 In this lecture, I will cover the subject of how electric charge responds to the presence of external electric fields. This will build on top of the ideas of electric charge, force, and field that we've been exploring so far. This may seem like a trivial concept that electric charge will feel a force due to some external source of electric field. But it's important to understand how that concept weaves into Newton's second law, and how that, as it always did in physics 1307, connects to the ideas of motion, equations of motion, and understanding motion. An understanding of how electric charge responds to electric fields will aid you in understanding technology that you may be exposed to or use in your everyday life. So for instance, e-book readers like the one shown here made by Amazon, or smartwatches that use e-ink displays like the Pebble smartwatch, these all rely on electric fields and charged particles and how those particles will respond when the electric fields are applied in order to make the technology possible. We'll explore in detail using a problem, this concept in class, but for the rest of the lecture I want to focus on the very basic ideas behind electric charge and how it responds to the presence of external electric fields. And we'll cover two subjects, point charges and how they behave in electric fields, and then electric dipoles, which are combinations of two point charges of opposite sign, and how they respond to external electric fields. Now that we have gone through the concepts of Coulomb's law and electric fields due to charges, it's important to combine these ideas of force and electric field, and think a little bit about the effect that this has on charged particles. So we're going to be interested in combining f and e and figuring out what this does to the state of motion of an electrically charged particle subjected to for instance an electric field. We can imagine creating a very long thin charged surface, and let me imagine that this surface is made entirely of positive charges. Each positive charge emits an electric field radiating outward from the charge, and at any point in space somewhere around this surface, the total electric field is the result of summing up all of the charges. So for instance, this could be like the line of charge that we look at in class, or this could be a flat plane of charge or a disc of charge like the book discusses in some detail. There are many ways that you can make a big surface with lots of charge, let's say positive charge uniformly distributed on that surface, and so the bottom line is that you can create electric fields that you can then use for other purposes. So if we imagine that this is a really long flat surface with a uniform distribution of positive charge, we can then get something out of this known as a uniform electric field. This is an electric field whose strength does not diminish with position in space. It doesn't increase and it doesn't decrease with position in space. So for instance, I can write this electric field as E, its magnitude times I hat. Let's imagine that this is the positive x direction here along the horizontal. So it's very simple to quickly write down a vector for this. And as long as I'm very close to the source of the uniform field and I don't wander too far away so that the size of this surface here becomes an issue. That is, I get so far away from the surface that it becomes tiny again and therefore acts more like a distant point charge. As long as I'm very close to the surface and as long as the charge is uniformly distributed on that surface, I can get a very good approximation of a uniform electric field. This will come in handy later. We'll look at this in some more detail later when we talk about devices called capacitors. But for now, suffice to say for the purposes of this lecture, we just need something that can create an electric field whose strength does not change with position. So to be very crystal clear about this, the strength of the uniform electric field does not change with position. It has the same Newtons per Coulomb anywhere in space. So I might imagine measuring the strength of this electric field at a point P1 or over here at a point P2 or down here at a point P3. And by definition, if it's uniform, the electric field at P1 is equal to the electric field at P2 and that's equal to surprise surprise the electric field at P3. So under those conditions, this is a very nice field to deal with. You always know what its strength is and you always know what its direction is at any point in space because it's the same everywhere. So starting from that building block, let's now consider what would happen if we dropped a charge into this electric field. So I'm not going to draw the positive charges that cause this uniform electric field any longer. It's just enough to know that somewhere in space, off to the left side of my drawing, somewhere over here, there is a surface with a net positive charge that is the source of the uniform electric field that I am drawing here whose strength is E and whose direction is I hat. Now into this, I'm going to drop a charge. So this charge is Q. It could be positive, it could be negative. I'm not going to specify that right now. We'll take a look at the implications of that in just a second. And what I'm going to do now is I'm going to combine the concepts of force and electric field and we're going to think about what's going to happen to this charge Q in the presence of this external uniform electric field. So this is referred to as the external field and its uniform in this particular example. And this is the charge acted upon by the external field. So we have a source of the force, the external field, and we have the thing being acted upon by that field Q. It could be a positive charge, it could be a negative charge, I'm not going to specify that right now. Let's think about what happens to this. Very generically, we know that electric field is by definition the Coulomb force acting on a charge Q divided by that charge Q. And that's so that we wind up with a charge independent quantity, something with units of Newtons per Coulomb. That doesn't depend at all on the charge we're looking at. But now we've dumped a charge into this problem. We've put Q into the problem. We have to handle that somehow. What's the force, what is the force F vector acting on Q? That's the question we want to answer. Once we know the force, we can do all kinds of things like find the mass of the particle, or if we're given the mass of the particle, figure out the acceleration, and so forth. Well hopefully some alarm bells are ringing at this point. It's just a simple algebraic rewriting of this equation to get to the one that we want. If we want to find out what is the force F acting on the charge Q due to the external electric field E, we just have to rewrite this equation. F will be equal to Q times E. Now in general, this electric field is at a specific point. So it could be, for instance, that we have a more complex field. Let me draw it like this. The field is strong here, and it's weak here. And into this, I drop my charge Q. And now let's specify what the sign of that charge is just so we can do a quick sketch here. I'm going to say that Q is greater than zero. This is a positive electric charge. Well let's think about what's going to happen. We have electric field lines. They start somewhere over on a positive charge. They end somewhere that we're not seeing in this drawing over on a negative charge. Q is a positive charge. So what direction is it going to move in the electric field that's present at this point where it's currently located? Well if you said it's going to move away from where the positive charge would be located, the source of these electric field lines, you're correct. I would expect that the force on this, we'll call this F1, is going to push the charge away from the strong part of the field and toward the weaker part of the field. Because that's the direction the electric field lines point, that's the direction that positive charges would move. Well, a moment later, it now winds up over here at a second point. So we'll call this point 1, and now we're over here at point 2. Well over here the field is a little bit weaker, so the force vector is not as long. And we have a force F2 that's still pushing it away from wherever the positive charges are down here someplace. Toward wherever the negative charges are over here unseen on the right. And so forth. This is a field that weakens with distance, and so every step the force acting on that charge particle diminishes a little bit. And you might imagine if this thing got all the way to infinity, an infinite distance away from wherever the source of this field is, then it would no longer feel a force at all. In Coulomb's law, the force strength drops off as 1 over the distance squared. And so if we imagine that that applies here, when R, the distance, is infinite, then the force will be zero. We're dealing with, however, in our example up here, a uniform electric field. So if I redraw that for a moment down here, here are my uniform electric field lines, all with the same length, all pointing along the same axis in space. Everywhere the electric field is E times I hat. I drop my charge into this, and it feels some force. So my charge Q, which I said was greater than zero, is dropped in here. And it feels a force pushing it to the right, following the electric field line arrows. That force is going to try to accelerate it to the right. And so a moment later, I would expect the charge to be someplace over here. Now at this point, because the field is uniform, it has the same strength everywhere, the force that it feels at that point is the same as the force that it felt at the first point. And this little dance continues. Every time it feels the same force, and if we consider uniform steps in time, we can look at how this particle would accelerate, given those uniform steps of maybe one millisecond, two milliseconds, three milliseconds, and so forth. But at each step, F has the same magnitude and the same direction, and so all we have to do is use the same acceleration, and we can figure out, for instance, what the velocity is of this particle, if we knew its mass. So this is a good point to remind you about Newton's laws. So we know that the force acting on a charged particle due to an external electric field looks like this. We also know from Newton's laws that you can relate any force to the corresponding acceleration by knowing the mass of the particle, and then bound up in the whole discussion of Newton's laws. Okay, so we have Newton's second law. We also have equations of motion that relate things like velocity, acceleration, position, and time. So for instance, if we want to know the average velocity after a time delta t has passed, we could use this equation of motion. The average velocity is going to be equal to the initial velocity plus one-half a delta t. If we wanted to know the position at that later time after delta t has passed, so if we were interested in position of the particle being acted on by this external electric field, then we might use this equation of motion that the final position is equal to the initial position plus any initial velocity times the change in time plus one-half the acceleration times the change in time squared. These should look familiar. These again are just equations of motion. So given a source of a force like an external electric field, we can figure out the force if we know the charge. Given the external force, we can figure out the acceleration if we know the mass. And given the acceleration, given a, we can get all kinds of things like v, x, etc. So there's a whole world of things involving motion and changes in the state of motion that now become accessible simply by specifying the relationship between a charge dropped into an electric field and the force that that charge feels due to the external electric field. And you can start to see now there's all kinds of things that you can start to think about involving charges and force and acceleration and so forth. And we'll explore some example problems in class. But one thing I can quickly sketch for you is a simple device that can be used to deflect charged particles. So imagine that I have two big long flat plates and on the top plate I put a whole bunch of uniform, uniform positive charge. And by uniform I mean there's no place on this plate where positive charge is clumped more than any other place. And similarly down here we have a uniform negative charge. Well, the electric field lines will go from the positive to the negative. So let me draw a few representative ones here. And here again we can get, if we do this just right, a uniform electric field. In this case it happens to have a strength E and it points down. So we could call that negative j hat. Into this I'm going to shoot a charge Q. And its initial velocity vector, v0, will be a straight line going to the right along the positive x direction. What happens when it encounters that first line of force from the external electric field? Think about that for a moment. Pause the video if you need to. Well if you've had a chance to think about it a little bit and you thought, well it encounters the external electric field, it must feel a force due to that electric field, that force will cause an acceleration, you're absolutely right. And so all we have to do is figure out just what the heck that acceleration is going to be. Well, let me specify that I want Q to be less than zero now. I want a negative charge Q to enter the system like an electron. Negative charges want to move toward where the positive charges are and away from where the negative charges are. So they will move against the electric field lines as they're drawn here by convention starting on the positive charge and ending on the negative charge. So the force that this electron or other negative electric charge will feel will tend to want to push it that way, vertically, against the flow of the electric field lines. And so what will happen is this charge will begin to bend. Much like a ball thrown in earth's gravitational field, it will follow a curved trajectory as it begins to bend away from the horizontal direction now that there's an acceleration in the vertical direction. And then because this is a uniform electric field, every step in the problem we take will yield the same force pushing it upward in this case. This is a uniform electric field, the force is always the same at every point in space and so it just feels a constant acceleration as it bends and so forth. If I send in a particle fast enough, so imagine I send in a second particle with a really, really big velocity, we'll call this v0,2. Maybe to be a little bit clear, we'll put the 2 in parentheses here. This might be moving so fast that even though it's accelerating in the vertical direction, this doesn't have enough time to deflect it significantly. So it will bend, but it might only bend like this and come out and continue moving on the trajectory that it was last on before the force stopped. So this would be v0,2 final. The final velocity will point in a different direction and also be larger because now it has a y component as well as an x component than the initial velocity. You can see already that this is a very useful thing. You can use electric fields, for instance, to bend charged particles of different masses by different degrees such that you can separate them. This is one way that you could imagine separating different species of electrically charged particles from one another. If you have a beam of particles and they're mixed up and you don't know at first where the solid dark particles are and where the unfilled lighter particles are and if they have different masses, m1 and m2, they will feel correspondingly different accelerations due to Newton's second law, even if the force on them is the same. So if they all have the same charges, q, q, etc., but different masses, it's possible then to create a situation where you have an electric field that's uniform and even though all these particles come in a big blob, some of them will bend a little bit and some of them will bend a lot because they have less mass, they're less able to resist changes in their state of motion so they have less inertia and they can be pulled out of this beam very easily. This is one way you might imagine separating contaminant particles that are electrically charged out of a beam of desirable particles that's otherwise mixed together. There are other ways to do this and we'll explore those later but this is one way you might imagine doing this and so this is a very effective thing to do. We'll cover an example of another place where this kind of idea is very useful in class but it has to do with separating the different components of a DNA strand using an electric field and agarose gel. Having considered generically what will happen to an electric charge that has been exposed to an external electric field, we can now go back and revisit our friend the electric dipole which is the next most complicated system of electric charges that we can study by hand. To remind you, the electric dipole consists of a negative electric charge which I've written here as having a charge minus q where q itself is a positive number so the minus sign enforces the fact that this is a negative charge and at the other end of the dipole is a positive charge plus q. The ends of the dipole that contain the charges are rigidly separated from one another. They are not free to move further apart, they are not free to move closer together and they are essentially joined in such a way that if they are exposed to an external force they will respond together to that force. You can think of an electric dipole as a long rod that is then subjected to forces along the length of the rod and this is going to sound very analogous to, for instance, lines of mass or rods under the influence of gravitational or other forces that you might have covered in 1307, the first semester course. This analogy will work very well and in fact it will serve us to understand what happens to the electric dipole when it is exposed to an external electric field and thus is exposed to an external force. Now the other key piece of the dipole that you need to recall is that its entire orientation can be described by something called the electric dipole moment written here as p-factor. It has a separation between the charges of d and the magnitude of the electric dipole moment is simply written as q times d and the direction p-hat that the dipole moment points in always goes from the negative charge in the dipole to the positive charge in the dipole. That's the convention for an electric dipole. So again another convention to be memorized but this one is specific to dipoles. Now what would happen if we were to suddenly immerse this dipole whose p-factor is tilted with respect to the horizontal to an external electric field that points entirely along the horizontal direction and is uniform in direction and strength. So here I have introduced an e-vector an external electric field with magnitude e and direction i-hat. We can think now about what is going to happen separately to each of these charges. Let me redraw the dipole here and remind you that this is the minus charge this is the plus charge that sort of like the large masses on the end of a weightlifting dumbbell these two are rigidly joined to one another. In the case of the hydrogen and oxygen atoms that make up a water molecule the rigidity is provided by the Coulomb force between the hydrogen and oxygen atoms that keeps them bound and keeps the hydrogen from being pulled away from the oxygen unless too much force is applied. Consider the case that the external force is not strong enough to actually remove these charges from one another's presence. Well we have an electric field acting on the positive charge and its magnitude and direction are the same as the electric field acting on the negative charge. So let's go ahead and draw the electric field represented by these arrows at both the negative and positive charges and again let me remind you that these are both the same vector they point in the same direction they have the exact same magnitude I'm just transporting this arrow whose tail is located here at the positive charge down here to the negative charge and transporting an arrow without changing its length or its direction does not change the vector. So now we can consider what the forces are going to be on each of the charges. I will write the force on the positive charge as f vector with a subscript plus and this is just going to be equal to the charge of the positive charge which is q a positive number multiplied times the vector e vector and we can go one step further and write this as qe e is the magnitude of that electric field i hat. Similarly I can write the force on the negative charge in the dipole as negative q times e vector now the only thing that's different between the positive charge and the negative charge is the sign of the charge because this is a negative charge it will tend to feel a force that points in the direction opposite that which e vector points in and that flipping of the force vector relative to the electric field vector is controlled entirely by the sign of the charge and we can go again one step further and write this as qe i hat with the minus sign in front of it and so we can start to ask questions like what is the net force felt by this dipole? Well the total force in a linear sense the total linear force that is felt by this electric dipole is just going to be equal to the sum of f plus and f minus the vector sum and we see already that we can write the pieces in here we have qe i hat minus qe i hat and we see that the total sum is zero the net force and let me emphasize that this is in a linear sense that is the net force along the lines on which the forces act is zero but does that mean that the dipole does not respond in any way to this external electric field? Well if you've had that experience in 1307 of looking at rotational motion you'll know that that was a leading question we know that when forces act at a point that is not the center of mass of an object such as the forces here that act on the ends of the electric dipole you have now a force acting away from the place that represents the central mass location of the system and this can induce a rotation a force acting along a line that is displaced from the center of mass of a system will generate a torque and that torque will generate a rotation so let's have a look at the torques that are generated by the forces acting on the positive charge and the negative charge to begin we have to remind ourselves what torque is torque is a cross product and this will be our opportunity to revisit the special vector product known as the cross product it is represented by the greek letter tau it is a vector and it is equal to the vector r which I'll explain in a moment cross product with the vector f so let's consider a system like a rigid rod with a uniform distribution of mass along its length if it has a uniform distribution of mass then the center of mass of the system will be dead center in the middle of the rod in fact this is how you can find the middle of a long uniform in a mass distribution rod you simply move your finger to where you think the center of the rod is and if the rod balances then you found the center of mass of the system if not, if it tilts to one side keep moving your finger around until you find that sweet spot and that will identify the center of mass the place where the net torques on either end of the rod cancel out and there is no more rotation if we were to exert a force on one end of the rod and another force equal in magnitude but opposite in direction on the other end of the rod so I have f and minus f same magnitude opposite direction this will induce a torque and that will take the place of the f that's shown here now what is r? well you have to start at the center of mass of the system and r is a vector that points from the center of mass of the system to the location where the force is being applied in the case of a body like a door where one end is rigidly fixed to the wall with a set of hinges r will go from the point of pivot which is the hinges to the location on the edge of the door where the force is applied so we're going to have to look at the problem carefully that you're given and figure out is this a problem about an object that's not bound in any one point in space and therefore is free to rotate for instance about its center of mass or is this an object that's fixed to a point in space and so if there is a torque it can only pivot about that point these are things that you have to consider when looking at problems involving rotation we're going to be looking at objects which are free to move about their center of mass the dipoles are not fixed on one end some place in space if a torque is exerted on them they can rotate about their center of mass so if we call this f1 and we call this negative f1 which I will write as f2 those are equal to one another then this can be written as r vector 1 and the corresponding r vector that points from the center of rotation the center of mass of the system to the other end where the other force is applied we can write as r2 and then we have two torques on the system we have torque 1 which is r vector 1 cross f1 and we have torque 2 which is equal to r vector 2 cross product with f2 now what is the cross product? I will only review the salient features that we are going to need for this exercise if you have two vectors a vector and b vector then the cross product yields a third vector which I will write as c vector that makes right angles with both a and b so here is c vector and if I were to denote the angle between a and c it would be 90 degrees or pi over 2 radians and then similarly the angle between b and c will also be 90 degrees or equivalently in radians pi over 2 radians and the equation for this is a cross product with b equals c now what are the consequences of this? well there are a few things since a and c are at 90 degrees to one another their dot product which tells you how much of one vector projects onto the other will be equal to a c cosine of theta where theta is the angle between them and then you have to ask yourself well since I know the angle is 90 degrees what is the cosine of 90 degrees and of course you should find that this is equal to 0 vectors that are at 90 degrees to one another have no mutual projections on one another and since the dot product tells you the degree of mutual projection of two vectors how much one's shadow lies on the other when they're at 90 degrees when they're at a right angle the answer is always 0 and similarly by the definition of the cross product b vector dotted into c vector will be a b cosine theta and that will also be 0 because the angle between b and c is also 90 degrees so that's the neat thing about the cross product the cross product of two vectors will give you a third vector that is perpendicular to both of the original two this is for instance very useful when thinking about axes in a Cartesian coordinate system for instance the cross product of i hat and j hat yield k hat in what is known as a right handed coordinate system which is what we typically use when we talk about x, y and z and the way you can see this is if this is the x axis and this is the y axis and this is a 90 degree angle between the two of them the cross product is calculated visually by first pointing the fingers of your right hand along the direction of x curling your fingers toward the direction of y and your thumb which should point either up or down will indicate the corresponding direction of the resulting cross product vector the so-called right hand rule or this is one variation on the right hand rule and what you'll find is that your thumb points up which is exactly what you'd expect in a right handed coordinate system and of course by construction because z is perpendicular to y and z is perpendicular to x you get the z axis so this vector i hat crossed into the unit vector that points along the y direction j hat will naturally yield k hat the unit vector that points along the z direction what about magnitude the magnitude of the dot product if you have for instance a dot c is ac cosine theta the magnitude of the cross product a cross b is equal to a b sin of the angle theta between them so up here in this picture this is theta it's the angle between a and b if that angle is zero then the sin of zero is also zero and the cross product does not exist if you have two vectors that are parallel to one another then there is no single vector perpendicular to both of them simultaneously and therefore the cross product does not exist as long as a and b have a slight angle to one another then the cross product will exist and its magnitude can be calculated from this formula a b sin theta so let's return to the torques that we were considering a moment ago and let's look at torque one for the electric dipole that we were talking about earlier we can write down the vector r almost with no trouble whatsoever if r points from the center of mass of the electric dipole and if the masses on either end of the dipole are equal then r one vector will start at the middle of the dipole halfway between the two charges and it will end where f one is applied on the positive end of the dipole so if I were to again resketch the system we had above here's the positive charge here's the negative charge I'm going to consider the special case where this has mass m and this has mass m so they're identical in mass and then the vector r one will go from the center of mass to the positive charge where f one is applied now you'll notice that the magnitude of r one is one half the length between the two ends of the dipole so we know already that the magnitude of r one is one half d what about its direction well for right now I'm going to simply write this as p hat the direction that the dipole moment points in and remember that's a vector that points from the negative charge to the positive charge it has a length of one and that's exactly where r one points in this case it points from the direction of the negative charge from the location of the negative charge to the location of the positive charge and now I'm going to cross product this with the force well what is the force the force is q e i hat that's the force applied by the uniform external electric field on the positive charge and so we can simply set this equal to a new symbol tau plus which is the torque that's being exerted on the positive charge and similarly we can figure out what tau two is again we're talking about r two which is a vector of length one half d but now it points in the negative p hat direction why because r two points from the center of mass toward the negative charge which is opposite the direction that p hat points so we can just put a minus sign in front of p hat this thing will definitely have length of one half d one half the length of the one half the length of the the dipole electric dipole and we're good to go and we're just going to take the cross product of this with negative q e i hat because that's the force that acts down here on the negative charge pointing in the negative x direction and again I can just change the notation here and call this tau minus the total torque is equal to pulling the numbers one half and d and q and e out in front of the cross product one half d q e p hat cross i hat and this is added to pulling again the numbers out in front of the cross product in the second term for tau minus we have one half d we have a minus one multiplying the p hat so we can pull this out as negative one half d and then we have negative q and e so we have negative one half d times negative q e yields a positive one half d q e and again we have p hat cross product with i hat and what you notice is you have two terms in the sum that are identical so you have one half of this term and one half of this term and you can write tau vector finally as just d q e p hat cross i hat now I'm going to take this one step further and I'm going to re-generalize this problem we have q times d which is the magnitude of the dipole moment which we've been writing as p p is equal to q d and we have e the magnitude of the electric field I'm going to take these numbers that multiply this vector cross product and I'm going to distribute them into the cross product I'm going to put d q in front of p hat I'm going to put e in front of i hat and when I do that hopefully you'll see what happens here we have p p hat cross e i hat this thing here is just p vector and this thing here is just e vector so the total torque on a dipole the total torque on a dipole due to a uniform external electric field is equal to p vector cross e vector that's the most general relationship between those things and you see here why it's so convenient to define the dipole moment if you have the dipole moment and you know where it points you just need the electric field vector you can calculate the torque for free from that and then once you have that you can figure out all kinds of things like angular acceleration and angular displacement and so forth now sketching this again we have a force acting here on the positive charge and that force is due to a uniform external electric field we have another force over here acting on the negative electric charge and we know that this is equal to negative f plus vector we get a nice rotation of the system so sort of drawing the sense of rotation you expect the positive charge to try to move to the right the negative charge to try to move to the left and so the whole thing kind of rotates like this in a clockwise direction and so what you get is you get the dipole attempting to line up along this external electric field to which it was subjected so if the dipole starts out oriented at an angle to the electric field after some time has passed we expect that the dipole will line up such that the positive charge is as far to the right as it can get in the electric field and the negative charge is as far to the left in the field as it can get so here's negative q and here's positive q and it's lined up along the field now the force pushing on the positive charge pushes it straight to the right the force pushing on the negative charge pushes it straight to the left they cancel each other out there's no linear motion and since the axis of the dipole along the electric field there is also no longer a cross product between r and f that yields a vector that's non-zero and so the result is when you reach this state you have no net torque in a uniform electric field and no net linear force in a uniform electric field and again this is for a uniform electric field now let's consider what we have learned and apply it and make a prediction about something that should happen in nature what if it were possible to create a non-uniform electric field and subject for instance some kind of tiny dipoles to that field like water molecules so imagine I have a line of charge so here I have a line of charge and it points out of the screen so this line of charge is coming out at you and it's disappearing off into the plane of the screen away from you so you're only seeing its end it's like looking at a dowel or a curtain rod or a pencil on one end oh you see it's a little circle from class we work through the problem of how to use calculus to figure out when this has a uniform distribution of charge what will the electric field look like and if the charge on here is positive so we'll say the Q distributed on this line is greater than zero then we expect the electric field to originate on this and radiate outward falling off in a linear sense away from the line of charge this is a non-uniform electric field it diminishes in strength as we go further from the line of charge so what would happen if I were to put a dipole in this and actually to be careful here let me be a little bit more cautious make this a little bit easier to see I'm going to put the positive charge up here and the negative charge here and here's the line that rigidly joins them here is the dipole moment pointing from the negative charge to the positive charge we have plus Q here we have minus Q here well now we have a situation where the electric field acting on the positive charge may not have the same direction or magnitude as the force acting on the negative charge the electric field strength up here which is a point that is somewhat further away from the line of charge is going to be weaker than the electric field strength down here where the negative charge is which is a point that's somewhat closer to the line of charge so these two will experience two different forces on either end of the same dipole you will therefore have no cancellation of linear force and you will have a net torque that is not equal to zero so what happens? well the net torque causes a rotation we expect that if there's a net torque the dipole will rotate to try to orient itself along electric field lines but if there's no cancellation of linear force then the dipole will move it will accelerate linearly a net force means a net acceleration and a net acceleration means motion so if we were to subject water dipoles to a non-uniform electric field what do we expect to happen? what's the prediction? we expect the water molecules to accelerate along a linear direction and to rotate now water molecules are very tiny we're very unlikely to notice they're rotating but if we get a strong enough electric field we're very likely to notice the water molecules translating especially if we put them all in a stream of water and subject it to an electric field so the prediction that we'll make based on this lecture and we'll test this in class is if you subject a stream of water to a strong non-uniform electric field we predict that it will move and that the water molecules will rotate water molecules are so small we probably won't notice the rotation but we are likely, if we make the field strong enough to notice the translation, the linear motion so I predict the stream will bend and we'll test that in class