Potential due to dipole (logical derivation)

imagine we have an electric dipole basically a negative charge and a positive charge of same value separated by some distance let's say we call it 2a traditionally we call that 2a our goal is to figure out what the potential due to this dipole is going to 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 um this is what it would look like if i were to actually show you this ins to 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 k q one by four epsilon naught which is called k so k q 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 the distance from the center so how do i what is this distance what is this distance that's not given to me so what i'll do is like 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 going to take i'm going to 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 going to 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 going to be k q divided by r2 so r2 and you will have due to the negative q you'll have minus k q 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 to a 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 um that means that this distance r sorry this distance r is way bigger than this distance 2a or hey whatever you want to 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 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 is r and r2 also as r what will i get substituency okay if i put both of them the new denominators become same numerators are also same when i subtract i will get 0 so let me just write that down if i were to directly 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 r2q 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 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 going to 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 as r i will get r squared r into r is r squared and i don't get 0 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 0 and again i'll get this as 0 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 will 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 um what happens when you approximate so what is the actual value what is the real value well that is 11 times 9 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 1 out of 99 so it's almost like saying 1 out of 100 so it's one percent error so we're saying this equation this this is one percent wrong with these values and i can say i'm okay with one percent 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 have to substitute the values over here 11 minus 9 is 2. so this is 2 in the numerator what if i substitute 10 here and here i get 0 that's the approximate value what's the error the error is also 2. now 2 out of 2 is 2 divided by 2 is 1 that's 100 error and that's not acceptable so if i were to approximate over here i'm getting 100 wrong answer and i don't know about you but i don't want to get a hundred percent wrong answer so when you get zero you end up getting a hundred percent wrong answer so hundred percent error and that's why that's why we can't substitute in the numerator but 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 it takes some time to soak in take something digest that so pause and then we can continue all right let's continue so we are now reaching an interesting point very very interesting point and 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 are 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 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 this real picture zoom in zoom in zoom in go close to my dipole and you notice something you notice that if the point p is very 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 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 why why does it work out that way well look at the look at this picture when i draw when i draw that perpendicular which you cannot see properly let me zoom in a little bit okay and 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 wonder 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 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 r 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 r2 minus r1 minus r2 delta r automatically becomes 2a 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 piece right in between these two distance from here to p and here to there would be zero would be equal and so there will be no difference so hopefully we can see that let me get rid of this that this delta r depends on this angle and so let's break 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 going to give you a clue it says trigonometry so pause the video and see final pause this is the final part all right so in this triangle notice 2 is the hypotenuse and this but what we need is the adjacent side so i'm going to use cos theta so from from here i can say cos theta equals delta r divided by 2a 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 2a cos theta let me write that delta r equals 2a cos theta and so if i substitute that i now have my final equation so vp is going to be k q into 2a cos theta to a cos theta divided by r square and that is our expression for the electric potential due to a dipole and so what you notice is that the potential at up 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 r1 minus r2 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 r1 is almost equal to r2 and the trick was to imagine that they are assume that they are parallel to each other and use trigonometry