Electric field due to dipole on the axis

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 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 now 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 you two 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 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 or you're considering from here or because 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 are from from here or here or from the center you should get the same answer and so i see an opportunity to you know 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 a coulomb of charge so if i keep one coulomb charge over here it gets repelled by this positive charge and if we 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 to the negative charge well 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 it 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 uh and because we're dealing with you know one dimension i don't have to worry about vectors i can just simply subtract them and that will give me the answer all right what is the electric field due to the negative charge we've seen the electric field to a point charge expression it's going to be k q by r square so it'll be again good idea you know 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 k q 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 k q 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 we'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 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 k q is common so i can take k times q out and what i get inside is 1 over r squared minus 1 over r plus d the whole squared if i simplify even further i will now get let's see if i had a common denominator in the denominator i get r square times r plus d the whole squared and in 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 2 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 1 over r to the power 4. what do i get in the numerator well i have a d square and i have a 2rd can you see that d square is way smaller than this term r times d here let me give an example let's say r was i don't know maybe 100 and d was i don't know maybe 0.1 then d square 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 it over here so i just gave this as 2 rd 2 r d and let's see what i get 1r cancels out and i get this to be 3 here and so if i write it down i end up with let's put 2 k out so i get 2 times k times q times d so times q times d divided by r cubed and so now can you tell why if we had directly neglected 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 but because we simplified we got a choice of just neglecting 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 depend on the r it's 1 over r cubed which means that the dipole field is falling much faster than the radial field which is 1 over r square 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 square 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 square and so one over r cube 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 what is whatever is the total charge if this was not a dipole then only depend on the charge but look at what the dipole field depends on the type of field depends on the product of the charge and the distance between them if i had another dipole or if i took this 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 2 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 you'll be pushing and so again if we go back to that you can see everywhere on the axis electric field is to the left so 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 a 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 right so when i'm looking at that field i'm 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 far away outside the dipole