# Tag Info

0

The example by CuriousOne is spot on - we know charged particle will accelerate in static external electric field, so generally, Poynting vector based on the external field does not need to be non-zero when charged particle gain kinetic energy. The apparent problem with local energy conservation is caused by using the wrong expression for the energy flux ...

1

Since $\nabla.B=0$ is true as long as we don't find monopoles one can always define a vector potential for magnetic field whereas $\nabla\times B=0$ is sometimes true ($J=0$ and $\frac {\partial E}{\partial t} = 0$ in the region of inerest) we can sometimes define scalar potential for the magnetic field. If you go by ampere's law you will find that ...

2

This is a more general answer. Atoms and molecules are quantum mechanical entities. This means that the "shape" of atoms depends on the solution of quantum mechanical equations, which give probabilities for locating in space the electrons that are bound to the nucleus of the atom with the electric potential provided by the protons of the nucleus. The ...

5

The water molecule is neutral on overall basis, i.e: the water molecule as a whole has no net charge. The water molecule is not linear rather it has a bent shape with two hydrogens on the same side. This happens because of the lone-pair-bond-pair repulsions. The oxygen has is a more electronegative element than hydrogen, i.e: oxygen has high electron-...

1

For simplicity I treat this as a two-dimensional problem. The diagram below shows a dipole of length $l$ and a Gaussian surface of radius R. At point $A$ the dot product $\vec E_{R1}\cdot \vec R$ is positive and at point $B$ the dot product $\vec E_{R2}\cdot \vec R$ is negative where $\vec E_{R1}$ and $\vec E_{R2}$ are the net electric fields at points $A$...

0

E=k*p/z^3. This is because usually, the Electric field is calculated at a point whose distance from the midpoint of the dipole(z) is very large as compared to distance between the two opposite charges(say a). That is z>>a. Thus, the original value of dipole at a point on the equatorial line=kp/(z^2+a^2)^3/2. as z^2+a^2=z^2.(z>>a). Thus, E=kp/z^3.

2

A dipole may be represented by a pair of nearby charges of the opposite sign (but the same absolute values), $+Q$ and $-Q$ at a distance $\vec r$ from each other. At the distance $R$ from the dipole (and I will only consider the case when we measure the field on the axis that includes the dipole itself), one adds $E$ going like $+1/R^2$ from $+Q$, and \$-1/(...

Top 50 recent answers are included