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12

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: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...

7

The magnetic analog of Gauss's law tells you that $$\oint B dA = 0$$ This says that he number of magnetic field lines entering and leaving any surface surrounding any configuration of magnets are always equal. So there is no configuration of equal and opposite poles which produces a monopolar field. Your configuration would neutralize the magnetic field ...

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

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 ...

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 ...

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 You'd need to consider the angle between the two antennas and the distance between them. Dipole antennas do not radiate uniformly into 4\pi. If you're looking up or down at the poles of the antenna, you will see no radiation (ideally). Looking at a direction transverse to this, the radiation is at a maximum. The angular dependence of the far (electric) ... 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 ... 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 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 dipole transition matrix element has a classical interpretation as the time Fourier series of the classical dipole moment of the Bohr orbit corresponding to one of the energy levels. The interpretation is only exact at high levels, at the correspondence limit, and the m,n matrix element is the m-n-th Fourier series coefficient for either orbit m or orbit ... 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 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} $$This makes sense, as the charge q also does not depend on distance - it’s effect on the electric field \vec E, however, does depend on said distance and the ... 2 The dipole moment of a system of charges q_i located at positions \mathbf r_i is defined as the vector$$\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 $Q_x$ 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 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

It's a matter of choice. You can set the potential energy to be any value at any angle. You don't even have to have a zero-value at all; you could make $U$ purely positive or purely negative if you're feeling adventurous. But the advantage for $U(\pi/2)=0$ is, as you said, the simple expression $U(\theta)=-pE\cos\theta = -\vec p \cdot \vec E$ instead of ...

1

this is basically a dipole type of problem. direction of dipole is from -ve to +ve charge in the direction of line joining the charge and it is p. torque is given by cross product of this p and electric field i.e. pXE. (please note that angle between dipole p and E). In this case it will be 150. had your p been N and N been P it would have been 30. So the ...

1

Short answer: in phase radiates more in the far-field. But read the long answer. Long answer: this actually depends on the frequency and the distance between the sources. For a wavelength significantly larger than the distance between the sources, the pressures emitted by each source simply add up when in-phase; and they cancel when out-of-phase. But what ...

1

I'm not completely clear what you are asking, but isn't this just an instance of elastic scattering from unbound charges? i.e. the dipole oscillates in phase with the electric field of the incoming wave and emits dipole radiation with the same phase. The example I'm familiar with would be Thomson scattering from free electrons, which oscillate like ...

1

The expression of the potential you started with has already assumed the dipole moment was in z direction. However the "correct" result you want to compare with is general for d in any direction. If you take d in z direction, it's the same as your result except the \delta function part, which represents the singular behavior at the exact position of the ...

1

$\def\vE{{\mathbf E}}\def\vd{{\mathbf d}}\def\vr{{\mathbf r}}\def\hr{{\hat r}}\def\hd{{\hat d}}\def\eps{\varepsilon}\def\l{\left}\def\r{\right}\def\htheta{{\hat\theta}}$ EDIT: The expected formula is corrected now in the question. So the following canceled comment is out-of-date. The formula you expect is false (see e.g., this answer or this wikipedia-page). ...

1

Dipole $\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac ...

1

If you take a permanent magnet, and place a sheet of paper over it. Now sprinkle iron filings on it, and you pretty much get this diagram. This has been the mainstay of field theory since Faraday's time. A test charge at rest will begin to move in the direction of the field line. Since there is nowhere that it can rest where there is more than one ...

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