Electric field due to dipole on the equator

i have a dipole which is basically two equal and opposite charges separated by some distance and i go very far away from it in fact i go so far away that it looks like a dot to me and the goal of this video is to figure out what is the strength of the electric field at some distance r very far away on the equatorial line what do i mean by equatorial line well equatorial line is the line that passes through the center of that dipole and is perpendicular to the axis of the dipole and this is a follow-up video to our previous one where we figured out the electric field on the axis so now we'll look at on the equator and see how is it going to be different than that of the axis and based on that we'll be actually able to figure out how the electric field is everywhere around the dipole all right so if you're as excited as i am let's begin so here we are we've zoomed back in and this is the point very far away from the dipole and that distance is r now again we could ask where what what distance exactly is r should i consider r to be from the point p to the center of the dipole well that's what your textbooks do and i think that's the safe way of doing this but just like in our previous video we can be a little bit more adventurous remember if i zoom out the dipole looks like a dot and so whether you consider that distance from here to here or the distance from here to here or from here to here they all should be pretty much the same we can say they're all almost all equal to r so i can say this is r or i can also say this is r and so what i'll do is and what we'll do is we'll be a little bit adventurous we'll deviate from the textbook a little bit we'll get the same answer it should and we'll call this distance as r the expression will become just a little simpler that's it no other reason and it's always fun to try to do things a little differently all right so how do we begin well i need to find the electric field here and i can use my superposition principle which says the field here is going to be due to this charge alone plus du field due to the discharge alone over here so why don't you pause the video and think about what the individual fields are going to be due to the plus q and minus q and also see if you can write the direction of those individual fields draw the direction of the individual's fields can you pause and try that yourself first okay so due to the positive charge electric field is always away from the positive charge so over here electric field will be away from it and along this line away so the field would look somewhat like this and if i consider the negative charge field is towards it again along the line joining so the field will be somewhat like this let's draw the arrow marks this is our electric field due to the negative charge and this will be the electric field due to the positive charge all right now we need to add these two vectors up and they're not in one dimension we need to be a little bit careful but the next thing is let's look at the magnitudes well notice the charge is the same magnitude of charge is the same and the distance is also the same so the magnitude of the electric fields must be the same right because it's k times q divided by r square so we can immediately say hey the electric field due to the positive charge should equal in magnitude the electric field due to the negative charge and that equals k which is the coulomb's constant 1 by 4 pi epsilon not 9 times 10 to the power 9 times q the charge is the same just the magnitude i'm looking at divided by the distance squared and distance is the same r so it's going to be divided by r square so we now have two equal vectors and we need to figure out what the resultant is going to be hmm how do we do that the physics is pretty much done now it's a math problem so the first thing we can do is because the two vectors are exactly equal in magnitude let's just give it one name let's just call it e so this will be e and this vector will also have the same magnitude e now how do we look how do we calculate the resultant one way is to use the parallelogram or the triangle of vector addition another way is to decompose the vectors and use perpendicular components and i like that because it's more conceptual think of it this way imagine this there is a ball over here and two people are pulling it like this with the same force what direction would the ball accelerate what what is the net force it will experience well because the two forces are exactly equal there is no reason for the ball to accelerate towards any of these forces so it will accelerate right in between them and so we know for sure the resultant has to be right in between them and so i can tell right from here the resulting electric field has to be right in between the two and from symmetry we could say the right in between has to be this way and so what we can now do is we can call this angle theta this angle would also be theta i'll just not write it i don't want to keep make it too crowded and then we can decompose each vector into this horizontal and vertical components and we will see vertical components subtract they they cancel each other horizontal components add up and from that we can figure out what the net electric field is going to be and to do this carefully here is a zoomed in version of whatever i've drawn over here i've drawn it better so why don't you pause the video and see if you can give it a shot first try figuring out what the net electric field is going to be by decomposing these vectors so pause and try let's start with this vector to decompose it because i'm decomposing it into a horizontal and vertical component i'll just drop a vertical from here to here and now i could say that hey this vector is the same as this vector this is the horizontal component plus this vector this is the vertical component and if this angle is theta then this is the adjacent side so if you use the trigonometry you can say hey this is e cos theta you can just uh you can just confirm that this length will be cos theta and all this length will be since this is the opposite side this length will be e sine theta and we can do the same thing with our blue vector drop a perpendicular from here to here and we could now say there's going to be a horizontal component and a vertical component and the horizontal component will be exactly the same it's going to be e cos theta and even the vertical component will be in magnitude exactly the same e sine theta and now look at what happens the vertical components are exactly equal and opposite they cancel out and so what remains is the horizontal component they add up so i get 2 e cos theta again it got a little crowded over here now i can finally write so i know my net electric field over here is going to be 2 e cos theta so let's write that so net electric field on the equator is going to be 2 i know the value of e is the field due to individual charges k into q divided by r squared times cos theta now the only thing i need to figure out is what cos theta is how do we do that well can i use these these triangles and use trigonometry no because i used angle theta to figure out what the side lengths are i didn't know the side lengths to begin with so the question is how do i figure out cos theta i can use a little bit of geometry and there's a triangle right over here so maybe i can figure out okay if this angle is theta in this triangle where theta is going to be and then i can use the sides of that triangle to figure out cos theta so it's the last part and again if you if you couldn't do it before no worries last part can you pause and see if you can complete this all right let's bring this home so if this angle is theta then this angle is a corresponding angle to it because this is a parallel line and so this is a straight line and you have these two parallel lines so this angle is theta and so i can from this triangle cos theta is the adjacent side divided by the hypotenuse i can write that so let's do that so the net electric field is going to be let me just write this equals 2 k q divided by r squared times cos theta which is the adjacent side what is this well the whole distance is d well this is half of d so it's going to be d over 2 adjacent side divided by the hypotenuse r and guess what we have gotten it we found the answer we have everything and so let's put it all together so 2 cancels out and so what i end up with is k into q into d divided by r cubed and so the net electric field is going to be the coulomb's constant k q times d hey what is q times d i think you know this q times d is a dipole moment we've introduced this before in previous videos and i'm pretty sure now you're familiar with this divided by r cubed divided by r cubed and i think we should box this this is the fruits of our hard work the last thing about this expression is what is what about the direction well over here the direction is to the right but let's compare it with respect to the dipole moment what direction is the dipole moment for a given dipole it's always from negative to positive charge this way this is the direction of our dipole moment and therefore you can see on the equator the electric field is in the opposite direction of the dipole moment and that's why we often write vectorially this is on the equator minus when you write it in vector form because it's in the opposite direction of the dipole moment i'm going to take a last couple of minutes to digest this and make sense of dipole in general so first thing i want to do is compare this with the electric field that we got on the axis if you remember that you'll find on the axis electric field was very similar but two differences it was 2 k p by r cube so twice the magnitude and the second is on the axis the electric field was in the same direction as the dipole moment and i want to try and understand that why has the strength become half and why has the uh you know why is it is in the opposite direction for that we can zoom out and re-look at the electric field lines first let's think about the strength see from an electric field lines point of view the feed lines are closer the strength is more and if the feed lines are farther away the the strength is weaker so if you take any two field lines let's say you take these two or maybe these two field lines you can see as they come towards the axis that's where they are the closest and as you go away from the axis near the equator notice the field lines are the farthest apart then again they become closer to each other so just from the field lines you can say ah the field should be maximum on the axis at a given distance and it should be minimum on the equator because the feed lines are farthest away and therefore we find the field on the equator which is the minimum is half that of the field on the axis which is maximum now let's look at the direction why is the dipole moment direction from negative charge to positive charge well look at how the field structure is it kind of looks like the dipole is pushing the field to the left and sucking the field from the right and so therefore it looks like the dipole is sort of oriented to the left and that's why we say dipole moment is this way and it happens to be from negative charge to the positive charge and so now if you look on the axis the dipole moment is exactly in the same direction but when you go on the equator because it is looping back notice exactly on the equator it's in the opposite direction compared to the dipole moment and that's why it's negative p you know when you look at it vectorially opposite direction of the dipole moment and now i can visualize how things change how the field changes when i go from the axis to the equator well you can see if i if i were to draw the field vectors you can see them turning when i go from axis to the equator it keeps turning turning turning and then strength becomes half and then it again turns turns turns becomes maximum turns turns turns becomes half turns turns turns becomes maximum and so now we have the complete picture of dipoles amazing isn't it if over here we had some other charge group of charges which was not a dipole then if you went far away would get just get boring radial field lines but as i keep telling dipoles are special and now we have demystified them