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17

Especially the hydrogen atom, with a proton in the nucleus and an electron revolving acting as a dipole This is a problematic way of understanding the hydrogen atom ─ it basically tries to insist on treating it within classical mechanics, and this is doomed to fail. Instead, the hydrogen atom must be treated within quantum mechanics. This introduces a bunch ...


4

Trivially, you could just have the field be uniform in a finite region around the dipole and not uniform elsewhere, so that the electric field as a whole technically isn't uniform, but this might not be the spirit of the question you're asking. Fortunately, you can just as easily construct situations in which: the electric field is non-uniform and smooth, ...


4

In a few words, it's because the vector potential is a polar vector, so it can't depend on the right hand rule. But if you give it a $\varphi$ component you have to pick one direction or the other, and the right hand rule is the only thing that can give you a preferred direction. More explicitly, being a polar vector means that it changes sign under ...


4

Flux is a measure of the electric field through a given surface. The field passes both ways, at different places, through the enclosing surface such that the total flux cancels out, precisely because the enclosed charge is zero. It does not follow that the field is zero at any given point.


3

Coherent versus incoherent is a red herring here. There isn't any fundamental difference between gravitational and electromagnetic waves, either. The facts are the following: For both coherent and incoherent waves, whether electromagnetic or gravitational, the typical intensity falls off as $1/r^2$, and the typical amplitude falls off as $1/r$. For a white ...


3

The quantity $\phi$ is his smooth test function. The dipole distribution itself is $T=-\delta'(x)$ with $$ \langle T|\phi\rangle\equiv -\int \delta'(x) \phi(x) \,dx= \int \delta(x) \phi'(x)\,dx= \phi'(x). $$ The integration by parts is really the definition of the derivative of $\delta(x)$ rather than a real integration by parts. Does not Schwartz ...


3

I prefer to think of dipoles as irreducible representations of the SO(3) group. Now you could have such representations over different fields, e.g. it could be electromagnetic field, it could be current density, it could be gravitational potential etc. For me, the most familiar one is the dipoles that occur in electromagnetism. So let us think about the ...


3

It looks like you are just missing the negative sign in front of the potential energy expression. Let me explain. Let us imagine the dipole to be at some angle in space with an electric field pointing horizontally to the right for simplicity. In this scenario, the torque $\vec\tau$ = $\vec r_1$ $\times$ $q\vec E$ + $\vec r_2$ $\times$ $-q\vec E$ where $\...


3

I don't know how you would orientate the x-y plane, however, the usual way would be to choose the direction of the dipole, i. e. the connection line between both charges, along the z-axis. A reasonable x-y plane with respect to the dipole axis =z-axis would be the symmetry plane perpendicular to the z-axis and symmetric to the charges so that the distance of ...


3

Dipole moment $p$ is defined to be $$p=\sum_{i} q_i\vec{r_i}$$ where $q_i$ is the charge, and $r_i$ are the charge and position vector measured of the ith particle measured from the origin. So dipole moment is a perfectly well-defined quantity even if there is only one charge. If you happen to choose the origin of your coordinate to be where the one charge ...


3

What would happen if the shield or the door was removed? The microwave oven would not work. The door is provided with interlock switches. They prevent the oven from operating with the door opened. So the same would be true if the door is removed. If both of the interlocks fails (or are defeated), I believe ovens are also provided with a back up system, ...


3

Recall that $\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}dV' $ hence for current density that is parallel with $\hat z$ the vector potential can have only one component and be parallel with the same, there cannot be a $\phi$ component. Furthermore, even if that were not the case, i.e., $\mathbf{J} \ne ...


2

In (3), the second step is obtained via integration by parts. Using the identity $\vec{\nabla}( f \vec{v}) = \vec{v} \cdot \vec{\nabla} f + f (\vec{\nabla} \cdot \vec{v})$, we have $$ \int_\mathcal{V} f (\vec{\nabla} \cdot \vec{v}) d^3 y = \int_\mathcal{V} \vec{\nabla}( f \vec{v}) d^3 y - \int_\mathcal{V} f (\vec{\nabla} \cdot \vec{v}) d^3 y = \int_\mathcal{...


2

No, it could not. The strong interaction has a different set of symmetries than the electric dipole moment. Specifically, the strong interaction is invariant under inversion of space, under reversal of time, and under conjugation of charge. These symmetries are known from observing which types of nuclear transitions are allowed or forbidden in strongly-...


2

Nonconductive materials like wood can get polarized by an electric field even though they don't conduct a current. Molecules within it can be polarized by an electric field even if they aren't willing to give up their electrons, and when molecules can line up with the field, they will be at their lowest possible energy. In the case of a speck of wood, it ...


2

The electric field of a dipole can be written in spherical coordinates (using your unit system) as $$\vec{E} =\frac{1}{4\pi r^3}\left( 2p\cos \theta\ \hat{r} + p\sin \theta\ \hat{\theta}\right)$$ $$ E^2 =\frac{p^2}{16\pi^2 r^6}\left(1 + 3\cos^2 \theta \right)$$ Integrating over a spherical volume (as you propose) from an inner radius $r_1$ to an outer ...


2

This has been done by many people. It is called the ac-Stark shift, and maybe you are able find programs in the internet, which are already for Lithium. Check out the phd theses in the ultra-cold atom community. I have done it for an atom which possesses only a $(LS)J$ coupling. In your case, you will have to adapt the methods, which will be painful -- I ...


2

No, this is an incorrect explanation. Even a moving charge in vacuum produces a magnetic field.


2

The two nontrivial things you need are $$\nabla r=\frac{\vec{r}}{r}$$ and $$\nabla(\vec{a}\cdot\vec{r})=\vec{a}.$$ You can show these using $$r=(x^2+y^2+z^2)^{1/2},$$ $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z},$$ and $$\nabla=\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}+\hat{z}\frac{\partial}{\partial z}.$$ The rest is just using ...


2

Dipole consists of two charges - one positive charge and one negative charge. Those two charges are not at the same point in space. That means that they give some nonzero electric field in total. You can apply Gauss law, but Gauss law tells you how much flux goes through your enclosed surface. You cannot always see the magnitude of electric field just ...


2

The interaction of atoms in a ferromagnet is typically modeled as an exchange interaction. The exchange interaction is related to the Pauli exclusion principle, which prohibits electron clouds of neighboring atoms from overlapping when the unpaired electron spins are aligned. This prohibition decreases the electrostatic potential energy of the spin-aligned ...


1

Gauss’ Law is a statement about electric flux. A nonzero field can have zero flux.


1

A magnetic field really is a consequence of relativity. I don't think you can explain the origin of magnetic field without at least knowing something about length contraction in a moving frame of reference.


1

Non metallic materials can consist of molecules that are polar in nature forming dipoles, or that can have dipoles induced by an external field. This is why insulators called dielectrics, which are polar in nature, are used in capacitors. When an electric field is applied between the capacitor plates, the dipoles line up with the field. In a capacitor, the ...


1

Normally, we assume the dipole just exists. By “energy of the dipole”, we’re not talking about the energy to assemble the dipole (the self energy), but rather the energy to orient the dipole in the external field.


1

The dipole is symmetric about its axis, hence the azimuthal symmetry. If the dipole is oriented along $\hat z$ (as it done usually) and one uses cylindrical coordinates with $\hat z$ as the axis of the cylinder, it is clear geometrically that points at constant height $z$ and constant $\rho=\sqrt{x^2+y^2}$, i.e. point in a plane perpendicular to the axis ...


1

Since the "mean positions" of the positive and negative charge are different, the magnitude of the force they exert on the other molecule is not the same. Hence they don't cancel out and the other molecule is acted upon by a net non-zero force. In the water molecule, the electronegativity of the $O$ atom and the $H$ atoms differ, oxygen being more ...


1

The dipole moment is a vector, and hence the magnitude of the total dipole moment will be the square root of the sum of squares of its components.


1

Yup faced the same confusion but see it is very logical and mathematical while it seems totally bizarre ; let me try to explain in a concise manner. let's consider a one dimensional case and for simplicity assume the dipole to be placed along the axis of variation of the field. Then the the force on the charges ( i.e. , $qE$ ) can be written as $ F = qE(x)...


1

You are correct in that the dipole field is present but the constant field dominates. $$ \lim_{r \to \infty}r = \infty $$ $$ \lim_{r \to \infty}\frac{1}{r^2} = 0 \quad \text{(or tends to zero)} $$ When matching conditions for large $r$ we need to match the dominant term. In a mode expansion approach to the potential (which seems to be what you are ...


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