New answers tagged electric-field
0
Ok, let's consider $\vec{r} \vec{dr}$, it is equal to $|\vec{r}||\vec{dr}|_\vec{r}$ where $|\vec{dr}|_\vec{r}$ is projection of $\vec{dr}$ on $\vec{r}$. If you draw $\vec{r}$ and a small (remember, you need infinitesimal!) $\vec{dr}$ you will notice that this projection is actual equal to $|\vec{dr}|_\vec{r} = d|\vec{r}|$, so $\vec{r} \vec{dr} = ...
0
Perhaps I did not understand the question correctly, but it seems to me that you cannot use a Gaussian shell in this case, because the intensity of the field $E$ would be different at different points of the shell. If you want the following expression to hold,
$$
\int E\cdot da = E \int da
$$
then $E$ must be equal to the same value all over the Gaussian ...
1
So I understand the electromagnetic spectrum -- electromagnetic
radiation is mediated by photons
Briefly, electromagnetic radiation is due to real (observable) photons; electric and magnetic force are due to virtual photon exchange.
The macroscopic electromagnetic wave phenomena we observe are due to an almost unimaginable number of photons, ...
6
Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates.
First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article:
$$
x=a\cosh\mu\cos\nu\cos\phi\\
y=a\cosh\mu\cos\nu\sin\phi\\
z=a\sinh\mu\sin\nu
$$
where ...
5
The exact solution is $${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$
Clearly the field inside the conductor (that is, for $r<R$) vanishes. Here $Q$ is the total charge on the disk. The field, for large values of $r$, looks ...
2
I'll suppose your disk (radius $R$) is uniformly charged with surfacic charge density $\sigma$ (your disk being conductor, any charged Q deposed on it will be distributed all over it so that $Q=\sigma\times\pi R^2$). The field outside the disk is necessarily radial so we can place the point $M$ where we would like to compute the field on the $x$ axis so that ...
0
If you are off by a factor of two, it's probably because the volume of a sphere is
$\frac{4}{3}\pi r^3$ and not $\frac{2}{3}\pi r^3$
0
You may recall that one way to solve the problem
A mass $m$ passes the table top moving upwards with initial velocity $v$ (in standard gravity). What is the maximum height above the table reached by mass?
is to use energy conservation: you equate the maximum gravitational potential energy relative the table top with the initial kinetic energy of mass. ...
1
Outside a current carrying conductor, there is, in fact, an electric field. This is discussed for example, in "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson, in American Journal of Physics -- July 1996 -- Volume 64, Issue 7, pp. 855 . To quote Norris W. Preyer quoting Jackson, "Jackson describes the three roles of surface ...
0
The direction of the electric force on a charged particle is constant, independent of the direction the charge is moving in. The magnetic force on the other hand is always at right angles to the direction the charge is moving in.
4
Magnetic field never does any work , as it acts perpendicular to the velocity and thus doesn't change Kinetic energy , however electric field is able to change the direction as well as magnitude of velocity.
Hence answer is electric field.
0
Your polarized sphere is spherically symmetric. Therefore outside the sphere, in the vaccuum, the solution must be spherically symmetric. However, the only spherically symmetric solution of Maxwell's Equations in vacuum is the electric field of a point charge. Since your sphere has zero net charge, it must be the electric field of a point with zero charge. ...
1
That equation is true for electrostatics. Inside an electrically neutral current-carrying wire, the electric parallel to the wire is zero. So outside the wire it's also zero.
More importantly, Gauss's law will tell you that the components perpendicular to the wire must also be zero.
So the electric field is zero everywhere for an electrically neutral ...
Top 50 recent answers are included
