# Tag Info

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: 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

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 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

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 ...

3

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 ...

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

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

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 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, $$\nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + ... 2 The Q_x and Q_y transitions are electronic excitations in the conjugated \pi orbitals of the Bchl a molecule. They involve two different sets of conjugated bonds. The Q_x involves a shorter chain of conjugated bonds so it occurs at a higher energy/frequency. I couldn't find a really good diagram to show which bonds are involved in the in the Q_x ... 2 The answer lies in the "polarizability" of the sphere. This relates the external field to the induced dipole moment. For a (ridiculously) rigorous treatment, a good book is "The Scattering of Light by Small Particles" by Craig Bohren. However, if you're looking for a simple result, the polarizability and the dipole are related like this: p = \alpha E ... 2 The dipole moment of a system of charges q_i located at positions \mathbf r_i is defined as the vector$$\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

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

You can find the answer by considering the multipole expansion of a charge distribution. The electric potential of any localized charge distribution can be expanded in terms of the distance (or inverse of it) from the origin to the field location in a multipole expansion. In this expansion, what you are actually doing is writing the potential of the charge ...

1

As mentioned in the answer by @Manishearth the potential energy of the dipole is not considered in simple considerations of the dipole. However, such considerations can't be neglected in the study of wave propagation and specific heat of the substance. You might remember even from the elementary study of specific heats in high school that at high ...

1

When we talk of a dipole, we usually assume that its configuration is fixed or the changes to the configuration are negligible. So the effect of an external field on changing the length between a pair of charges is not considered when we are taking them to be a dipole (usually we consider the charges to be rigidly bound).

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

The dipole transition matrix elements $\mathbf{d}_m$ and $\mathbf{d}_n$ are complex-valued vectors which are relatively easy to define. Their "direction" is a mathematical convenience, and it is essentially given by the vector divided by its modulus, for an appropriate interpretation of the latter. Consider first the case of a single molecule, with a ...

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

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

I had learnt that the dipole moment is defined for 2 point charges only with equal magnitudes but opposite character. Actually, that is not the case. You can calculate a dipole moment for any charge distribution. In fact, the dipole moment is just one of a whole series of multipole moments which can be used to describe a charge distribution. The ...

1

I’m not an expert in electromagnetism, but if one of my students (general physics, BSc level) asked, I'd say the following: the rotating proton indeed generates an magnetic field, and it behaves like a magnetic dipole for these purposes (and at large enough distances). However, that only concerns the field created by this proton, and its interaction of this ...

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