4
It the question it is said:
"[it] is easy to see that the flow through a small region in the image
above, not containing the origin, is not equal to zero"
but if you look at the diagram, what you see, for any region not containing the origin, is field lines coming in from one side of the chosen region and going out the other side. And if you ...
2
The limit of the cylinder as its length $\,2L\,$ approaches zero is two disks one on the other but with opposite normal orientations $\,\mathbf n_{\texttt{left}}\,$ and $\,\mathbf n_{\texttt{right}}\,$ as shown in above Figure-01. You must choose exclusively one of the normals to have one oriented surface (disk).
Hint :
In general the flux through an ...
2
One can make a 1D situation by extending all the charges into infinite parallel planes or sheets, just as one can make a 2D situation by replacing the charges with parallel lines. Since everything is the same if one translates along the extended directions, the field does not depend on the added dimension(s).
Yes, the result implies that in an 1D universe ...
2
For what its worth, Gauss's law in 1+1D reads
$$ \frac{dE}{dx}~=~\frac{\rho}{\epsilon_0}\qquad\Leftrightarrow\qquad E(b)-E(a)~=~\frac{\int_a^b \!dx ~\rho}{\epsilon_0}.$$
(The boundary terms live on a zero-sphere $\mathbb{S}^0\cong\{a,b\}$.)
For EM theory in various spacetime dimensions, see e.g. my Phys.SE answers here and here.
1
Electrostatics in less than 3D is a fascinating subject. However, one should be aware that the extension to spaces with dimensionality different than three is not unique.
On the one hand, there is the most obvious choice of introducing a smaller dimensionality as the effect of dealing with special 3D charge distributions that require less than three ...
1
As @Uyttendaele comments(1)
\begin{equation}
\boxed{\:\:
\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\dfrac{\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}}{\:\,\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert^{\bf 3}}\right)\boldsymbol{=}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)\:\:\:\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}
\tag{A-01}\label{...
1
You need to be careful when calculating the divergence in this case. In particular, notice that the function is undefined or infinite at the origin.
If you calculate the divergence in any area not containing the origin you should be able to convince yourself visually that there is an equal number of field lines entering a region as there are exiting.
If you ...
1
you can calculate the flux through the surface of the cube, by finding the electric field then calculating the flux through the face of the cube.
First, calculate the electric field
the electric field for a point charge is
$$\underline{E}=\frac{Q}{(4\pi\epsilon_0)(\sqrt{x^2+y^2+z^2})^3}*(x\widehat{\underline{i}}+y\widehat{\underline{j}}+z\widehat{\underline{...
1
Gauss' law states that the net electric flux through a closed surface equals the enclosed net charge divided by the permittivity of the space. There is no charge within the closed surface formed by $S_1$, $S_2$ and the surface of the cone connecting them, thus the net flux is zero.
The electric flux is not flux density. The electric flux through an area is ...
1
according to inverse square law, the density of flux lines is inversely proportional to the square of the distance from the source. So according to this statement, shouldn't $\Phi_1>\Phi_2$ be the correct answer, since density of flux lines is greater in $S_1$?
The flux is defined as the product of the flux density (a.k.a. the density of flux lines) ...
1
The formula $E=\frac{k Q}{r^2}$ is the electric field of a point charge in radial direction at a distance $r$ from the charge. This means, especially, that the field is not homogeneous, i.e. not the same at every point in space. Compare this to your "the magnitude of the electric field is" and you'll find that we are either missing information or ...
1
The formula $$E=\frac{kQ}{r^2}$$ should be used for point charges or objects with spherical symmetry and uniform charge distribution.
Using it for arbitrarily shaped uniform (or non-uniform) charge distributions will not allways give the right answer since the question as to what value you would use for $r$ will not be clearly defined. And I think that is ...
1
If you apply an electric field E' on a material. Then polarisation happens inside that material. This polarisation will create a new electric field that will be equal to the field caused by $\rho_b \, and \, \sigma_b$(That's why we are able to find those two factors from polarisation only.). Now if you try to measure the electric field practically what you ...
1
Suppose I have a conductor sphere, and a point charge at a distance x away from the center of the sphere.
The sphere has no net charge I assume?
The sphere will be polarized, but what about it field lines of that sphere? Is zero?
The net flux from the sphere will be zero with any increase in one side being associated with a decrease elsewhere. But that ...
1
The fundamental idea of the Gauss law is the connection between charge density $\rho$ and the electric field $E$: the charge content is always specified by the electric field at its boundary. This is the heart of the Gauss law and holds in any dimension. Actually current experiments, called quantum simulators, try to implement the quantum version of ...
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