Recent content by Avalanche_

Oh, sorry, I didn't see that circles on top of your picture. And about that last part, you already did that. If P is in between wires [ a) part], you calculated the magnetic field, and just look at your circles in between the wires, both pointing in the same direction, so that got to be the...

Everything is correct in a) part. About b) part, that's true that each wire generates magnetic flux in opposite direction so it will cancel, but you got to calculate that just like in a) part. B coming from (lets say) S wire will have + sign and B from R wire will have - sign (depends on...

Hi. I'll start from the beginning. If u have a density operator, call it $$\hat{D},$$ and some eigenstate, call it $$|\psi>.$$ You can simply make a projector to that eigenstate: $$ \hat{P}= | \psi> <\psi|.$$Then the probability of finding your system in that eigenstate is defined as $$ Tr(...

OK, i don't really get your notation, use latex next time :D . I'll just write how you compute this and I hope it will be clear for you. So, if we define that the potential at infinity is zero, then : V(r)-V(\infty)=V(r)=- \int_{\infty}^{r} \vec{E} d\vec{l} , by the definition. Here d\vec{l} =...

True, flux is integral of dot product of electric field and \vec{da}, that's an infinitesimal element of a surface. But if you change shape of your surface, let's say you move some part of surface away from charges, the flux through the part of your surface that you didn't touch stays the same...

Look at that problem from mathematical point of view. Let's see some more complicated example, when you are integrating some vector field along a line that is not straight. \int \vec{E} d\vec{l} . Now you can't integrate then do a scalar product, because you are integrating a component of that...

You can see that just some of the free electrons will come to the surface and not all of them from Gauss' law: \nabla E=\frac{\rho}{\epsilon} . We know that electric field is zero in a perfect conductor, so the right hand side of that equation also has to be zero, hence, \rho = 0 . Now, if all...

Simple explanation: in electrodynamics when you want to write (for example) charge density function for point charge in the origin. You would write that as ρ = q \delta (x) \delta (y) \delta (z) . You will write it in that way so that ∫ \rho dV over the whole space gives you the result q...