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So here we have a familiar picture. Well, blue particle with charge Capital que No red particle with a charge Q one that's at some displacement vector are from the blue particle, and we know by columns law that the red particle will feel a force from the blue particle of K capital. Q Q one over R squared in the direction of our hat. Let's ask ourselves, what happens if we remove Q one? Well, there's no particle there for the blue particle to act on, so this force has to go away. So what's happening here at this point? Well, we can put Q one back, and we know that force will come back. Alternatively, we can put a different particle there, say, with the charge of Q two, and we know we'll have a force in the same direction this time with Q two in its magnitude instead of Q one. We also know we can move Q one. Stay down here at some displacement are to well, we know this will feel a force from the little particle of F K Capital Q Q. One over our two squared in the direction of our two in fact, we know we can place charged particles anywhere in the vicinity of the blue particle, and each one will feel some force from that blue particle. And so just the mere presence of the blue particle is doing something to the space around it. And we call that something the electric field, and it gives us a new way to think about electric force. We've always said that Q two feels a force from Q. But a better way to think about that is que creates an electric field. Q. Two feels the force from that electric field, and so we say that this blue particle charge capital Q grades an electric field in space. But how do we quantify that field was considered this yellow particle here, which we know feels this force from the electric field created by the blue Particle? I'm going to rewrite this a little more suggestively like that. Recall are Red Particle was originally at this point, and it felt this force in that direction, and we could say that this particle here has a charge of Q three and move it there, and we know by cool arms law the force it feels would be equal to that, and right away we'll note that all of these forces have that in common. So part of the electric force is determined by the charge of the particle feeling that force from the field. But the rest is determined by the charge that is the source of the electric field, its distance from that source and, of course, the direction from that source. And so we can rewrite the electric force on a charge. Q. As Q times the electric field. At that point, we're now e of our It's simply the force divided bike or cake over R squared our had Remember this. Q is the source of the electric field, and this is true for all point charges. Cool. Now what's interesting to note is that, yes, that is the value of the electric field here. But our is arbitrary. We just called that are we could call this our and we'll have a value there for the electric field. We can call this our and we have an infinite amount of ours because space is infinite. And so there are unfinished amount of Decter values possible from the electric field and So we called the electric field of Vector Field. We'll know two things about the electric field. The first is its units. This is the expression we used to derive an expression for the electric field. Of course, ah force has units of Newton's and a charge has Newtons of units of cool arms. So the electric field has units of Newtons per kula, and this makes sense. Um, electric field applies a force upon a charge. The second thing will note is that the electric field obeys the principle of superposition. So say we have a space that has a bunch of test charges. Q one through Q I and we're interested in a point here that will call point P well if we put a test charge there at point P with charge. Q. We know that by the principle of superposition the net force on that charge, it's simply the some of the vector forces on it from the individual charges around it. What if we take you out and just let focus on playing P? We know we can divide this side by that charge, which will give us a vector some of electric fields and so the electric field at that point P is given by the vector sum of the electric fields from all the charges affecting Point P to demonstrate the superposition of electric fields will look at a very common example, and that's the electric dipole on electric. Dipole is simply a system of two particles with equal and opposite charges separated by some distance. So here we have a particle in blue with charge, plus Q and a particle in reading might with charge minus que They're both on the X axis, and they're both separated from the origin by a distance D. And for this example, we want to find what the field is at some point p that we can arbitrarily place on the Y axis. Since it's arbitrary, we could just call this distance. Why? Well, the field created at that point by the blue particle, which will call E plus simply equal to cake over R squared in the art direction. Where are is this displacement vector? But from the geometry of the problem, we have a right triangle, and we know R squared is equal to D squared, plus y squared for the red particle. Our electric field is equal to cake over this displacement our prime. But from symmetry Lino, our prime has the same magnitude as our. And so our prime squared is also equal to D squared plus y squared. And this is the our prime direction. Now here you'll know I've drawn the electric field vector for the blue particle away from the particle and for the red particle toward the particle. This is very common. Let's consider what happens if we put a particle with charge at Point P. Well, we know that force equals Q times, the electric field vector. So if we put a positive particle at point P, these force vectors will point in the same direction as the field. Factors in a positive particle will be repulsed by a positive particle and attracted to a negative particle, which we expect. However, if we point put, a negatively charged particle appointee will flip the direction for the field vector for the four. Specter's and a negative particle will be attracted to the positively charged particle and repulsed by the negatively charged particle. And that's what we expect. So we always draw field vectors as pointing away from a positively charged source and toward a negatively charged source. And now it's simply a matter of finding our components and adding those together. And what will note is that these feel vectors have the same magnitude. We'll also note that they make the same angle with the horizontal, but in opposite directions. So by symmetry, we know that the white components are going to be equal and cancel each other off. So the net in the Y direction it's simply zero. Similarly, we know they're going to have the same X component. And so we can say that E. Net X is simply this same as two times e plus X or two times E minus X, and we only have to calculate one. So what is the X component? Well, e minus X or e plus X. It's simply equal to cake over de squared, plus y squared times, the cosine of that angle theta and from our triangle, we know the coastline of data is simply the adjacent side over the opposite side, the adjacent side of being D over the high pot news and the high Patna is being squared of d squared plus y squared and so we see that plus X equals K Q D over de squared plus y squared to the three halfs. And, of course, that's in the X direction. So we met at that point. P is two k Q d over de squared plus y squared to the three halves hold racks. We'll notice that as why goes to zero. So as we move this point down to the origin, the Nina simply goes to K two cake over de squared in the extraction, which we've seen before, is the correct expression for a point between two equally charged particles at its center.

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