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11

The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: \begin{equation} U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...


6

Dipole moment is a vector and can be calculated using formula $$\vec{p} = \sum_i q_i \vec{r}_i.$$ It can be shown easily using the formula above that in case of two charges separated by distance $d$ $$\vec{p} = q \vec{d},$$ where vector $\vec{d}$ goes starts at negative ends at positive charge. ...


5

I'll give you the derivation from my book which includes a nice way to see how the delta functions arise: .............................................................................................................................................................. We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and ...


4

It is true that there is no (electrostatic) force between an electrified body and a body not electrified. (Let's ignore gravitational force for now.) It is also true that all bodies (in earth or earth-like environment) are electrified or will be electrified if approached by another electrified body. But in general, not all bodies can be electrified. For ...


4

Because the black area is half the box below. To explain: move the dipole from an area of no field to an area of field strength E. As you do, there's a force proportional to the dipole moment and to the gradient of E. For a fixed dipole, this force depends only on the gradient (horizontal dashed line). But for an induced dipole, the dipole moment depends ...


4

Classically a non-pointlike spinning charged object possesses a magnetic dipole moment due to the fact that charged particles in the object are spinning around some axis. In contrast, the electron has a dipole moment that arises from its intrinsic spin angular momentum. As you point out, the electron has no internal structure, so the spin does not refer to ...


4

Here's one way to think about it (though it isn't mathematically rigorous). From very far away the dipole would appear to have zero charge and thus there wouldn't be an electric field at all. However, you also know that the electric field falls off as $1/r$, so from very far away you'd expect the electric field to be small. The additional charge ...


3

The force on a dipole placed in an electrical field is given by $\mathbf{F} = (\mathbf{p}\cdot \nabla)\mathbf{E}$ (see, e.g., Griffiths, 3rd edition, eq. 4.5). Recall that, $$ \nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + ...


3

The potential energy in this case should be $U=+\vec{m}.\vec{B}$, hence the potential energy is minimized, as it should be. Here is the explanation: Let’s look at the derivation of interaction energy between magnetic dipole and magnetic field carefully. The dipole energy $U=-\vec{m}.\vec{B}$ is derived using principle of virtual work with an assumption ...


3

If the magnetic dipoles in a material are ordered, the material has a lower entropy because there are many fewer ways how the spins may be oriented if most of them (or all of them!) are required to be aligned. Such an alignment also reduces the heat capacity because before the dipoles got aligned, the orientation (direction) of each dipole was a degree of ...


3

The electric dipole moment is defined as $$p = \int r \; dq$$ In the case of a pair of charges for which both charges are of the same magnitude, the choice of the origin turns out to be irrelevant: $$ p = \mathbf{r_1} q - \mathbf{r_2} q = q(\mathbf{r_1} - \mathbf{r_2}) = q\mathbf{d}$$ where $\mathbf{d}$ is the distance between the charges. However, when ...


2

One must distinguish two conditions: whether the eigenvalue of $|\vec J|^2$, the squared total angular momentum, is changing; and whether the whole vector $\vec J$ is changing. The latter is guaranteed in a dipole transition: one can't keep the whole vector constant. At most, you may satisfy the former condition: the length of $\vec J$ may stay constant so ...


2

The position dependence comes in via the $R$ in the following expression¹: $$ \vec E = \frac{3 \vec p \cdot \vec R}{4 \pi \varepsilon_0 R^3} \vec R - \frac{\vec p}{4 \pi \varepsilon_0 R^3} $$ This makes sense, as the charge $q$ also does not depend on distance - it’s effect on the electric field $\vec E$, however, does depend on said distance and the ...


2

Nope. The electron dipole moment is defined as its intrinsic property – a property that may be imagine as a consequence of the particle's internal structure or, more generally, "something that happens inside it". So you may imagine that the relevant wave function for the "center of mass" is $\delta^{(3)}(\vec r)$. It's not literal because one can't localize ...


2

There are two misconceptions present in your explanation of the problem. $N$ is not number of dipoles, but their volumetric density $Q$ is not total charge, but equivalent charge at boundaries of the dielectric. The idea is that (a) dielectric of the area $A$ and height $L$ polarized homogeneously along its height and (b) two plan-parallel plates of the ...


2

The $Q_x$ and $Q_y$ transitions are electronic excitations in the conjugated $\pi$ orbitals of the Bchl a molecule. They involve two different sets of conjugated bonds. The $Q_x$ involves a shorter chain of conjugated bonds so it occurs at a higher energy/frequency. I couldn't find a really good diagram to show which bonds are involved in the in the $Q_x$ ...


2

The dipole moment of a system of charges $q_i$ located at positions $\mathbf r_i$ is defined as the vector $$\mathbf d=\sum_i q_i\mathbf r_i.$$ If you have a single charge $q$ at $\mathbf r=d\hat{\mathbf e}$ then $\mathbf{d}$ has magnitude $qd$ and points along the unit vector $\hat{\mathbf e}$. Usually, however, this is introduced for two charges of equal ...


2

The answer lies in the "polarizability" of the sphere. This relates the external field to the induced dipole moment. For a (ridiculously) rigorous treatment, a good book is "The Scattering of Light by Small Particles" by Craig Bohren. However, if you're looking for a simple result, the polarizability and the dipole are related like this: $p = \alpha E$ ...


2

When an electric dipole is placed in a uniform electric field making an angle with the direction of the field as shown in the figure. Force on charge $-q=-q\overrightarrow{E}$ (opposite to $\overrightarrow{E}$) Force on charge $+q=q\overrightarrow{E}$ (along $\overrightarrow{E}$) Thus, electric dipole is under the action of two equal and unlike ...


1

Equation (2.5) expresses the velocity field in function of the stream function. It's not clear to me it really should be presented at this stage in the process, I guess it's useful to impose the conditions at infinity. Equation (2.6) expresses that the two pieces of the total solution, the one inside the disk $\psi$ and the one outside $\psi_1$, have to ...


1

When introducing the stream function, the steps that you usually take are as follows. Replace $u$ and $v$ by the streamfunction. Derive the horizontal momentum equation (for $u$) with respect to $y$ and the other with respect to $x$. Eliminate the pressure term, to end op with a single equation in $\psi$.


1

The following is relevant: V.V. Meleshko and G.J.F. van Heijst, "On Chaplygin's investigations of two-dimensional vortex structures in an inviscid fluid," J. Fluid Mech. 272, 157-182, 1994 . While their outline may be less than "full", their bibliography looks comprehensive.


1

In addition to joshphysics's answer, the gyromagnetic ratio of an electron which may be measured experimentally proves that it is not the rotating dipole. Dipole momentum of an electron may also be measured experimentally. And, with a good precision is zero.


1

There are two common types of dipole vector fields in physics: The "divergence-free" dipole field: $\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} + \frac23\vec{\mu}\,\delta(\vec{r})$ The "curl-free" dipole field: $\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi ...


1

Actually, if you ask about radial dependace of time-avaraged Poynting flux vector, then it depends on where the observer is (far field or near field). If you assume that the observer is far away (more than $\frac{2D^2}{\lambda}$, where $D$ - is the maximum size of your source and $\lambda$ - is the walength (I see you use monocromatic case)) than the ...


1

The force applied to a point dipole with dipole momentum $\vec{p}$ is $$ \vec{F} = (\vec{p} \cdot \vec\nabla) \vec{E} $$ In Cartesian coordinates that is $$ F_i = \sum_j p_j \frac{\partial}{\partial x_j} E_i $$ But in spherical coordinates it is not the same. There is no field components along $\vec{\theta}$, but there is a gradient of field components ...


1

A dipole is formed of two opposite charges. By bringing them ever closer together, all the while increasing their charge but keeping the product of the charge and the separation $p = q \times d$ constant, we can form an ideal, elementary dipole. The field on a line in between the charges goes as $~ 1/d^4$ and thus goes to infinity as $d \to 0$. This forms ...


1

The origin of the problem is the special point $\mathbf{r}=0$. In the usual derivation (when using easily-derived formula for the potential $\varphi = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$): $$E_{\alpha} = -\frac{1}{4 \pi \epsilon_0} \nabla_{\alpha} \left( \frac{p_{\beta} x_{\beta}}{r^3}\right) = -\frac{1}{4 \pi \epsilon_0} ...



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