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GL
University of California, Berkeley

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Electric Dipole Field

In physics, a dipole is a pair of equal and opposite point charges (of magnitude Q) separated by a distance (of magnitude d). The electric field of a dipole consists of two components, one pointing from positive charge to negative charge, and the other pointing from negative charge to positive charge. In a vacuum, the electric field of a dipole is always non-zero.

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Video Transcript

in this video will derive an expression for the electric field to produced by a dipole. In a previous video. We actually determine this field for some point anywhere along the Y axis. So here we'll begin with a point anywhere along the X axis. At some distance, say X away. Note. This X will be positive or negative. It could be greater in D over to or less than the over to such that it's inside or outside of the dipole. It doesn't matter, but not the geometry of our dipole. The two charges are on the X axis centered at the origin, and each charges a distance d over to away from the origin. So we know the total electric field of the dipole, which since we're on the X axis and both charges on the X axis, will be on the in the X direction. It's simply the superposition of the fields created by the two charges. So for the positive charge, we have cake and the distance between them, which is X minus D over to squared. For the negative charge, we have negative k que since the charges negative over X plus D over two squared. That's this X distance, plus this deal or two. And of course, as we mentioned, this is in the X direction so we can factor out a common term and what I'll also do. Aside from the carrot cake term, this factor out term or X squared, this will become apparent in a little bit as to why I do that. So we have one minus D over two X square, minus cake hoops minus one over one plus D. We're two x squared in the extraction. And so we have an expression for the electric field that a die pole position along the X axis creates on the X axis. But now, let's asks, are Let's ask ourselves what happens if we move very far away from the dye poll? In other words, X is much, much greater than D, and we'll call from Taylor Expansions in calculus. But the expression one plus ra minus X squared in verse can be tailor expanded as such where we get higher and higher order terms. X squared X cubed, we'll notice here X is much, much less than one. So X is very small, and as we square it and cubit, the number gets smaller and smaller, so we're going to disregard everything except for the excellent year term. Since we know X cubed X squared X to the fourth will all be much smaller Now notice in our expression here for the dipole field instead of extra dealing with D over two X, which is very small in this case here. And so our expression becomes approximately for the first one. We have one plus to the two D over two X minus one minus two d over two X in the extraction, and we'll note that these ones cancel. We're left with the double of two D over to x or D over X. And so we get tu que que de over x cubed in the extraction, and we'll know in here that que de it's simply the dipole moment and cutie in the X direction is actually the vector dipole moment. So we have an expression for the electric field produced on the X axis very far away from a dipole positioned along the X axis. So now we've taken this type opposition along the X axis and determined the electric field better produces at a point on the Y axis and at a point on the X axis. But what about a general point in space? Some displacement vector are from the origin. Now the algebra on this becomes a little tricky in a little bit cumbersome, so we're gonna define a few terms to help us with this. The first is a displacement vector from the positive charge to the point of interest that we call our plus. The second is similar, except for the negative charge, and we'll call that AR minus instead of buying. These are it's simply a general point in space X Y r plus. Well, the X component is merely the full X minus this d over to We have X minus D over two, and the white component is the same as for our. Our vector and similarly AR minus is equal to x plus t over to and why. And to continue with some definitions, recall that the electric field is always proportional to one over some displacement squared where the displacement squared is simply the inner product or dot product of a vector with itself, which we can write as the sum of squares of its components. And if we went to three dimensions, we could add a Z term. Tow R squared is merely X squared plus y squared and are plus or minus squared is X minus or plus D over two squared. Noticed the change in signs from plus the minus plus y sward. So now we have the magnitudes and the vector positions we need in order to calculate this. But this problem still a little bit cumbersome. So I wanna break it into X and y components. And for that, I'm gonna define untangle fatal. Plus that the R plus factor makes with respect to the horizontal and similarly theta minus for the R minus factor. So I could break these in the components now where e plus our Sorry, The magnitude of the total electric field in the X direction is simply the superposition of the magnitudes of the electric fields in the X direction for each charge. So we have cake you over our plus squared Times co sign data Plus for the X direction minus cake over R minus squared co sign data minus. And now we can substitute in our terms that we've defined here. And cosine theta is simply going to be the X components, which is X minus D over to for the positive particle times. Again, another factor of this total vector size magnitude to the one half, and we'll arrive at something similar for the negative charge and simplifying this equation. Pull out a factor of care, Rick. You. So we have the total X component of the electric field generated by the Styx pool. Now leave you to do the work, but we'll see that the white component can be written as such. So now we have the electric field anywhere in space produced by a disciple centered at the X axis with the charges along the X axis. Let's double check this. What happens if we're on the X access again? In other words, why equals zero? Well, e x becomes cake over X minus d over to We'll notice these wives just go to zero and the square terms times three halves give us que terms cubed terms which, of course, reduces to So we recovered. Uh huh. The X component that we had Dr earlier in this video for the electric field created by a dipole along the X axis just to double check for ey. Well, why goes to zero, which turns this terms of zero, which turns the entire white component to zero. So we recovered what we expected when we're measuring the electric field along the X axis.

GL
University of California, Berkeley
Top Physics 102 Electricity and Magnetism Educators
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Cornell University

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