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## Hot answers tagged dipole

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 ...

5

I don't think you need quantum mechanics to understand what's going on in dipole-induced dipole interaction. The basic mechanism is quite simple and just the details of the calculations change by switching to a quantum description. Polarizable molecule in an external field So first things first. Let us consider a simple model of polarizable molecule as ...

4

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} + ... 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 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 ... 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 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 Very simply, the field of the positive and negative elements of the dipole "almost" cancel out - but not quite. It is because they are some small distance away that there is a residual (third order) term. You can see this by taking two charges +q and -q at a distance 2d, and look at the field a distance r from the center of the two (on the same ... 3 The electric dipole moment is defined asp = \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 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 ... 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) ... 2 Your expression "the angular momentum is m_j \hbar" (where \hbar = h/2 \pi) is incorrect. This quantity is the projection of the angular momentum on the z-axis; it represents the direction of the spin. This is why it corresponds to the m quantum number, not the \ell quantum number in the | n \ell m \rangle basis. Which angular momentum to use ... 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 Born approximation is when you technically have an extended body but you ignore the scattering from the object itself. An example would be if you had material that was almost transparent, imagine some stained glass that is only ever so super lightly stained. You can basically treat each part as if it saw the normal unchanged light, technically the ... 2 The dipole has its least potential energy when it is in equilibrium orientation, which is when its momentum is lined up with Electric field (then \tau = 0) It has greater potential energy in all other orientations. We are free to define the zero potential energy configuration in a perfectly arbitrary way , because only difference in potential energy ... 2 The binomial expansion says that (1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + .... This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret {n \choose k} correctly for negative numbers; for our purposes, we just need to know {n\choose 1}=n always. For very ... 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 Normally the transition amplitude is calculated with e^{i\vec{k}\vec{r}}. For "small" product \vec{k}\vec{r} one expands the exponential in Taylor series and one leaves only the "dipole" term in the transition amplitude calculation: \propto\vec{k}\cdot\langle\psi_1|\vec{r}|\psi_2\rangle . So it is a "dipole" part of the transition amplitude, which in ... 2 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 ... 2 Hint: Interaction energy of two dipoles :U=\frac{1}{4\pi \epsilon_0r^3}\left( \mathbf{p}_1.\mathbf{p}_2-3\left ( \mathbf{p}_1.\hat r )(\mathbf{p}_2.\hat r\right) \right)$$2 Nope. The electron dipole moment is defined as its intrinsic property – a property that may be imagine as a consequence of the particle's internal structure or, more generally, "something that happens inside it". So you may imagine that the relevant wave function for the "center of mass" is \delta^{(3)}(\vec r). It's not literal because one can't localize ... 2 The position dependence comes in via the R in the following expression¹:$$ \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 ...

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