 Imagine we have an electric dipole, basically negative charge and a positive charge of same value, separated by some distance, let's say we call it two A, traditionally we call that two A. Our goal is to figure out what the potential due to this dipole is gonna be at some point P, let me write that, show that at some point P, far away from that dipole, at some distance are far away. These dotted lines are showing far away. And just to show you the real picture, this is what it would look like if I were to actually show you this in two scale, imagine it this way, that this distance R is way bigger, way bigger compared to A. That's what I mean when I say I want to calculate the distance potential very far away. All right, so let me just keep this picture somewhere over here. Now the question is how do I do that? Well, I already know how to calculate potential due to a point charge. We've seen the expression for that. It's KQ, one by four pi epsilon naught, which is called K, so KQ divided by R. So all I have to do is figure out what's the potential due to this charge over here, what's the potential due to this charge over there, and then just add them up. But you could say, hey, the problem is I don't know what's the distance of that point P from this charge and from this charge. That R is a distance from the center. So how do I, what is this distance? And what is this distance? That's not given to me. So what I'll do is like in most cases in physics when something is not given, well you draw that and you assume something. So let's do that. So I'm gonna take, I'm gonna draw some lines from here. Let's call this distance from here to here. Let's call that R1. So this is my R1 and from here to here, let's call that distance to be R2. And now in terms of R1 and R2, we can write down what the electric potential is gonna be. It's due to, it's due to this plus due to this. So can you pause the video and write down the, what the expression for the electric potential P turns out to be in terms of R1, R2, Q and minus Q? Pause the video and give it a shot. All right, let's do this. So the electric potential at point P is going to be the potential due to a positive Q at point P plus the potential due to negative Q. You might be thinking here there's a negative sign, right? Where the negative sign comes here itself when I substitute. All right, so if I do that, if I substitute, due to plus Q, it's gonna be KQ divided by R2. So R2 and you'll have, due to the negative Q, you'll have minus KQ, the minus sign comes from the negative charge here, divided by R1. Divided by R1 and we're done. Now comes the question, what do I do further? This is not done because we have to find out the equation in terms, expression in terms of R, that's given to us, not R1 and R2. So what do I do is the question. Well, I know the condition is going far away. And what that means is that this distance R is way bigger than that distance 2A. So let me write that down. So that's our condition. So we are calculating this far away and we're doing that, what does that mean? That means that this distance R, sorry, this distance R is way bigger than this distance 2A, whatever you wanna call that. Okay, so how do I bake this in over here? How can I use this to somehow simplify this? Well, when R is very far away, something very interesting happens when you compare R1, R2 and R. Let me show you, let me draw R1 and R2 here also. And you will see something really interesting. What you now find is that the length of, the length of all the three lines are almost equal to each other. Can you see that? Because P is so, so far away, this looks like almost a point, these three lines will almost have the same value. Which means because it is far away, I can now say this means, this means R1 is almost equal to R2 which is almost equal to R. Okay, they're almost equal. So now comes the question, what do I do? Can I somehow use this over here? And my first thought would be, since we want an approximate value, let me just go ahead and substitute and see what happens. And why don't you give it a shot? You don't have to actually, you can do it in your head. What will happen if I substitute R1 as R and R2 also as R, what will I get? Substituency, okay? If I put both of them, the denominators become same, numerators are also same. When I subtract, I'll get zero. So let me just write that down. If I were to direct the substitute, so direct substitution, what does that give me? We get potential at point P to be zero. And that is wrong. Why is that wrong? Why can't that be an approximate value? Because although they are approximately equal to each other, I know from the diagram, I can see that R1 is slightly bigger than R2. Slightly bigger than R2. Q plus Q is slightly closer compared to minus Q. Therefore, the potential over here must be slightly positive. So there is some slight small value, which is not zero. And my goal is to figure out the expression for that small value. Are you getting that? Okay. So how do I figure that out? How do I figure out that expression for that small value that you're gonna get? So what I learned whenever I'm doing this is if direct substitution doesn't work, you simplify that one more step, let's say, and then try substituting and see what happens. So let's go ahead and simplify this. Now, this is no longer physics. We're just doing algebra, simplifying. I can take common denominator. So let me go ahead and do that. So if I take the common denominator, I get VP to be equal to, let's see. I can take KQ out. And when I take common denominator, I get R1 minus R2. So I get R1 minus R2 divided by R1 times R2. So that's gonna be R1 times R2. And now let's see if we can substitute this over here and see what we get. If I were to substitute in the denominator, R1 is R and R2 also is R, I will get R squared. R into R is R squared and I don't get zero. So yeah, I can substitute that in the denominator. So what I'll do is in the denominator, I'll call this R squared. And what about in the numerator? If I call this R and if I call that R, now R minus R is zero and again, I'll get this as zero and so I can't substitute. So in the numerator, I will not substitute. And at this point, I'm sure you may be thinking, what is this business of somewhere you will substitute and sometimes you will not substitute. What is this? What is going on? So let me take, let me walk you through this. It took me some time to realize but now finally I understand what's going on. So let me take some numbers, it'll make sense. So imagine R1 was say 11 and let's say R2 was a little smaller, so let's call it nine. And let's say R somewhere in between the two numbers is 10, okay? Let's write that down over here as to what happens when you approximate? So what is the actual value? What is the real value? Well that is 11 times nine, that's 99. And what's our approximation in this case? It's 10 square, which is 100. And so what's the error we can say, right? When you approximate, you get an error. So what's the error over here? The error is one. There's a difference of one, positive one but it's okay, you get one. And compare the error with the real value. It's one out of 99. So it's almost like saying one out of 100, so it's 1% error. So we're saying this equation, this is 1% wrong with these values. And I can say I'm okay with 1% wrong, no problem for me. But now let's do the same thing for the numerator and see what happens. So what is the real value if I were to substitute the values over here? 11 minus nine is two. So this is two in the numerator. What if I substitute 10 here and here? I get zero, that's the approximate value. What's the error? The error is also two. Now two out of two is two divided by two is one. That's a 100% error. And that's not acceptable. So if I were to approximate over here, I am getting 100% wrong answer. And I don't know about you but I don't want to get a 100% wrong answer. So when you get zero, you end up getting a 100% wrong answer. So 100% error. And that's why we can't substitute in the numerator where we can substitute in the denominator. And pause this if it didn't make a lot of sense. It didn't make a lot of sense to me initially as well. Take some time to soak in, take some time to digest that. So pause and then we can continue. All right, let's continue. So we are now reached an interesting point, a very, very interesting point and an important point in this derivation as well where we need to find what the difference between R1 and R2 is. We need to find that under the condition that they're almost equal to each other, okay? How do you do that? How do you say that they're almost equal to each other and somehow find the difference between the two? Now there are multiple ways to do that, but the way I like to do it is zoom in on this real picture, zoom in, zoom in, zoom in, go close to my dipole and you notice something. You notice that if the point piece way far away, then all these three lines look parallel to each other and that's the secret we're going to use. Look at them. They look so parallel to each other. So I'm going to go back, I'm going to redraw this and this time I'm going to draw R1 and R2 parallel to each other and the point P is really, really far away. So how does that help us in figuring out what this is? Well, remember what I need to figure out is the difference between these two lines and one of the ways to calculate the difference between two lines which are parallel to each other in this case would be, let me show you, to drop a perpendicular from the shorter line onto the bigger line. So let me try that one more time. All right, there you go. So this is a perpendicular and this is a perpendicular. You may ask, how does that help? Well, it helps because now, if I call this point or something else, the point P is at the same distance from here to here. From here to P and here to P distance is the same which means that extra distance R1 minus R2 is this. Let me just quickly write that down. This is that extra R1 minus R2. So that is our delta R. And if you're wondering, why does it work out that way? Well, look at this picture. When I draw that perpendicular, which you cannot see properly, let me zoom in a little bit, okay, when I draw that perpendicular, notice we end up with a giant triangle and in that triangle, this angle is almost zero. And these two angles are almost 90 degrees, okay? You may be wondering, how can you have a triangle like that? Well, think of it as 89.99, 89.99 and 0.0001 or something like that, okay? So look at that triangle. These triangles are equal. That means this is an isosceles triangle. And as a result, this side is exactly equal to this side and therefore that extra distance should be R1 minus R2. So this should be R2, so this should be R1 minus R2. Does that make sense? So this is that extra distance, okay? The final question we have is, what does that extra distance really depend on? And if you look carefully, you can see that that really depends upon the angle between this R and the dipole. Let me show you that. I have an animation over here, let me show you. So here you go, this is our delta R and notice what happens as this angle decreases and this point P comes on the equator, oh sorry, on the axis, notice what happens. Look at that, look at that, look at that. Can you see? When the point comes on the axis, hopefully you can see that R1 minus R2 delta R automatically becomes two A, maximum. All right, notice what happens as this now comes off this particular axis and comes back. What happens? This will become smaller and smaller and smaller. And if it now comes somewhere over here, notice, eventually becomes even smaller and finally, when that point P is right in between these two, that difference becomes zero, which makes sense, right? If that point P is right in between these two, the distance from here to P and here to there would be zero, would be equal and so there'll be no difference. So hopefully you can see that, let me get rid of this, that this delta R depends on this angle and so let's bake that angle in. So let's call that angle as theta, okay? And so if that angle is theta, we're assuming that all lines are parallel. This angle also becomes theta. And so now comes the question. Using theta and using this right angle triangle, can you figure out what delta R is going to be from this? I'm gonna give you a clue, it has trigonometry. So pause the video and see, final pause. This is the final part, all right? So in this triangle, notice two is the hypotenuse and this, but what we need is the adjacent side. So I'm gonna use cos theta. So from here, I can say cos theta equals delta R divided by two A. And so what is delta R? So I can substitute that directly over here. This is our delta R, okay? So what is delta R? Delta R is just two A cos theta. Let me write that. Delta R equals two A cos theta. And so if I substitute that, I now have my final equation. So VP is going to be KQ into two A cos theta, two A cos theta divided by R square. That is our expression for the electric potential due to a dipole. And so what do you notice is that the potential at a due to a dipole not only depends on the charge, but also depends upon the distance between the two charges. Not so surprising because we've seen that before when it comes to electric field as well. And so this is what we call, the product is what we call the dipole moment. But notice, here we also see that it's dependence on this angle theta. And we saw why that is the case because this R over minus R two, the difference between the two, that depends on that angle. And so the most important step over here was to figure this out in the limit that R one is almost equal to R two. And the trick was to imagine that they are, assume that they are parallel to each other and use trigonometry.