 The macroscopic world is not made from singular electric charges, but rather from collections often quite large collections of electric charges. And even when only a few electric charges that are unbalanced by their opposites are present, nonetheless their effects in the space around them can be large and add up. Let's look at how these effects add up, starting with distributions of just two charges and building up to much larger distributions of electric charge. In this lecture we will look at both simple and complex distribution of charge and the associated electric fields that go with those charges. We'll use the basic principles of the point electric charge field, and we will build up to larger and larger structures, starting with very simple additions of electric field vectors and concluding with a general discussion of how one employs calculus to handle more complex distributions. Since like force, electric field is a vector, we must add electric fields that are present in the same region of space but caused by separate sources together as vectors. So if we ever want to know the total electric field at some location in space, let's say a point P which I'll denote here on this graph paper, if there are two electric fields present, for instance one that points this way and is denoted E1 vector, and another one that points this way with a different length and is denoted E2 vector, then the total electric field here is simply the sum of E1 vector and E2 vector, and that means that you have to add the x components and the y components and the z components of each of the electric fields separately and then that resulting sum component by component gives you the total vector. To put this in a sort of generic algebraic notation, let's imagine that we have the vectors pictured here, so E1 could be written as E1x in the i-hat direction plus E1y in the j-hat direction, and E2 could be written similarly as E2x in the i-hat direction along the x direction and E2y along j-hat the positive y direction. Now E1y, E1x, E2x and so forth, these could be positive numbers or negative numbers so they could flip the directions of the unit vectors along the coordinate axes, but for right now I'm just going to leave them as algebraic symbols that represent numbers which could be either positive or negative, and if we have to add these together then of course we wind up with a relatively complicated looking thing here in E total, so now if I try to write E total symbolically it's going to be the sum of the x components which are E1x and E2x and that sum will be in the resulting x direction or i-hat unit vector direction and then I will move down to this line just to keep this nice and clear, but then we have E1y plus E2y and that sum then also points in the j-hat direction, and now E1x and E2x may be numbers with different signs so they might add up or they might cancel out, you might get a resulting positive number or a negative number from doing the addition of the components, it's the sign of that number that will tell you whether or not you have to flip from the positive i-hat direction to the negative i-hat direction to get the final x component, so quite generically this is how one sums up the electric fields and even more generally if you have an arbitrary number of electric fields so imagine a point P that is subject to a whole bunch of different electric fields, there is our point P right there, each of these might be a different electric field with different magnitude and different direction, you can see this is getting quite complicated now, well it's a finite number, it's only 5 of its individual electric fields that one has to consider at this point and generically in algebra when one sees this, one denotes now the total as the sum from i equals 1 to 5 of E sub i vector, so i is an index, it's a place holder in this so called summation notation, the big Greek sigma symbol means sum over and the index tells you its minimum and its maximum on the top, so we want to make the index go from electric field number one all the way up to electric field number five and we want to sum each of them together, so this written out represents E1 plus E2 plus E3 plus E4 plus E5, you can see why this summation notation is so convenient, if one had to write this out as an actual sum by hand every time it would get quite clunky to have to do this and in fact what if I had 100 electric fields here that I had to add, I'm not going to write E1 plus E2 plus E3 all the way up to E100, I'm going to write this more compact form and I'm going to take the index and say it starts at 1 and I'm going to tell the index that it ends at 100 and then I can have that nice compact symbol that I write down on my piece of paper, saves a lot of space, having motivated generally how one adds together electric fields from different sources, we can now consider a simple combination of charges and start to think about what the electric field at any point in space around that set of charges is going to look like. In the original lecture on electric field I considered really just single point charges and in the simulation that I showed you I considered what happens to the electric fields of two point charges when they're put in proximity to one another and I let the computer calculate all of that but I'd like to go back and actually take a pair of charges one with a positive sign one with a negative sign and consider at any point in space what's the electric field going to look like. This is going to look quite complicated but we're going to consider a simplifying case of this and actually by doing this exercise we're going to motivate a very important piece of chemistry and biology which is present in the natural world all the time and that is something called the electric dipole. So let's consider a negative charge and a positive charge that are separated by some distance. So I will denote this as a line between the two and we will call that distance d. You know d could be measured in meters, centimeters, millimeters, nanometers, something like that but it's essentially in meters. The charges are equal in magnitude but opposite in sign. So the positive has charge plus q and the negative has charge minus q. This is a special case of course I could have picked any two charges to be here but I'm going to motivate the electric field sum from this particular configuration and then use it to talk about something called the electric dipole. You could instead replace the positive charge with a negative charge for instance that has twice the magnitude of the first negative charge that I drew. That's a very different situation. That's not an electric dipole. I'm going to consider this very special case. Now the final thing that an electric dipole specifically has is rigidity. It is impossible for the charges to move toward each other or away from each other. It's as if they were fixed on the ends of a rod and held in place. Now that's not a necessity for what I'm about to do. That's a necessity later when I talk very specifically about the electric dipole but I'm just going to mention that feature now and I'll come back to it again later. Now we already have a sense of what the electric fields from these two objects are going to look like. I mean a guy you can hand sketch the electric field from the positive charge for instance. I can draw a few lines of force for this and actually let me draw a total of eight lines of force just to get a nice clean picture here. So I would expect my electric field to start on the positive charge. Positive charges are the sources of electric field by convention and I would expect them to radiate outward. Now I'm only considering the electric field from the positive charge. I don't care what the electric field from the negative charge is doing to change this shape that I've just drawn. We'll come to that. Now let's consider the negative charge. Well again ignoring the positive charge and thinking only about what the electric field of a single negative charge is going to look like. We already know from the pictorial representation from the sketching that we did in the original electric field lecture that all of the lines of force should point in and again to be symmetric between the two to be fair and equal between the both of them I'm going to draw eight lines of force. Now from this picture we can already see graphically vectorally what we expect to happen at different points in space. So let's pick a few different points in space and think about what the electric field lines are going to do there. So for instance I might pick this point right here in between the two charges. We'll call this P1 and we'll come back to that in a moment. I might pick this point down here P2 and I might pick this point up here P3 just for convenience sake. Let's consider what's going on at P1. Here if I sketch this out I have the electric field for the negative charge pointing to the left and I also conveniently have the electric field for the positive charge pointing to the left and let me put a little minus sign above this and a little plus sign above this to remind you which field line goes with which charge. And we see they point along the exact same direction. They're already neatly lined up head to tail and so we already know what vectors head to tail are going to do. They're going to very simply add. All you have to do is add their lengths and the direction that the resulting vector points in is the same. So if these vectors were both pointing along negative I hat and let me go ahead and sketch this we're going to call this E1 negative and we're going to call this E1 positive. I could just declare that the horizontal direction is the x direction and so I can immediately write these in unit vector notation. This one is equal to E1 negative and I'm going to put a magnitude around that so that we know that that's exactly a positive number times negative I hat. And we know immediately what E1 positive is going to look like. It's going to look like E1 positive and I'm going to again I'm going to put these absolute value lines around it to indicate that this is a positive number. I'm going to put all the sign here in the unit vector and this also points in the negative I hat direction. So to add them together is quite simple and let me go ahead and do that. So I would expect that the electric field total at point one which is the sum of the electric field from the negative charge at point one and the electric field from the positive charge at point one. This is a very easy sum. It's just the sum of these two positive numbers. The magnitude E1 minus plus the magnitude E1 plus and that's all going to be in the negative I hat direction. So that's not so bad. We could now for instance plug in for the magnitudes of point charges electric fields. So just to sketch this out we would expect that the magnitude of the point charge electric field E1 is going to be equal to KQ. Remember for the negative charge the magnitude of the charge is just Q and because this is a magnitude there are no signs left over. So to be quite explicit about this I could put the magnitude signs around everything I'm about to do and put the minus sign in front of the Q there. We're going to have the distance from the negative charge squared and because this is a magnitude that unit vector that would normally set out in front of the full electric field vector equation we take the magnitude of that and the magnitude of the unit vector is one. So we don't have to write it that one is implied here we're multiplying this whole thing by one. Now I've been explicit about putting the magnitude absolute value lines around this so I'm just going to go one more step and say that this is equal to KQ over r minus squared. So this is just the constant K 8.99 times 10 to the 9 Newton meter squared per coulomb squared times the charge magnitude of the negative charge which is just Q divided by the distance that we are from the negative charge squared. So if I was going to go back here and draw this if this is the negative charge over here then we have this is r minus it's the distance starting from the negative charge going to the point P where we're measuring the electric field it's that distance vector and of course we're taking the magnitude of it and squaring that so it's just r minus squared and you can repeat this now you could take e1 plus and write that down okay so why don't I go ahead and do that continuing over here our exercise from point P1 we have that the magnitude of e1 minus is equal to KQ over r minus squared and similarly if you work through it the magnitude of point one of the positive electric field is going to be given by KQ over r plus squared and if we now want to get the total electric field at point one all we have to do is sum these coefficients so we have KQ over r minus squared plus KQ over r plus squared the distance from the positive charge squared all times negative i hat and we can pull out the common factor KQ and we're just left with 1 over r minus squared plus 1 over r plus squared and in this case the direction is the negative i hat direction that's about as far as we can take it without knowing some numbers for instance how far exactly are we from the negative charge how far exactly are we from the positive charge what is the magnitude of the charge of either of these two Q what is Q we don't know that and again keep in mind that this is a very special case we're right on the line that connects the two charges and I've declared that to be the x-axis so I've really simplified my my vector here I've I've said it lies on the x-axis and so all distances along this axis will point either in positive i hat or negative i hat let's consider point p2 so point p2 is a little bit harder if I draw the electric fields at point p2 the negative charges electric field points up into the left if you were to extend the original drawing I did and the positive electric charges electric field points down and to the left and where those arrows meet had to tail that's point p2 so now we have a situation where the vectors do not lie along the same line now you have to write all their components out with the i hat and the j hat can't just drop one of them anymore because we're clearly not working on a single coordinate axis here and in fact if I were to sketch coordinate axes like this this is x and this is why you see immediately that you're going to have to do some trigonometry get the components of the positive electric field vector and the negative electric field vector and then add those components together to get the resulting total vector this one's a bit more complicated and so in fact what one has to do here is something that's quite a bit more complicated but I'm gonna basically write the answer down for you and what I welcome you doing is making sure that what I'm about to do makes sense can you get to the same answer that I'm going to get to just as a symbolic algebraic equation so I will argue that if you work through the vector algebra here using the fact that the electric field and I'm gonna go ahead and write this out in its full glory the electric field from the positive charge is going to look like this so here's the positive electric field now I can't drop the r hat anymore because I'm talking about full vectors and then I'm gonna add to it the negative electric field at this point P2 now I have to leave the minus sign and when I'm talking about magnitudes and we're talking about the full symbols with all of their signs so the negative electric charge has a negative electric charge and then we just have the distance to the negative electric charge from the point P2 and we have the unit vector r minus hat well there are some common symbols here so we can pull the K and the Q out but we really can't get this much simpler and so we're left with something that doesn't look terribly pretty but in many ways this is the most generic thing you could write down actually let me go ahead and put the minus sign from the charge in between the two of them here like this okay so this right here is the total electric field at this point P2 in a very generic notation in fact I could use this very same equation for the total electric field from this positive and negative charge system really at any point in space as long as at that point in space I figure out what's r plus what's r minus what's r plus hat in vector notation and what's r minus hat in vector notation so I can pick a pretty arbitrary point in space write down my coordinate axes decompose my vectors use trigonometry to figure out the components and then write all that down explicit to that point in space if somebody were to tell me exactly where in space I was doing the calculation I could put some real numbers in here okay but I would argue that you could always start from this nasty looking formula here do the trig use your coordinate system use the numbers that you're given and then write down the total electric field really anywhere around this particular pair of charges so let's take what we have just established as a basic framework for doing calculations of electric fields for this system the dipole and let's apply it to a special case so the special case that I would like us to consider is essentially sketched in the picture in the upper right here and that special case is consider a point P that lies along the line that connects the two charges and I'm going to write that line is the x axis so what we would like to find is what is the dipole field specifically along this connecting line this connecting axis between the two charges basically the line denoted with length D for this system so to figure this out what we're going to do is we're going to take this very general formula up here and we're going to try to homogenize it and simplify it to this very specific case and to give you a few useful bits here of course we're going to be summing vectors so the first thing we have to do is we have to establish a coordinate system and I've drawn a conventional coordinate system that's used for this problem where the y-axis and the x-axis meet at an origin that is exactly halfway in between the two charges such that the distance along the x-axis to either charge is negative D over 2 to get to the negative charge and positive D over 2 to get to the positive charge so all I've done is taken the distance between the two cut it in half and said that that halfway point is where zero is on the x-axis and then of course I've drawn my y-axis straight through that that zero at 90 degrees to the x-axis that's my coordinate system now the point P is a distance x from the origin and that's along the x-axis itself so the point P lies on the x-axis it has no y-coordinate y is equal to zero for P and the distance however though we don't know what it is it's just some variable x so it's some distance x it could be positive number so it could be to the right of the positive charge it could still be a positive number but lie to the left of the positive charge but ahead of the origin you know it could be a negative number that lies to the right of the negative charge or it could be a negative number that lies way to the left of the negative charge we don't know and that's okay we're just going to use x as a placeholder to set this question up of exactly what this electric field will look like so the next thing that we need to do is we need to use this information this coordinate system and this new variable x that denotes the distance that P is from the origin we need to use this information to try to solve for some of the things that we don't know well what do we know well we know k and we know q or at least we're going to assume that we were given a charge at some point but q is just some number so we don't have to figure that out that would be given or not it's a symbol or it's not but in terms of the information we have been given we do have to figure out things like our vector so for instance r plus vector and r minus vector we don't know what those are and we have to figure those out in terms of our coordinate system and these would then allow us access to things like r plus r minus r plus hat and r minus hat the corresponding unit vectors that just indicate the direction that our r plus vector and r minus vector point so really getting any pairs of this information will help us already if we can get r plus and r plus hat for instance we can get the full r plus vector or if we can get r plus and r plus vector we can get our plus hat they're all connected to each other and similarly we hope to do the same thing for the negative charge let's focus on the positive charge right now so let's just focus on quantities that we could derive for the positive charge from the picture that I've drawn in the upper right here well for instance it's actually not as hard as you think to figure out what the distance r plus is so r plus that's the distance from the positive charge plus q to the point p so this right here is r plus vector it goes from the source of the electric field to the place where the electric field is being measured by convention again something to memorize that's the full vector we can get its magnitude using information we've been given we know that the point p is a distance x from the origin so we know that using green we know that this distance here is x and we know that the distance that the positive charge is from the origin is d over two and so we can very quickly use this information to figure out what exactly r plus is we can see that x which goes from the origin all the way out to point p minus d over two is going to be equal to just the pure distance from the positive charge to the point p so again this is the distance x but we're interested in the distance from the positive charge to the point p so to get that we have to subtract off this little piece of distance between the origin and the positive charge so x minus d over 2 equals r plus we're done we figured out a piece of information using other stuff that we were given in the problem so that's great let's keep going we now need the unit vector well this one isn't so bad right I mean let's look at this all we need to know is what is the length one vector that indicates the direction from the positive charge here to the point p here where we're measuring the electric field well all the action takes place on the x-axis and we see that the point p is going to just you know quite naively it's going to be somewhere to the right of the positive charge so that's in the positive x direction so we can already know about really doing any work at all that the unit vector r plus hat is simply i hat it indicates a direction positively along the x-axis because we are to the right of the positive charge and that is a direction that is positively along the x-axis and that's it we just need to know the direction we don't have to know anything else but what's great about this is we can assemble these pieces of information and we can get the full r plus vector although we don't really need it we need our hat we need the magnitude of the distance let's just go ahead and write it down so we have that the r plus vectors x minus d over 2 i hat done we've got all three pieces of information so you see how getting any two of these will result in giving you the third pretty much for free and that's because they're all related to one another awesome well we can repeat this success for the negative charge and I'm just gonna write down the answers our minus is equal to x plus d over 2 and if that confuses you rewind a little bit look at that picture get some pen and paper see if you can figure it out our minus hat is also equal to i hat and again if that's confusing rewind the video look at the picture get some pen and paper see if you can figure it out we don't need it but I'm just gonna be complete and I'm gonna write down our minus vector that's just x plus d over 2 i hat and now we can go back and have a look at this formula generically for the dipole electric field anywhere in space around the dipole now we're specifically interested in turning this into the electric field along this axis to the right of the charges for the dipole so all we have to do to adapt this generic formula into this more specific thing we're looking for is plug in the specific things that we have learned so let's go ahead and do that e dipole along the axis this is a very special electric field is k q now we have 1 over r plus squared times the unit vector r plus hat so we have 1 all over x minus d over 2 all squared times i hat and then we have minus 1 over r minus squared which is x plus d over 2 all squared again i hat what we see some common things here the unit vectors are the same on both of these terms in this difference inside the square brackets so that's another common factor that can be pulled out in front so let's go ahead and do that remember these unit vectors are just algebraic symbols they're independent from the other things that are being drawn in here but if you see two of them in a sum you can pull them out as a common multiplicative factor in front of the sum there's nothing that says you can't do that it's just that you can't mix i hats and j hats and k hats they're distinct from one another and so if one of them multiplies one term and a different one multiplies the other term you don't get to pull both of them out that's bad but we can do this here because it's the same unit vector i hat and now we're just left with this beast which we can simplify now i invite you to add in the missing steps for what i'm about to do but i'll give you a hint all you have to do is find for the stuff inside the square brackets a common denominator for these two terms and this will yield new numerators and you will add those numerators together and then try to simplify the result and what you will find is the following equation k q i hat to d x all over x minus d over two squared and x plus d over two squared and I'm going to leave it at that this is the electric field including magnitude and direction of the dipole as one moves along the x-axis that connects the two charges this is a special case this is not a generic electric field this is not necessarily what the electric field looks like anywhere off the x-axis but along that axis along that line connects the two charges this is what the electric field looks like in all of its glory now you'd have to be given the distance x you'd have to be told the separation of the two charges d you need to the charge magnitude on either of them q in order to do a numerical calculation with this but nonetheless all the ingredients are here to do something like that you could be given numbers and plug them in and get an answer and it would be glorious let's go one step further now an asterisk we're going to use is when one has a small number added to a big number you can effectively ignore the small number as long as the difference in sizes between the two of them is great enough so for instance if I have one plus 0.01 I can write that as 1.01 if I have a thousand plus 0.01 I could go ahead and write that as 1,000.01 but 1,000 is already a very big number compared to 0.01 and so it's okay to say that this is approximately just equal to a thousand that is adding a tiny number to a big number doesn't really significantly change the value of the big number especially if this is like one times ten to the nine plus 0.01 it's pretty safe to say that that's approximately just one times ten to the nine so a billion is approximately a billion even if you add a hundred to it so we're going to take advantage of that and look at a special special case and that is what happens if we move our observation point P so here's the positive and negative charges here's the distance D between them so what happens if we move our observation point P very very very very far away from the dipole system itself so this is our distance X and this indicates a big gap in distance occurs along this axis in other words this axis is not a linear representation of distance for all we know a whole three orders of magnitude of distance could be hidden in this little gap here along the line so I'm using this intentionally to mean that there's a jump in distance that I'm not showing here and this point P could be very far away and in fact the way we denote this mathematically as we say consider the special case where the distance X is much much greater than the distance D then anytime you have a term like this X plus D you can say well that's just approximately X if I have X minus D that's also approximately X I'm just taking a very small number D and I'm either adding in adding or subtracting it from a very great number X and that doesn't really change X all that much and if I do anything to the number D to make it even smaller like for instance X plus D over 2 that's also really going to be approximately X or X minus D over 2 anytime I see a term like that I'm not really changing the value of X in any meaningful or significant way and so I can go ahead and write it just as X so let's go back to our electric field the electric field so let's go back to our electric field this is the electric field of the dipole along this axis and it was written exactly as K Q I hat times 2 D X all over this seemingly nasty little product of stuff that I didn't even bother to write out a little bit there that's the exact formula now consider the case and I'll put this in parentheses here where X is much much greater than D well in that case we see we have terms in the denominator X minus D over 2 X plus D over 2 and what did I say about small numbers well if you've got a small number added to a big number or a small number subtracted from a big number you basically don't change the big number so we can approximately write this as K Q I hat 2 D X all over X squared X squared again X minus D over 2 is approximately X X plus D over 2 is approximately X so these sums collapse into just X all squared and X all squared this is nice because you see I have X to the fourth in the denominator and an X in the numerator and this simplifies one more step to 2 D over X cubed this is a special special case if you have a tiny dipole whose separation is infinitesimal compared to your observation distance you get to use this approximation as long as you're looking at the electric field along the axis that separates the two charges and there are real cases of this in the real world and in fact I'll talk about one of those in just a bit but this is a greatly simple a simplified formula this is much easier to work with in the nasty general thing that we wrote down up here which is already a special case of the dipole electric field so all you have to know here is the separation between the two charges and the distance that you're observing from as long as that distance is far greater than the separation you can use this formula no problem but if your observation distance is comparable to the size of the separation between the two charges you do not get to use this formula so keep that in mind special special case now one last bit of nomenclature I'd like to introduce here for the dipole which is what we've been exploring in this video which is really in many ways the next most simple charge distribution in nature it's a very common charge distribution in nature as you'll see electric dipoles are distinguished as I said earlier by the fact that you have equal magnitude but opposite sign charges separated by a distance d now you'll notice from the way that I wrote my coordinate axes before when I set up this problem that in the formula I just wrote up here where we are observing at some distant point p away from the dipole itself that we have the direction I hat so where does I have point I have points this way it points to the positive x direction and we have this quantity q times d which appears in the equation so we have q and d this quantity appears in dipole problems all the time and in fact it's given a very special name it's the dipole moment and in fact to be quite specific it is the electric dipole moment or at least it's its magnitude so q times d is a special quantity it's defined only for dipoles so when you have equal magnitude opposite sign charges rigidly separated by a distance d and you can use it to quantify very quickly the properties of the dipole itself and I'll show you why in this case from the formula above for this special special case where x is much much greater than d we can write now using the dipole moment a simplified version of the formula that I just had here so we have 2p over x cubed k I hat so I've just substituted for q times d with p the electric dipole moment magnitude it's possible to define the electric dipole moment vector it's one more step p vector is equal to the magnitude of p times a unit vector that points in the direction that the dipole moment is pointing this is not a surprise any vector can be defined this way as the magnitude of that vector times the direction unit vector in which it points well we've already identified what the magnitude is it's q times d so I have to do is figure out what is p hat well we were given p hat p hat in this problem this special special case is equal to I hat now let's look schematically at what I hat is it's a unit vector denoted here and let's think about what direction it points from and what direction it points to yes it points in the positive x direction but in a dipole system it quite specifically points from the negative charge to the positive charge I can transport this vector over here without changing its direction and it's still the same vector and it points from the negative to the positive so quite generically p vector is a vector with magnitude equal to q times d the magnitude of the charge of either of the dipole charges and the d the distance separating them both positive numbers the points from the negative charge to the positive charge from negative q to positive q that is the most general definition of the dipole moment the electric dipole moment that is the full vector that is the whole electric dipole moment and what's great about the electric dipole moment is once you specify q and d and that direction from the negative charge to the positive charge you can tilt the dipole any way you want and all you have to do is continue to specify q and d and the direction that points from the negative to the positive charge and you know the full orientation and separation and magnitude of the charges of the two and in fact dipole moments are a whole lot easier to measure than individual charges and distances in a dipole especially when those dipoles are very tiny here in all of its glory is a schematic of the dipole electric field this is far better than I could ever draw it this is a rendering straight from the textbook and it gives you an idea of using just a handful of electric field lines of how complex the dipole electric field can actually be case of the dipole electric field the electric field strength along the axis connecting the two charges and then we've looked at an extra special case where we take the approximation that we go very far away from the dipole and look at its electric field we haven't looked at any random points around this so you get now from this picture a sense of just how rich and very the structure of this electric field can be it turns out that this is one of the most important electric fields in nature where do you find dipoles in nature well pretty much everywhere as a very macroscopic example here is a schematic of a thunder cloud you're very used to the idea north texas of storm systems moving in and you get lightning strikes well lightning strikes are caused when you have lightning strikes are caused when you have water droplets in the clouds rising up brushing past ice crystals in the clouds the friction between the droplets and the crystals basically transfers charge right so you get the triboelectric effect a charge induced due to friction charges are transferred from droplets to the ice crystals which do drop low in the clouds and you get a charge separation up in the cloud system this induces through strong electric field lines from the bottom of the cloud a corresponding opposite sign charge in the ground so for instance you might get a buildup of negative electric charge in the ground due to the electric fields from positive electric now when those electric fields build up to a specific minimum strength which I covered in an earlier lecture the air molecules themselves can no longer retain their integrity the electrons and because they're being ripped apart by these external electric fields from the cloud and ground and what happens is that the for instance nitrogen molecules rip apart this causes a big separation of charge which essentially turns air into a perfect conductor and you get this massive transfer of charge from either the ground to the cloud or from the cloud to the ground you get different kinds of lightning depending on how the transfer occurs so until that breakdown point you have a dipole you have a big positive charge up in the base of the clouds and you have a corresponding equal but opposite sign negative charge down in the ground and as long as the electric field strength between those two doesn't exceed a certain minimum value everything's going to be fine but when they do you get an electric field breakdown of the air charges transfer the system becomes briefly electrically neutral until the tribal electric effect generates more electric charge in the clouds and the whole thing repeats itself again with a subsequent lightning strike and then another lightning strike and so forth now one of the most important places that dipoles occur are in molecules themselves this is a cartoon of a water molecule it consists of two hydrogen atoms bonded to a single oxygen atom now because of the nature of the bond the hydrogens lie on one side and the oxygen which is much larger has a many more electrons sort of occupies the other side of the molecule and so you get this Mickey Mouse head shaped molecule with a separation of charge the hydrogen atoms bond to the oxygen atom by sharing electrons with the oxygen atom and so the electrons that normally would be orbiting the hydrogen atoms uniformly are spending somewhat more of their time over in the oxygen atom and this causes the naked protons inside of the hydrogen atom to set on one side of the molecule and then the electrons add a net negative electric charge to the other side of the molecule you get a charge separation you get this naturally occurring electric dipole molecule you have more positive charge on one side more negative charge on the other and so as a result of that water dipoles as we'll explore in a future lecture tend to line up such that the negative charges on of one water molecule align closely with the positive charges of a neighboring water molecule and this whole process repeats this causes a force between water molecules it's not a extremely strong force but it's the water molecule and it's dipole bonding to other water molecules is one of the first challenges that we as human beings have to overcome when we're born many infants are unlucky enough to have been born significantly premature this is a fairly common things that that happens and as human beings we've developed medical practices that allow us to deal with early births births that are early enough that the final stages of human development particularly lung development have not finalized if a six to eight week old premature infant is born one of the things that they're lacking in their lungs is a chemical called a surfactant that relieves surface tension in the lungs now where does that surface tension play an essential role well the lungs are basically filled with tiny sacks called alveoli and when those sacks expand we take in air and thus get oxygen the oxygen is exchanged and co2 the waste product of a processing oxygen is expelled and when you compress the alveoli you exhale a breath and the waste product co2 leaves your your body that's the way it's supposed to work but on the inside surface of every alveolar sack is a thin layer of water and it's so thin that it's strongly affected by surface tension surface tension is just the molecule to molecule bonding between the water dipoles in the volume of water right on its surface now the bond is normally not very strong but for an infant with underdeveloped lungs lacking this chemical called the surfactant that can relieve surface tension this is essentially a death sentence because while the infant may be able to draw their first breath when they leave the womb once they exhale the their first waste product co2 the alveoli will collapse and the surface tension of the water is so great that the sacks cannot reinflate unless that surface tension can be read can be reduced in some way so without that surfactant which is naturally produced in the very last stages of infant development that first breath is also the last one so normally what happens when you're born significantly premature is that the attending physician or nurse in the room will take an inhaler and they'll put it in the baby's mouth and inject a small amount of surfactant into their lungs that surfactant is recycled so it will stay in there for many days or weeks until the body is able to naturally produce its own surfactant this is also why premature infants need respirators it's just too difficult their lungs have not fully developed and so they can't really breathe on their own so within just a few weeks they're typically off respirators and then they can go home they just have to be home monitored but it's that first encounter with the dipoles specifically the water molecule dipole that can that can be the last encounter that one has with a molecule outside the womb if one doesn't know that surface tension which is caused by the dipole force between two adjacent dipoles can be a strong and very difficult to overcome thing for underdeveloped lungs so I hope you get a sense of how the dipole it plays an essential role in the world around us and why it's important to understand where it comes from and how to define its parts and how to describe it in space and how to how to picture its electric field we're going to use the dipole concept a lot in this course but you see how it connects to the world around you how does one handle far more complicated distributions of charge for instance I could imagine having some mashed potatoes shaped amorphous blob of material imagine this thing is actually three-dimensional I've sketched it only in two dimensions but I'm not going to put the effort into making this look nice and 3d but imagine this is some 3d blob of material that can hold a net electric charge on its surface how then does one calculate the electric field due to something like this well the shorter answer is you use a computer but in order to use a computer to do this you have to understand the basic principles that underlie the calculations that the computer is going to do and that basic principle at its heart is really just the electric field of a point charge this blob could be represented as a sum over Avogadro's number of atoms so that's going to be 6.02 times 10 to the 23rd and we'll write this as AI where each atom contains a nucleus and electrons that orbit the nucleus and Avogadro's number is a typical number of things you find in the terrestrial sized object pen refrigerator magnet cup of coffee that all roughly has about Avogadro's number worth of stuff inside of it and at the heart of all that stuff are point charges as we know from looking at the the atomic theory of nature we have a situation where we have a central nucleus with a positive charge being orbited by a large number of electrons each carrying a negative charge and for all intents and purposes and as far as experimental science has been able to determine the electron truly is a point charge protons are not but for our purposes they will effectively behave like point charges so we can imagine that this blob is made from a huge number of points all added up to give the illusion of a solid blob of material it's an excellent approximation and it's consistent with the atomic theory of matter so we're going to use that and we're going to employ calculus in order to calculate total electric field so let's imagine that we're over here at some point P and we would like to know what is the total electric field at this point P equal to I'm going to sketch schematically how this is going to work we're not actually going to calculate the electric field of this randomly shaped blob again I would need a computer to do that but what I can do is I can give you the basic ideas and then we can work problems together that demonstrate how you apply these ideas so let's begin this total electric field is going to be equal to the sum from I equals one to N where N could be Avogadro's number of electric fields from individual point charges so all I have to do is sum up K QI over R I squared R hat I for each of those Avogadro's numbers worth of things so what is R well if I pick a random blob and label it I so here here's a here's a nice blob a piece okay this is a little piece of the blob all right I can begin to write down all my little conventions I know that this thing's going to have some little charge and I'm going to denote that little piece of charge that is part of the whole charge of the entire object in calculus notation as D Q sub I for now I'm going to drop the eyes in a minute when I actually do the calculus in this this is a differential it's not two symbols D and QI it's a single symbol representing a tiny little piece of charge a finite sized piece of charge might be written as delta QI so if this is some finite sized thing not big but a piece that's big enough that you could sum up the pieces by hand you might that write this is just delta QI but if the pieces are so tiny and so numerous that you could never imagine writing the sum by hand then we write this as a differential and this is an infinitesimal size charge so for instance it could be an elementary charge there's no way you could add up by hand all the elementary charges in an object that would take more than your lifetime to do that okay well that's the little piece of charge of that little piece of the blob and now of course we have to write down vectors so the convention is that the vector R I goes from the source of the force or the source of the electric field in this case to the thing that's feeling the force or to the point that's where we're trying to measure the electric field so there's our R sub I for that little piece of charge D Q I and of course this is going to have a corresponding magnitude R I and a corresponding unit vector R I and that's about as far as we can take this just with one piece now this could be any point we would have to then figure out the radii for each and every single piece of this blob over here and then add up all of those pieces that's why it's important to use a computer to do this because you could actually break this into a million points and let a computer add this up for you by hand without having to use calculus calculus is a shortcut that we're going to use with simple shapes where the relationship between where the D Q is and the dimensions and geometry of the object are very easily relatable to one another so lines spheres circles those are all very well-defined geometries they're very regular and you can handle those fairly easily using the techniques of calculus but for something that's quite amorphous like what I've drawn here it doesn't make sense to attempt to do this with calculus because there's no well-defined relationship between the geometry the shape of this thing and how far you are from different parts of it okay so all we can really do schematically at this point is turn this into a calculus problem so let's go ahead and do that so now I'm going to convert this into a calculus problem and the way I'm going to do that is I'm going to remind you that in calculus you do this sum in the limit that the infinitesimal things have a size that goes to zero okay so you can go back and you can look up the raw definition of an integral remember the integral is the anti-derivative so it undoes what the derivative does and the derivative also involves taking small pieces adding them on to the coordinates of a function solving as much as you can and then sending those pieces to size zero to solve the problem so the total electric field in our case will be an integral from some minimum value to be determined to some maximum value to be determined of our constant K our differential dq and since we're doing the sum implicitly now I don't have to put that little subscript I on it anymore the distance from that dq to the point where we're measuring the electric field and that's squared and then finally the unit vector that points along the direction of that vector now how the heck do we go to the next step with this well it's very hard to do that with the block again I would just in a computer now that I know that all I have to do is sum up a bunch of point charges I would break it into a bunch of uniformly sized point charges maybe a million of them and then let the computer out of the electric fields for me with all the vectors and all that stuff no problem we're going to do problems in class however where we actually look at simple shapes like a line of charge and see how you can assess that using the rules of calculus and get the total electric field without having to manually break it up into a million points and then add them up by hand or program a computer to do that so we'll illuminate the next step in person rather than doing this in the video and then you'll get some hands-on experience with doing integrals to get electric fields as well