16

Yes it is about the partial charges on the atoms but the dipole is a vector not just the charge distribution which is positive in the center and negative on the O's but the vectors cancel


14

The explanation of the force as due to oscillating dipoles is only an approximate description and should be regard only as a guide for students. The actual explanation is that the wavefunctions of the two atoms overlap and the system now needs to be described by a larger wavefunction that includes the electrons in both atoms. In principle this new ...


12

The difficulty might be due to nomenclature : How is polarity defined? In physics the electric field of any distribution of charges can be decomposed into a series of electric multipole moments. The most fundamental is the monopole which means that the resultant electric charge is non-zero. Chemists use the term ionic. If the distribution is overall neutral ...


5

Linear B-A-B molecules cannot be polar. i.e. cannot have a permanent electric dipole for symmetry reasons. Stated plainly, the electronic displacement from A atom towards the two B atoms (or the other way around if A is more electronegative than B atoms) must be symmetric with respect to the central atom, ending up with a molecule without a net electric ...


3

As drawn you have taken $\theta$ to be positive in the anticlockwise direction? The torque is in the clockwise (negative) direction so should be written as $-Fd \,\theta$.


3

There are many things here. $\vec{F}=m\vec{a}$ , therefore in space a force vector would always be larger in length then the acceleration vector right? No! Not neccesarily. That would only happen if $m>1$. It is well known that multiplying by something smaller than 1 (between 0 and 1) is actually redicing the value. In other words: $0.1\times a < ...


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

The "monopole moment" is the magnetic analogue of electric charge: the strength of the source of magnetic field. In nature, objects with a magnetic monopole moment have not been observed, though they are predicted by some Grand Unified Theories. "Magnetic moment" refers to magnetic dipole moment, which is not the same as monopole moment. In general, a ...


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 curl of the polarization is a tricky beast to consider in this situation, because $\mathbf P$ is a discontinuous function, and you need to tread very carefully with its derivatives. To deal effectively with $\nabla\times\mathbf P$ for this problem, the easiest route is to do two things: Use the cylindrical-coordinates version of the curl, \begin{align} ...


3

It is a convention that the dipole moment of a system of one positive charge and one negative charge points in the direction from the negative one to the positive one. Instead of seeing this as being opposite to the direction of the field along the axis between them, see it as indicating the direction the field sprouting out of the “top” of the dipole, past ...


2

There is no tangential force. Both forces due to the central charge on the circumferential charges are exactly radial. So they have no effect on moving the charges on the ring.


2

In the chemistry dialect polar means having a non zero electric dipole moment. That is, if you put your molecule in a uniform electric field, it will orient itself aligning its dipole moment to it. Let's consider an arbitrary molecular structure with no symmetry: it is likely that it will show some dipole moment. But if we stick that to its mirror image in ...


2

In $\text {CO}_2$ the central carbon atom is $sp$ hybridised and hence has a linear shape. Now the $ \text O$ atoms on both side have higher electronegativity and hence the electron density is partially shifted towards $\text O$ atoms. This leads to two separate dipoles of equal magnitude and opposite direction in parts of the atom but the molecule as a ...


2

Nothing. The atomic cloud ( electron distribution) is probabilistic in nature. Untertainty in charge distribution causes polarization


2

You are right, that the medium is made of quantum systems which have to be treated accordingly. When you consider every individual molecule/atom in your material the induced polarization at that individual molecule/atom is given by the induced dipole moment $d(t)$, which is well defined via $$ d(t) = \langle \psi(t) | \hat d | \psi(t) \rangle, $$ where $\...


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

Yes, there should be a measurable magnetic field. Consider the expectation value of the magnetic field between one-electron states (with some given spin and momentum) $\langle e^-|\vec{q}\times\vec{A}(q)|e^-\rangle$. Through the LSZ formula this is related to the vacuum expectation value of the $A$ field and two electron fields. But calculating this vacuum ...


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

I agree that in an external field they will recieve opposite forces and try to seperate but if the distance between the electron and the nuceleus increases, the couloumb interaction between them weakens. How can they settle to a new equlibrium? They are talking about molecules that are not normally polar (zero dipole moment). The electric field induces some ...


1

The electric field inside a uniformly charged sphere increases with $r$. It is a simple exercise in Gauss Law to prove that Electric field inside a uniformly charged solid sphere of charge density $\rho$ is given by $\rho r/3\epsilon_0$. Thus when the positive charge moves with respect to the center of the negative charge, it experiences an increasing force.


1

There is no reason to compare vector magnitude values of vectors of different units. Units are (usually) arbitrarily created by humans. We might have decided that the base unit for force is the "glerf", which is the force required to accelerate the Sun to half light speed in 1 femtosecond. In that case the force vector magnitude on a baseball over one second ...


1

In polar molecules specifically, consider the structure of the atoms making up the molecule. Using H2O as an example, as PhysicsDave said, the electrons of the hydrogen atoms exist in a low-energy, stable configuration in which they are around the oxygen atom more often, which results in the oxygen atom having higher electron density and being more ...


1

In all points of your spherical conductor potential is the same, outer surface included. External potential is only determined by boundary conditions: $V=\rm const.$ on the outer surface, $V=0$ at infinity. You can see that outside field has spherical symmetry. Then its flux being zero implies field is zero everywhere (outside conductor).


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