Now we know what an electric dipole is, so now I want to show you how this dipole behaves in an electric
field so we will calculate to you to force and to talk acting on the electric dipole.
So this is what we have found out.
This is the dipole.
So you have two charges, a negative and a positive one.
And they are connected by such a vector, eh?
So let's think about what will happen when we have a dipole and we apply a homogeneous electric field.
So this is not really space dependent.
Yes, I have written down e0.
It's it's everywhere.
It's not just here and here.
It's really everywhere.
And it affects the whole dipole everywhere in the same manner.
So here you can see that the dipole and also this vector eight here, they are not aligned with the
electric field.
So you can see here we get an electrostatic force.
So F is equal to the charge multiplied by the electric field.
So you can see it in this case, the force acts to the right and in this case, the force acts to the
left.
So what will happen is the dipole will be aligned with the electric field.
However, its center point will stay where it is, which means that the total force on this dipole is
zero.
Both of these individual forces, blue and red, have compensated.
So we have no force, no motion, but we have a reorientation, which means there is a talk that has
acted.
So now let's calculate the force and to talk to see that this is correct.
So here you see again, our dipole with this connecting Vector A and here's our zero position, and
we want to consider here the middle point of the dipole as our point of reference that is labeled here
by this position Vector R, so now we calculate the force, which is the sum of these two individual
forces.
So we have one force that is basically given by charged times, the electric field, and this is at
the position R plus one half a piece.
So that's that's one here.
And then with the negative force we have this one here.
So there the position or the argument of the electric field is R minus one half a.
So now what we can do is we can do a Taylor expansion.
So this means we take the first term, which is just east at the position of R, and then the second
term is basically proportional to the degree to the derivative.
So the gradient in this case of the electric field and then we multiply it by one half a, of course
there would be other terms, but let's forget them for now and not write them down.
Then for the other term we can do the same thing again.
We do the Taylor expansion and if we would consider all of the infinite number of terms, this would
be correct in any case.
Now you can see that these two terms cancel out so they have the opposite sign, but the same magnitude.
And these two have have the same sign actually, because we have here minus and minus.
And so they add up.
So what we get here for these four terms is just a single term.
The charge then here is vector eight times gradient or NUPE operator.
And this acts on the electric field at the position R, which is the center position here.
So if we now truncate this Taylor expansion and just take this term here, we will find that the force
is given by the electric dipole moment queue times.
And then the scalar product with the Knobler operator and everything acts on the electric fields.
And especially what this means is when we have a constant field or, you know, a homogeneous field
where it does not depend on R and it's just a constant, then this gradient acting on E will be zero,
which means that the total force is zero.
So this is why in a homogeneous magnet, homogeneous electric fields, we do not observe a net motion
of the dipole.
So the middle point stays where it is.
However, we have seen or we have figured out that this whole dipole rotates.
Therefore there must be a talk.
So classically, the talk is given by the vector product of the position vector and the force.
But here we have two individual forces and we should really consider them individually.
So if we calculate the talk with respect to the center point, then this vector becomes just a divided
by two at minus eight, divided by two.
So this is just one here and that one.
And here we have two positive and a negative charge.
So this is why I wrote here.
Plus because we have here minus sine and here, minus sine and then as the force we have, as I mentioned
already, to the charge and multiplied by the electric field and so does electric field is the same
as on the left hand side.
And once again, we can see that is tailor expansion.
So we write down the electric field as the first term and then this term here.
And we get, of course, more terms.
But these we will ignore now here for the other term, we get two more terms.
And this time because we have here this additional minus sign of this vector A which compensates the
minus sign from the charge we get now here for the first terms, the same signs they add up and for
the second terms, we get opposite sides.
So these will cancel in this case and therefore for the talk we get just as first order term.
So it's really just this electric dipole moment, which is a few times a and then the vector of product
with the electric field.
And as you can see here, even for a constant electric field, we get a torque which allows us to reorient
our electric dipole moment.
So when we start from such an oh, yeah, misaligned electric dipole moment with respect to the electric
field, then we will see that reorients.
But there is no net motion of the center point of this electric dipole moment.