Recent content by Tarti

I hope I understand your question correctly that you are trying to find an intuition for Noether's theorem. As you state, I would always fall back to something that you really know, e.g. charge conservation in electrodynamics and how you can derive it from gauge invariance, not directly from the...

Oh come on russ_watters... there are so many stories about nuclear power plants that have minor failures (not to be reported) causing children to be disabled. I also disagree that coal pollutes. There are a lot of technologies available that make it basically clean, more info on wiki...

Re-think your question from the perspective of charge conservation: If a certain amount of charge enters the resistor, it cannot go extinct, also not by collisions to nuclei (if we disregard nuclear reactions for the moment). Of course, the current has a cause: an electric field aka a voltage...

Hey rikinmathur, just think in terms of charges, maybe this is easier for you: A and B are positively charged (in a magnetic sense) and C is negative. So, which wire-pairs will experience an attractive and which a repulsive force?

Dear vst98, I do not understand your question. Nanoantennas can in principle be used to probe their near-field for sensing, spectroscopy etc. But if you have a statistical ensemble of small nanoantennas, you may use the Clausius-Mossotti relation to define an effective refractive index.

I would not use the vector potential here. Just take the formula of a magnetic dipole, shift it by -L/2 and do the same for just another one at +L/2 to find the field by superposition. Really, there is no difference to electric dipoles from the perspective of the fields.

The charge distribution of a pure (mathematical) dipole in terms of the \delta-distribution is given by \rho_D(\mathbf r )=-\mathbf{p}\cdot\nabla\delta(\mathbf r - \mathbf r_D ) , which is not hard to show if you solve Gauss's law for the electrostatic potential - can you do that...

Ehild and QuarkChamber are exactly right! If you connect points of equal potential, you end up with an equivalent current of this form: (image taken from here: resistance cube solution) I guess you can solve the resistance cube now on your own :)

Hi Ipos Manager, TSny was exactly right with her/his hints. It is really of vital importance for you to understand the right hand rule. Once you understand this rule, you will be able to handle most electrostatic problems since you will have understood Ampere's law. You can verify the right...

I would describe it a little bit different. For me, the main point is that the tangential component of the electric field on the plate has to vanish. Otherwise, there would be movement of the electrons at the surface of the metal. So, the metal's surface is an equipotential surface of the...

The situation is very easy if you use assume really an infinitely long wire. Then, the electric field can only depend on the distance to the wire, say ρ in cylindrical coordinates. Gauss's law \int_{\partial V}\mathbf{E}d\mathbf{A}=\frac{1}{\varepsilon_0} \int \rho dV becomes: 2\pi \rho...