an answer but we do know what the electric force is, we do know the distance is. So, we can find the charges in Coulomb's law. Remember, that Coulomb's law relates three things, distance and charge to force, if we know any of the two we can find the third, in this case we know the distance and we know the force. So, we can find the charge, let's write out Coulomb's law in. Coulomb's law says that the electric force is k, q1, q2 over r squared, in our problem both of the charges of the same which we called q. So, let's plug that in, that's k, q squared over r squared which is 10 Newtons. Remember, that we know this, we know this. So, we can absolutely find q, okay? k is just a constant. So, we need to do is we need to rearrange this equation and say that q squared is going to be 10 Newton's times r squared because we have to multiply it up divided by k because we have to divide k to get it to the other side, so this is going to be 10 Newton's times r which is 5 centimeters or 0.05 five meters squared divided by k, you have to remember this constant guys, and that whole thing equals about 2.8 times 10 to negative 12, okay? So, what we need to do is take the square root and our final answer is 1.7 times 10 to the negative 6 coulombs, okay? Coulomb's law is incredibly important in electricity and you're going to need to know it very well before continuing. So, make sure that you watch this video as many times as you need to understand it and make sure you do all of the practice problems that follow and watch the solutions if you get stuck, okay? Thanks for watching.

Problem: If the force between two charges is F when the distance is d, what will the force between the two charges be if they were moved to a distance of 2d

3m

Example: Charges In A Line (Find Zero Force)

9m
Video Transcript

Hey guys. Now we're going to do an example on Coulomb's law, okay? Where should we put 1 Coulomb charge so that the net force on it is 0, okay? So, that the electric force due to the 2 and the 3 coulomb charge cancel each other out, okay? There are basically three places we can put this charge, we can put it to the left of the 2 Coulomb charge in between the two or to the right of the 3 Coulomb charge, okay? So, where should it go? Well, hypothetically, if we were to put it to the left of the 2 Coulomb charge. Notice that the force between the 2 Coulomb and the 1 Coulomb charge which is repulsive because they're alike charges it's going to point to the left in the force between the 3 Coulomb and the 1 Coulomb charge which is also repulsive because they're like charges is also going to be to the left. Now, is there any position that we can put this 1 Coulomb charge so those two forces cancel? Can two forces that point to the same direction ever add up to 0? No. the only way the two forces can add to 0 is if they point In opposite directions. Now, what if we were to place the 1 Coulomb charge to the right of the 3 Coulomb charge? Okay, well, in this case those forces are still repulsive because everything is still alike charge, the force of the 2 Coulomb charge is still to the right and the forks are the force due to the 2 Coulomb charge is to the right, it's still repulsive and the force of the 3 Coulomb charge is also to the right because it's still repulsive can those two forces ever cancel out? No, two forces pointing to the right just like two forces pointing to the left can never cancel, okay? So, you cannot put the charge on the left of the 2 Coulomb charge or on the right of the 3 Coulomb charge, you have to put it somewhere between, okay? So, there's going to be some position between where the force of the 2 Coulomb charge which is repulsive and points to the right it's going to exactly balance the force of the 3 Coulomb charge, which is also repulsive and points to the left, okay? These two forces which do point nn opposite directions will balance out somewhere in between the two, okay? So, let's say that at the position where they equal each other, where they balance out.

The one Coulomb charge of the distance x from the two Coulomb charge, if we call 10 centimeters d which is the total distance between the two then this 1 Coulomb charge will be a distance d minus x from the 3 Coulomb charge, okay? Now, our condition for solving this problem is that these two forces have to balance each other. So the condition of solving the problem is going to be that F2 equals F3, okay? That's our condition, that's what we have to apply to solve this problem. So, what is F2, what is that F3 those will be given by Coulomb's law, okay? Now, remember Coulomb's law is extremely important, you have to remember that equation, well, F2 is going to be k Coulomb's constant times the 2 Coulomb charge times the 1 Coulomb charge divided by the distance between the two which we called x squared, F3 is going to be k times the 3 Coulomb charge times the 1 Coulomb charge times the distance between those two which we said was d minus x squared, okay? Now, an algebra note, something is very, very important to remember here is that you cannot just distribute this square, you cannot say, oh, well, this 2 can be applied to both of these which would say that was d squared minus x squared, this is absolutely not true these are not equal do not do this, okay? So, let's set those two equal, that's our condition. Remember, for solving this problem. So, we'll say that k times 2 over x squared equals k times 3 over d minus x squared. Now, what we're planning on doing is isolating the x and solving for it. So, we got to multiply this x squared up and we're going to multiply that d minus x squared up. So, we can say this is equivalent to 3 x squared, right? The x squared becomes paired with the 3 that equals to d minus x squared, that d minus x squared becomes paired with the 2. Now, we can take the square root of both sides and this becomes square root of 3 times x equals square root of 2 d minus X, okay? Continuing with our algebra, we want to distribute this square root of 2 to both of those terms. So, this becomes equivalent to square root of 2 d minus square root of 2 times X, okay? So, let's rewrite our equation square root of 3 x equals square root of 2 d minus square root of 2 times x. Now, what we want to do is we want to group the x's on one side. So, I'll take this and I'm going to add it to the left side, right? It's negative on the right becomes positive on the left. So, it's become square root of 3 x plus square root of 2 x equals square root of 2 times d, alright? Now, both of these share an x in common. So, I can factor that x out, this is essentially adding the coefficients of x together this becomes square root of 3 plus square root of 2 times x, once again, let's rewrite our equation. So, we have square root of 3 plus square root of 2 times x equals square root of 2 times d, what do, I have to do now, to isolate the x? All I have to do is take this coefficient here and divide it over to the other side that gives us x equals the square root of 2 divided by the square root of 3 plus the square root of 2 times d. Now, plugging this into your calculator you find that this whole mess here, square roots is actually just 0.45 and d, if we remember, looking at your worksheet, is 10 centimeters multiplying those two together you just get that x is 4 and 1/2 centimeters. So, what does this mean? Well, remember what x was with the 2 Coulomb charge with the 3 Coulomb charge and we had the 1 Coulomb charge at distance of x from the 2 Coulomb charge, sorry about that, a distance of x from the 2 Coulomb charge so that means that the 1 Coulomb charge is 4 and 1/2 centimeters from the 2, this distance was the rest, what was left over d minus x which is 10 minus 4 and 1/2 or 5 and 1/2 centimeters. So this is our answer, that is the position where you have to put in 1 Coulomb charge so that the net force on the 1 Coulomb charge is 0, okay?

Problem: In which direction will the – 1 C charge move? If it has a mass of 10 g, what will its initial acceleration be?

6m

Example: Charges In A Triangle (Rank Force Pairs)

3m
Video Transcript

Hey guys. Let's do another example using Coulomb's law. Rank all the possible pairs of charges in the following figure by which pair has the greatest electric force, okay? So, we know Coulomb's law said that the electric force depends upon two things, it depends upon the multiplication of the charges, right? q1, q2 and the distance squared, okay? Those are the two things that it depends upon. Notice that for all the possible pairs of which we have 3, we have the 2e into e charge, we have the 2e and a 3e charge and we have the e and the 3e charge, those are our three possible pairs, okay? For all of those pairs the distance between the two charge is d. So, we can ignore the distance between them and all we need to do to figure out which force is going to be greater is figure out the first thing that the force depends upon which is the multiplication of the two charges, okay? So, if we look at this first pair here, this is 2e times e which is 2e squared, okay? We look at the second pair here, this is 3e times e which is 3e squared and if we look at the fourth pair, I'm sorry, the third pair 2e times 3e, this is going to be 6e squared, okay? So, which force is going to be greatest? Well, clearly this one, it's going to be k times 6e squared divided by d squared, okay? So, the 2e, if we rank them, this is the 2e, 3e pair is the highest the next highest is the 3e pair and the lowest is the 2e e pair, so that is the rank.

Example: Charges in a Plane

10m
Video Transcript

Hey guys. Let's do another example with Coulomb's law. Now, this is a bit more complicated than any of the ones that we've done so far in the sense that this is going to involve angles, okay? This isn't going to be everything in a line and we can't exploit symmetry in this problem to say right away in which direction our force is going to point, okay? So, this is going to involve some trigonometry, this is going to involve some algebra and a little bit of more complicated math but don't worry I'll take you through this step-by-step se we can all understand it, okay? Now, first there going to be two forces put on this 3 Coulomb charge and we want to find the net force, the force due to the negative two Coulomb charge is going to be attractive and I will call this F1, let's start here, okay? What is F1 going to be? Well, Coulomb's law tells us is going to be k, q1, q2 over r squared where k constant is what? Remember, got to memorize this 8.99 times 10 to the 9, q1 will say is two, two q2 will say is three and the distance is 8 centimeters or 0.8 squared multiplying that all together, we get 8.43 times 10 to the 12 newtons, okay? Very, very simple. Now, what we want to do is find the magnitude of the force between the one Coulomb charge in the lower-left and the three Coulomb charge, this is going to act across a greater distance, okay? And because both of these are positive charges it's going to be a repulsive force which I will call 2. Now, what's the magnitude of F2, well, it's still going to be given by Coulomb's law k, q1, q2 over r squared. So, let's start plugging in our numbers I'll say that q1 is 1 in this case q 2 is 3 doesn't matter which is which, the question is what is this distance, okay? Now, this distance we're going to have to find using the Pythagorean theorem because we know the two sides of this triangle so that distance is going to be the square root of 10 centimeters squared plus 8 centimeters squared which is 12.8 centimeters so this is 0.128 squared which is 12.8 centimeters in meters the SI unit that we need and so the whole thing becomes 3.29 times 10 to the 12 Newtons, okay? Now, finding the net force is not as simple as adding these two numbers together which is what we could do if everything lied on a single axis but they don't. So, first, we need to draw ourselves a Freebody diagram, okay? Now, I'm going to draw F2, F1 and a coordinate system in order to sum F1 and F2, I have to do the vector sum which means, I have to find the components of F2. F1 already lies on the X axis. So, I don't have to worry about its components, this is going to be theta and this is going to be F2 Y, this is going to be F2 X. So, we can see from here, that X component of F2 is cosine because it's the adjacent edge the one that touches the angle and that 2y is the sign because it's the opposite edge, the one that doesn't touch the angle. So, we know right away F2 X is F2 cosine theta and F2 y is F2 sine of theta, the question is what is cosine and what is sine of theta? The angles weren't given, all we know are the dimensions of the triangle 10 centimeters, 8 centimeters and 12.8 hypotenuse. So, let's draw this triangle and figure out what our trigonometric equations are, the triangle as is looks like this 10, 8, 12.8 hypotenuse, the triangle that we need is this triangle, this is the one with this angle, okay? That is given in my Freebody diagram, this triangle right here, okay. Notice that this is eight centimeters and this is 10 centimeters. So, we can pretty easily construct our trig equations, cosine of theta is just going to be the adjacent edge which is 8 over the hypotenuse which is 12.8 and that's going to be 0.63, sine of theta is just going to be the opposite edge 10 divided by the hypotenuse 12.8 and that's going to be 0.78, okay? So, this is our cosine, this is our sign. Now, we can find the x and y components, right? These components of F2, okay? Let's do that. Now, F2x which is F2 cosine of theta is going to be 3, sorry about that, is going to be 3.29 times 10 to the 12 times 0.63 three which is 2.07 times 10 to the 12 Newtons, F2y is going to be F2 sine of theta which is still 3.29 times 10 to the 12, that's just F2 sine of theta is 0.78. So, this whole thing is 2.57 times 10 to the 12 Newtons, okay? Now, we have our X and our Y components, we're still going to need F1. So, just take a second make sure that you've written F1 down because we're going to need to reference it and let's move on, okay? So, now once again our free body diagram look like this, we add F2y F2x and F1, where F1 was 8.43 times 10 to the 12 Newtons and that F2x is here and F2y is here. Now, we need to construct our net force, the net force in the x-direction is just going to be the difference between these two x forces. Now, the larger one is F1. So, I can just say, that's the positive one because the net force is going to point in that direction anyway. So, I can say, this is F1 minus F2x which is going to be 8.43 times 10 to the 12 Newtons minus 2.07 times 10 to the 12 Newton's which is 6.36, sorry, times 10 to the 12 Newtons and F net in the y direction which is going to just be F2y is 2.57 times 10 to the 12 newtons, okay? Now, finally, we have the X in the Y component of the net force, all we need to do is apply Pythagorean theorem to find the net force and then we're done, this is a really long problem and it's a lot of tedious math but we're at the end.

So the net force is just going to be the Pythagorean theorem of our x and y components, okay? And that is going to be 6.86 times 10 to the 12 Newtons, okay? This is going to be one of the more complicated Coulomb's law problems that you're going to encounter and you can't exploit symmetry, not everything lies on an axis, this is going to involve angles which means it's going to involve trigonometry, okay? And it's going to involve quite a bit of tedious math. Alright, so watch us a couple of times if you're unsure of any of the steps. Alright, I hope this helps.

Example: Exploiting Symmetry

5m
Video Transcript

Hey guys. In this example we're going to be talking about how to exploit symmetry, okay? This is a very common phrase you guys are going to hear in physics, a lot of problems are going to involve the summing electric forces for instance on a Freebody diagram, the net force is going to depend upon the vector addition of all the forces that go into it and there will be a lot of problems that due to the symmetry of the arrangement of charges some of the components of the forces will always cancel regardless of the magnitude of the charges, okay? So, these are two quick examples that illustrate how to exploit symmetry to quicken your calculations, let's look at the first triangle the one on the left, okay? We have a 2 Coulomb and a 2 Coulomb charge they're each put in a force on a 1 Coulomb charge and we want to know the direction of the net force on that 1 Coulomb charge, okay? So, let's look at the direction that each of these charges put a force on the 1 Coulomb charge the 2 coulomb charge on the left here is going to put a force in this direction, right? Repulsive, along the line connecting the charges. The 2 Coulomb charge on the right is going to put a repulsive force in this direction. Now, the thing here to realize is, if I were to draw a Freebody diagram for this 1 Coulomb charge, we have these two forces, because of the symmetry of the problem that we have a 2 Coulomb charge on the right, a 2 Coulomb charge on the left both of which are the same distance away from the charge with the 1 Coulomb charge directly in between them, we know that all of the components are going to be the same magnitude, okay? So, if we break this up, we have the X components for each force which are the same and the Y components for each force which are the same, since both X components are the same and they're pointing in opposite directions they have to cancel and the two Y components are in the same direction, they just sum, so the net force in this case simply points upwards and it isn't pointing in the general direction of up it points exactly up as in a 90-degree angle from the horizontal, okay? So, symmetry tells us exactly the direction of the net force on the 1 Coulomb charge alright.

Now, let's look at the problem on the, right? We have a 2 and a negative 2 Coulomb charge, the same thing applies, we're going to have two forces here, due to the 2 Coulomb charge a repulsive force is going to pull on the 1 Coulomb charge, okay? Due to the negative 2 Coulomb charge an attractive force is going to be put on the 1 Coulomb charge. Now, if I draw the free body diagram, I have these two forces and the same logic applies because this problem is symmetric all of the components of the forces are going to be the same, we have the same vertical components and the same horizontal components, the big thing here is that the two vertical components point in opposite directions. So, they cancel. The two horizontal components point to the same direction so they sum and that means that our net force is going to be pointing horizontally, once again, not just in the general direction of to the right but exactly horizontally with no angle between that in the horizontal axis, so this is what we would be referring to as exploiting symmetry, we're going to use the fact that these problems are symmetric in order to make problem-solving easier, okay? And this is something that appears commonly in physics in general. Alright, I hope this helped.

Problem: What is the direction of the net force on the charge at the center of the square in the following figure?

3m

Example: Electroscope (Find Charge)

9m
Video Transcript

Hey guys. Let's do an example using Coulomb's law. Two identical charges at the end of an electroscope leaves, each have a mass of 50 grams, if the electroscope leaves are deflected by 30 degrees as shown in the figure, what is it charged at the end of each leaf, okay? What an electroscope is, is it's just two conducting leaves that can accumulate a charge, okay? Now, normally, when these leaves are uncharged they hang, okay? Because their weight just keeps them in equilibrium pointed down. Now, when you put a charge on the electroscope it's going to spread evenly throughout the conductor. So, each leaf takes up an identical charge, okay? Because they're like charges the force is going to be repulsive and these leaves are going to push away, how far they push away depends upon how strong the charge is relative to their mass, okay? The heavier these leaves are the smaller the deflection, right? The lighter they are the larger the deflection, the same is true for the strengths of the charge or the strength of the electric force, the larger the charge is stronger the electric force the more they repel, okay? And viceversa is true for a smaller charge. Alright, so what we have here is two charges, I'll call them both q, they're identical and they're each putting an electric force on each other, that's repulsive, okay? So, here we have a repulsive electric force here, we have a repulsive electric force across some distance r. So, what's the equation for that electric force? that's just Coulomb's law that the electric force is going to be k q, q over r squared which we can just say is k, q squared over r squared, right? They're both identical charges q times q is q squared. Now, we want to find q, okay? This is what we're looking for right here, but we don't know the electric force so we can't solve for q that equation up there two unknowns the electric force and q, we can't use it, okay? So, let's look at what other forces are here and see if we can use Freebody diagrams to figure out the magnitude of the electric force, whenever you see that things like this that are in equilibrium, this charge is not moving the other charge is not moving either, okay? They're still. So, all the forces we know have to be balanced the best thing to do is to do Freebody diagrams to analyze the forces, okay? So, we have the next force is just a weight downwards, they both have a weight and they both have a tension in the leaf, okay? That's what's preventing them from just flying apart, okay? From just ripping away from the electroscope and flying apart, is the fact that these leaves are solid, okay? You can't pull the leaf apart, okay? So, these are the free body diagrams for each of the charge, let's look at one of them and I'll choose the right charge, the one right next to me, this charge has a tension in this direction t, a weight downwards and the electric force to the right and if I were to be able to find this angle theta then I would know the x component, right? Is just the cosine because it's the adjacent edge so this is t cosine of theta and the vertical component would be T sine of theta because it's the opposite edge, okay? So, our first question is what's beta, without theta we can't break the part of the tension, okay? Look at this triangle here, both of these edges are the same length and this total angle is 60 degrees that means that this is an equilateral triangle okay, if both of those edges were the same but the angle was not 60 degrees, let's say it was bigger or smaller than 60 degrees then it would be an isosceles triangle, only two of the edges would be same, okay? But because those two edges are the same length and that angle is 60 degrees all the other angles have to be 60 degrees and this length has to be half a meter because it's an equilateral triangle all the lengths are the same, okay? So, we know right away that this theta is 60 degrees, okay? Now, the question is what do we know, what do we not know, we know that mass. So, we know the weight, we know the angle but we don't know the tension and we don't know the electric force, however looking at the vertical component, if we can find the weight, right? Then knowing the angle, we can find the tension then knowing the tension, we can find the electric force by balancing the horizontal direction, okay? So, that's exactly, what we're going to do. So, let's start with the vertical direction, we're going to say that m, g the weight equals T sine of theta, solving for tension what we have to do is divide sine of theta over. So, T equals m, g over sine of theta, the mass is 50 grams which is 0.05 five kilograms gravity is 9.8 and theta is 60 degrees, okay? And the tension is 0.57 Newtons. Alright, let's give ourselves a little bit of space here.

Now, that we know the tension, we can look at the forces in the horizontal direction, in the horizontal direction it tells us that T cosine of theta equals Fe, okay? That means that Fe is just T which is 0.57 Newton's times cosine of 60 degrees cosine of 60 is half so this is about 0.9 Newtons, okay? There's another thing that we know, that was the last bit of information that we needed to know to now find q because remember the electric force is K, q2 squared over r squared. So, all we need to do now is divide this k over multiply this r squared up and that isolates q squared q squared q squared is now r squared Fe over k then all we have to do is take a square root of both sides to isolate q, lastly all we have to do is plug everything in. So, this becomes the square root. Remember, that r we found was half a meter 0.5 squared at e, we found was 0.29 Newton's divided by k which you guys should have memorized by now, 8.99 times 10 to the 9 plugging all this into our calculator, we get 2.84 times 10 to the negative 6 coulombs, okay? And remember times 10 to the negative 6 is micro, okay? So, we can say that this is equivalent to 2.84 microcoulombs, both of these answers are absolutely correct whether or not you put it in microcoulombs, if the problem specifically asks you to put it in microcoulombs don't forget to convert, okay? That wraps it up for this example, thanks for watching.

Coulomb's Law (Electric Force) Additional Practice Problems

An alpha particle with charge +2.0 e is sent at high speed toward a silver nucleus with charge +47e. What is the electric force acting on the alpha particle when the alpha particle is 4.2 × 10−14 m from the silver nucleus? The value of the Coulomb constant is 8.99 × 10  9 N·m2/C2.

1. Unable to determine

2. 12.2639 N, repulsive

3. 13.0467 N, repulsive

4. 12.2639 N, attractive

5. 13.0467 N, attractive

6. None of these

Watch Solution

Three charges are positioned as indicated in the figure. What are the horizontal and vertical components of the net force exerted on the +15 µC charge by the +11 µC and + µC charges?

Watch Solution

Three point charges +3 μC, +4 μC, and -7 μC are placed along the x and y axis with the values and positions shown. What is the net force, magnitude and direction, acting on the +4 μC charge?

(a) 198 N, 37° above +x axis

(b) 298 N, 47° above -x axis

(c) 342 N, 53° above +x axis

(d) 305 N, 53° below +x axis

Watch Solution

Two identical spheres A and B carry charges Q and 2Q, respectively. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:

(1) 5F/16

(2) F/4

(3) 3F/8

(4) F/8

(5) 7F/16

Watch Solution

Five charges are equally spaced along the x axis. Each charge has the same magnitude e, but some of the charges are +e and some are −e. Four different configurations of charge are labelled A, B, C, D in the figure. Rank the magnitude of the force on the middle charge for the different configurations with largest first and smallest last.

(1) C,D,A,B

(2) C,A,D,B

(3) D,C,A,B

(4) A,C,D,B

(5) C,B,D,A

Watch Solution

A positive test charge q is released near a positive fixed charge Q. As q moves away from Q, it will move with 

A) constant velocity.

B) constant acceleration.

C) decreasing acceleration.

D) increasing acceleration. 

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A charge q= -18.0 nC hangs from a string. It has a mass of 2.9 g. A second charge of q= 53.7 nC is brought to a distance of 2.3 cm from q 1. Find the angle θ assuming the system is at rest as shown in the figure.

A. 15°

B. 21.3°

C. 30°

D. 45°

E. 47.3°

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A positive charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released?

(a) As it moves farther and farther from Q, its acceleration will keep increasing.

(b) As it moves farther and farther from Q, its speed will decrease.

(c) Its speed will be greatest just after it is released.

(d) Its acceleration is zero just after it is released.

(e) As it moves farther and farther from Q, its speed will keep increasing.

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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be

(a) 1/2 F

(b) 1/16 F

(c) 1/8 F

(d) 1/4 F

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Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F ′ , if charge 1 is increased to  q ′1 = 5 q1, charge 2 is decreased to 2 = q/ 2, and the distance is decreased to  ′ = / 2? Choose one.

1. ′ = 5/2 F

2. F ′ = 5/4 F

3. F ′ = 20 F

4. F ′ = 25/4 F

5. F ′ = 50 F

6. F ′ = 25/2 F

7. F ′ = 100 F

8. F ′ = 5 F

9. F ′ = 25 F

10. F ' = 10 F

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Two charged particles of equal magnitude (−Q and +Q) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge +q is placed at a third corner. What is the direction of the force on the test charge due to the two other charges?If F is the magnitude of the force on the test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?

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Two charged particles of equal magnitude (−Q and +Q) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge +q is placed at a third corner. What is the direction of the force on the test charge due to the two other charges?

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Three charges are placed at the following (x, y) coordinates: charge + 6.0x10  -6 C at (0, 0.5 m), charge + 2.0x10 -6 C at (0.7, 0 m), and charge - 4.0 x 10 -6 C at (0.7 m, 0.5 m). Calculate the electrical force on a point particle with charge -1.6 x 10-6 C at the origin (0,0).

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An oil droplet with 4 excess electrons is held stationary in a field of 1.27x10  4 N/C. What is the radius of the oil drop? (The density of the oil is 824 kg/m3, e = 1.60 x 10 -19 C.)

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Four charges −5×10−9 C at (0m, 0m), −8×10−9 C at (1m, 5m), 9×10−9 C at (3m, 1m), and 7×10−9 C at (3m, −3m), are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters).

Find the magnitude of the resulting force on the −5 nC charge at the origin [coordinates, (0m, 0m)].

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Two identical small charged spheres hang in equilibrium with equal masses as shown in the figure. The length of the strings are equal and the angle (shown in the figure) with the vertical is identical.

The acceleration of gravity is 9.8 m/s and the value of Coulomb’s constant is 8.98755×10 9 N m2/C2.

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Four charges are arranged as shown in the figure below each with a positive (red) charge or a negative (blue) charge. Find the net total force (magnitude and direction) exerted on the charge with + 6.0 micro coulomb by the other three charges.

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A positive test charge q is released near a positive fixed charge Q. As q moves away from Q, it will move with

A) increasing acceleration.

B) constant velocity.

C) decreasing acceleration.

D) constant acceleration.

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A point charge Q = -12 μC, and two other charges, q 1 and q2, are placed as shown. The electric force components on charge Q are Fx = +0.005 N and Fy = -0.003 N. Find charge q1 and q2.

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Electrical and gravitational forces follow similar equations with one main difference:

A) Gravitational forces obey the inverse square law and electrical forces do not.

B) Electrical forces obey the inverse square law and gravitational forces do not.

C) Gravitational forces are always attractive but electrical forces can be attractive or repulsive.

D) Electrical forces attract and gravitational forces repel.

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What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.2 g that has been charged to -3.0 nC?

A) 7.2x106 N/C

B) 8.1x105 N/C

C) 1.5x106 N/C

D) 2.2x106 N/C

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Two equally charged spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate at 779 m/s2. What is the magnitude of the charge on each sphere?

A) 160 nC

B) 130 nC

C) 190 nC

D) 100 nC

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At what separation will two charges, each of magnitude 6.0 μC, exert a force of 0.70 N on each other?

A) 1.1 x 10-5 m

B) 0.23 m

C) 0.48 m

D) 0.68 m

E) 1.4 m

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Three identical point charges, Q = 3μC, are placed at the vertices of an equilateral triangle as shown in the figure. The length of each side of the triangle is d = 0.15m. Determine the magnitude and direction of the total electrostatic force on the charge at the top of the triangle.

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Three charges, q1, q2 and q3 are placed left to rigth in a straight line. The distance between q1 and q2  is twice the distance between q 2 and q3. Charge q1 is negative. If the force on charge q3 is zero, which of the following statements about q 2 is true?

A) q2 is negative with a magnitude twice that of q 1

B) q2 is negative with a magnitude half that of q 1

C) q2 is positive with a magnitude one fourth that of q 1

D) q2 is positive with a magnitude one ninth that of q 1

E) q2 is positive with a magnitude equal that of q 1

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Two point charges of +60.0 μC and -12.0 μC are separated by a distance of 20.0 cm. A +7.00 μC charge is placed midway between these two charges. What is the electric force acting on this charge because of the other two charges?

A) 4.53 N directed towards the positive charge

B) 453 N directed towards the negative charge

C) 4.53 N directed towards the negative charge

D) 453 N directed towards the positive charge

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The force of attraction between a -40.0 μC and +108 μC charge is 4.00 N. What is the separation between these two charges?

A) 1.13 m

B) 3.67 m

C) 3.12 m

D) 2.49 m

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Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes of their charges are doubled and their separation is halved, then what happens to the electrical force between these charges?

A) It increases by a factor of 16.

B) It is quadrupled.

C) It is doubled.

D) It stays the same.

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Two small beads having charges q1 = +9Q and q2 = -Q are fixed on a horizontal insulating rod. The distance between q1 and q2 is d = 1.5 m  The third bead with charge q 3 = +Q is at the right hand side of the second bead and is free to slide along the axis of the rod. At what position x is the third bead in equilibrium?

Watch Solution

An equilateral triangle  ABC. A positive point charge +q is located at each of the three vertices A, B, and C. Each side of the triangle is of length  a. A point charge Q (that may be positive or negative) is placed at the mid-point between  B and C. Is it possible to choose the value of Q (that is non-zero) such that the force on Q is zero?

A) Yes, because the forces on Q are vectors and three vectors can add to zero.

B) No, because the forces on Q are vectors and three vectors can never add to zero.

C) Yes, because the electric force at the mid-point between  B and C is zero whether a charge is placed there or not.

D) No, because the forces on Q due to the charges at  B and C point in the same direction.

E) No, because a fourth charge would be needed to cancel the force on  Q due to the charge at  A.

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Two positive point charges Q and 2Q are separated by a distance R. If the charge Q experiences a force of magnitude F when the separation is R, what is the magnitude of the force on the charge 2Q when the separation is 2R?

A) F/4

B) F/2

C) F

D) 2F

E) 4F

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Two identical,  0.25 μC charges hang from the end of electroscope leaves. If, in equilibrium, the electroscope leaves deflect from one another to an angle of 40° from the horizontal axis, as shown in the figure, what is the mass of each charge?

 

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What is the net electric force on the 1 C charge? Consider the 1 C to placed directly in the center of the arrangement.

 

 

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What is the difference between the electric force on a charge q when moved from a distance d to a distance 3d. Consider the initial electric force to be F0.

 

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Find the magnitude and direction of the net electrostatic force on q 1 due to q 2 and q3 as shown in the figure. The charges are q 1 = +4.0 μC, q2 = -6.0 μC, q3 = -5.0 μC.

Watch Solution

A dipole with negative charge of -5.0 x 10 -6 C on the x-axis at x = -3.0 cm, and a positive charge of 5.0 x 10-6 C on the x-axis at  x = + 3.0 cm. A negative point charge of  q = -4.0 x 10-6 C is placed on the y-axis at  y = -4.0 cm.

a) Draw the two forces acted on the negative point charge  q = -4.0 x 10 -6 C (in arbitrary scale, but correct DIRECTION).

b) How big is the TOTAL eletrostatic force acted on the -4.0 x 10 -6 C charge? What is its direction? (i.e., the angle with respect to the x-direction).

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