 Suppose I have an electric dipole with me, which is basically two equal and opposite charges separated by some distance. And imagine I go far away from this dipole. So I zoom out, zoom out, zoom out. In fact, I zoom out so much that the dipole looks like a single dot to me. Then the goal of this video is to derive an expression for the electric field somewhere far away on the axis of the dipole. An axis basically means a line passing through both these charges. Before we begin, a quick question. Why are we doing this? Simple answer, it's in your syllabus. But a better answer would be, we've seen in a previous video that if you have any other group of charges, then the electric field far away is simply radial. Very similar to the field that you get due to a single point charge. But the field due to dipoles are special. Now these field lines do help us in visualizing what the field looks like everywhere. But if you want to figure out, for example, how does one dipole interact with another dipole somewhere far away, let's say? Something that could be super important chemistry because a lot of chemical molecules are dipoles, then we need to look at this mathematically. And that's why in this video, we're focusing on what would the mathematical expression of the field look like on the axis. So let's zoom back in and we've got to figure out what the electric field is somewhere on the axis. So let's choose a point somewhere over here. But we're going very far away, okay? And that's why I put the dotted lines over here. And that distance from here to the dipole, we're calling that as R. Now one immediate question is, where should I consider R from? Should I consider R from this charge to this charge? Is this my R? Or should I consider it from this charge to this point? Or should I consider it from the center? Most textbooks and derivations do that. It's a very safe way of doing things. But I like adventures. So I'm not going to consider R from the center. I'm not going to follow the textbook. I'm going to consider R from here. Why? My argument is, look. When we're going far away, it really doesn't matter whether you're considering R from this charge, whether you're considering from here, or whether you're considering the R from the center. Because from far away, this looks like a dot to us. And so the final answer should not depend upon where you're considering R from, from here or here or from the center. You should get the same answer. And so I see an opportunity to be a little adventurous and do things a little differently from the textbook and see if we get the same answer. Sometimes you might make mistakes and that's okay because you'll learn something new. So let's begin. How do I, how do I start? Well, I want to know the electric field over here. And there are two charges. So I can use the superposition principle, which says you find the electric field here due to this charge alone. Find the electric field here due to this charge alone. And then add them vectorially. So first let's draw the vectors. What would be the direction of the electric field at this point due to this charge? Well, remember, electric field is basically force acting on one a coulomb of charge. So if I keep one coulomb charge over here, it gets repelled by this positive charge. It would go to the right. And so the electric field due to the positive charge would be to the right. So let's call that electric field as E plus. And the electric field due to the negative charge where it pulls the coulomb. So the electric field due to the negative charge would be this way. So that would be E minus. And so the total electric field would now be a vector sum. But since they're in the opposite direction, we basically have to subtract them. And so the immediate next question for you is, what do you think will happen when we subtract them? Do you think the two fields will cancel out or not? Can you pause and think about it? All right, my first instinct would be to say that, hey, point P is so far away from the dipole that it's pretty much the same distance from minus Q and plus Q. And if the distance is the same, then the electric field values are the same, which means they should cancel out. But that's wrong because if that was the case, then the field everywhere would be zero. And we know the field everywhere is not zero. There is an electric field somewhere, which means this tiny distance matters. And we need to consider that. So if I consider that, we can see that point P is slightly closer to the negative charge compared to the positive charge. And therefore the electric field due to the negative charge is slightly bigger than the electric field due to the positive charge. And as a result, we will find that the net electric field would be in this direction. Does that make sense? This should be the total or the net electric field. All right, so now let's go ahead and figure out what that net electric field is. So from superposition principle, that net electric field will be, because this is bigger, I'll say this minus this. So it's going to be the electric field by the negative charge minus the electric field due to the positive charge. And because we're dealing with one dimension, I don't have to worry about vectors. I can just simply subtract them and that'll give me the answer. All right, what is the electric field due to the negative charge? We've seen the electric field due to a point charge expression. It's going to be KQ by R square. So it'll be again, good idea to pause and see if you can substitute yourself and see what you end up with. Okay, let's do this. So if I consider the field due to only the negative charge, that would be KQ divided by R. What is the R over here? Well, the distance is R. From here to here is R. We consider that as R, so it's going to be R squared minus what is the electric field magnitude due to the positive charge? Well, that's going to be KQ divided by this distance squared. The distance from here to here is R plus D. So it's going to be R plus D, the whole squared. And what's interesting is, remember, D is much, much smaller than R, right? Because you've gone very far away. And so can I neglect D over here? Can I say R plus D is pretty much the same thing as R? Well, you can't because if we did that, then you'll get the same problem. These two will cancel out, you'll get zero. So you have to be careful when you're approximating. That's what makes this interesting. So this means we need to simplify it a little bit further before we start approximating. That's what it means. So let's simplify a little bit further. What I can do next is KQ is common, so I can take K times Q out. And what I get inside is one over R squared minus one over R plus D, the whole squared. If I simplify even further, I will now get, let's see. If I take common denominator, in the denominator, I get R squared times R plus D, the whole squared. And the numerator, I get R plus D, the whole squared minus R squared. Okay, what do I do next? Well, let's see. I can expand this. If I expand this, I'll get R squared plus D squared plus two times RD minus R squared divided by R squared times R plus D, the whole squared. So R squared cancels out. And now let's see if we can approximate. We can use this. If I do that in the denominator, I could say, hey, let's ignore D because R plus D is pretty much the same as R. Then I get R squared times R squared, which is R power four. And I can do that because that's not giving me zero. As long as you're not getting zero when you approximate, that is valid. So when I approximate in the denominator, I get one over R to the power four. What do I get in the numerator? Well, I have a D squared and I have a two RD. Can you see that D squared is way smaller than this term, R times D? Here, let me give you an example. Let's say R was, I don't know, maybe a hundred and D was, I don't know, maybe 0.1. Then D squared would be 0.01. But R times D would be 10. 100 times 0.1 is 10. And so if you compare, 10 is much larger compared to 0.01. And so you can see, I can neglect this. And therefore, I can now say that the net electric field is going to be, okay, let me write that over here. I can write that over here. So I just give this as two RD, two RD. And let's see what I get. One R cancels out and I give this to be three here. And so if I write it down, I end up with, let's put two K out. So I get two times K times Q times D. D divided by R cubed. And so now can you tell why if we had directly negated D over here, we would have gotten zero because we would have neglected this term and this term. And so the whole numerator would have been zero. Because we simplified, we got a choice of just negating this term and we considered this term. This is how we approximate. Beautiful, right? And here is our mathematical expression for the field on the axis of the dipole. But let's see if we can make sense of this. That's the final thing we'll do. The first thing I want you to focus on is see how it's dependent on the R. It's one over R cubed, which means that the dipole field is falling much faster than the radial field, which is one over R squared. Does that make sense? It kind of does. If we go back and look at our dipole visualization, you can, if we compare this with the radial one over R squared field, then you can see from the field lines itself that the dipole field, as you go further away, look at that, it goes farther away from each other much quicker, much quicker than the radial field. And so it makes sense that it should die out much faster than one over R squared. And so one over R cubed kind of makes sense. The second major difference we are seeing between the radial and the dipole field is that the radial field only depends upon the total charge, whatever is the total charge. If this was not a dipole, then it will only depend on the charge. But look at what the dipole field depends on. The dipole field depends on the product of the charge and the distance between them. If I had another dipole, or if I took the same dipole, and let's say I doubled the distance, but I made the charge value half, the product would stay the same. And so when I go far away, it feels like it's the same dipole. So the identity or the strength of the dipole is not in their charge or in the distance, but it's the product. The product decides the strength of the dipole the product decides how strong the field is going to be. And that's the reason this product for a dipole is given a name, it's called the dipole moment, which is represented by P. And so finally, finally, we can now write the expression for the electric field on the axis is two times K, which is the Coulomb's constant times P divided by R cubed. And what is the direction of this electric field? You can see over here everywhere to the right, the electric field will be towards the left. What if I go towards the left of this dipole? The electric field would still be towards the left because positive charge will be dominating and so it'll be pushing. And so again, if we go back to that, you can see everywhere on the axis, the electric field is to the left. Let's come back over here. And it's for that reason we said, let's say the dipole moment is also to the left. The direction of the dipole moment is also taken to be towards the left. So the dipole moment is from negative charge to the positive charge. And so we can say everywhere on the axis the electric field is in the same direction as the dipole moment and so vectorially, I could say they have the same direction. Of course, one small detail is if you were to look at the field inside the dipole, then it would be from positive charge to the negative charge. It would be in the opposite direction of the dipole moment. But remember, this is the field far away from the dipole. So when I'm looking at that field and not looking at the field inside the dipole, I'm looking at the field far away from the dipole. So I'm always looking at the field outside the dipole. So remember, this is the expression for far away outside the dipole.