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Now we know what an electric dipole is, so now I want to show you how this dipole behaves in an electric



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field so we will calculate to you to force and to talk acting on the electric dipole.



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So this is what we have found out.



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This is the dipole.



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So you have two charges, a negative and a positive one.



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And they are connected by such a vector, eh?



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So let's think about what will happen when we have a dipole and we apply a homogeneous electric field.



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So this is not really space dependent.



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Yes, I have written down e0.



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It's it's everywhere.



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It's not just here and here.



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It's really everywhere.



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And it affects the whole dipole everywhere in the same manner.



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So here you can see that the dipole and also this vector eight here, they are not aligned with the



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electric field.



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So you can see here we get an electrostatic force.



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So F is equal to the charge multiplied by the electric field.



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So you can see it in this case, the force acts to the right and in this case, the force acts to the



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left.



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So what will happen is the dipole will be aligned with the electric field.



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However, its center point will stay where it is, which means that the total force on this dipole is



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zero.



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Both of these individual forces, blue and red, have compensated.



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So we have no force, no motion, but we have a reorientation, which means there is a talk that has



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acted.



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So now let's calculate the force and to talk to see that this is correct.



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So here you see again, our dipole with this connecting Vector A and here's our zero position, and



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we want to consider here the middle point of the dipole as our point of reference that is labeled here



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by this position Vector R, so now we calculate the force, which is the sum of these two individual



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forces.



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So we have one force that is basically given by charged times, the electric field, and this is at



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the position R plus one half a piece.



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So that's that's one here.



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And then with the negative force we have this one here.



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So there the position or the argument of the electric field is R minus one half a.



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So now what we can do is we can do a Taylor expansion.



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So this means we take the first term, which is just east at the position of R, and then the second



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term is basically proportional to the degree to the derivative.



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So the gradient in this case of the electric field and then we multiply it by one half a, of course



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there would be other terms, but let's forget them for now and not write them down.



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Then for the other term we can do the same thing again.



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We do the Taylor expansion and if we would consider all of the infinite number of terms, this would



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be correct in any case.



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Now you can see that these two terms cancel out so they have the opposite sign, but the same magnitude.



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And these two have have the same sign actually, because we have here minus and minus.



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And so they add up.



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So what we get here for these four terms is just a single term.



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The charge then here is vector eight times gradient or NUPE operator.



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And this acts on the electric field at the position R, which is the center position here.



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So if we now truncate this Taylor expansion and just take this term here, we will find that the force



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is given by the electric dipole moment queue times.



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And then the scalar product with the Knobler operator and everything acts on the electric fields.



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And especially what this means is when we have a constant field or, you know, a homogeneous field



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where it does not depend on R and it's just a constant, then this gradient acting on E will be zero,



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which means that the total force is zero.



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So this is why in a homogeneous magnet, homogeneous electric fields, we do not observe a net motion



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of the dipole.



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So the middle point stays where it is.



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However, we have seen or we have figured out that this whole dipole rotates.



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Therefore there must be a talk.



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So classically, the talk is given by the vector product of the position vector and the force.



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But here we have two individual forces and we should really consider them individually.



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So if we calculate the talk with respect to the center point, then this vector becomes just a divided



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by two at minus eight, divided by two.



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So this is just one here and that one.



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And here we have two positive and a negative charge.



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So this is why I wrote here.



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Plus because we have here minus sine and here, minus sine and then as the force we have, as I mentioned



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already, to the charge and multiplied by the electric field and so does electric field is the same



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as on the left hand side.



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And once again, we can see that is tailor expansion.



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So we write down the electric field as the first term and then this term here.



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And we get, of course, more terms.



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But these we will ignore now here for the other term, we get two more terms.



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And this time because we have here this additional minus sign of this vector A which compensates the



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minus sign from the charge we get now here for the first terms, the same signs they add up and for



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the second terms, we get opposite sides.



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So these will cancel in this case and therefore for the talk we get just as first order term.



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So it's really just this electric dipole moment, which is a few times a and then the vector of product



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with the electric field.



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And as you can see here, even for a constant electric field, we get a torque which allows us to reorient



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our electric dipole moment.



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So when we start from such an oh, yeah, misaligned electric dipole moment with respect to the electric



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field, then we will see that reorients.



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But there is no net motion of the center point of this electric dipole moment.