Recent content by Argelium

It doesn't, but forgive me but I do not understand at all what does that mean for my problem?

If I recall,##\dot q_i## is a constant of motion if ##\frac{\partial L}{\partial q_i}=0## and that is not the case here.

However nothing is a constant of motion in this problem, right?

Yea, I wasn’t very explicit on the original post. But the constraint equation is $$f(x,z)=a-z-bx^2=0$$ Using that and plugging them on the EL equations (of the first kind) we arrive at the last two equations I typed

Homework Statement Consider a particle moving over the curve ##z=a-bx^2## under the force of gravity. If the particle starts from rest at point ##(0,0)## (I'm guessing it means point ##(0,a)##), tell if the particle ever separates from the curve; if yes, find the point at which it does...

So basically what I had (and you yourself, although I have a factor of ##h## which does not appear in yours), but with the caveat that the charge in the heads is not the same charge as in the lateral (which should have been obvious). This yields a not so pretty solution. Anyways, as you said the...

Ok well, I'm on mobile so that's why I wasn't very detailed (and messed up the notation, the current is ##i##, whoops) As for the expression for the current, i don't see why it is necessary since the current is already predetermined by the charge (well ,the surface charge density and the...

Homework Statement A cylinder with radius ##R## and height ##h## which has a distributed charge on its surface with density ##\sigma## spins over its axis with angular velocity ##\omega##. If the cylinder has a mass density ##\rho##, find the relationship between magnetic momentum and angular...

Thanks for all the help! :)

About the first part, are you sure? The polarization as it is is calculated as: $$P = \frac{Q}{4(3R)^2}(\epsilon_0(1-\frac{3R}{r}^2))$$ At ##r=3R## polarisation is 0.

Alright, so talked to the prof and used used @TSny hints and arrived at this solution: The electric field at ##r## is as follows: $$\vec{E}=\frac{Q}{4\pi\epsilon r^2} \hat{r}$$ Using Gauss' Law the displacement at ##r## is $$\vec{D}=\frac{Q}{4\pi r^2}\hat{r}$$ Now, if ##E## is constant, we...

But at ##r=R## there's no bound charge is it? Also, I thought the relation you just proposed is just valid in linear dielectrics which this clearly isn't, otherwise the oermittiperm would be constant

Wow, so I don't know, I'll ask my prof what he meant later in the day, but in the meanwhile ##\vec{D}=\epsilon_0\vec{E}+\vec{P}## So in ##R## and ##3R##, polarisation is zero, and we can solve for ##\epsilon## in both those points (boundary conditions) Then we know that ##\epsilon(r)r^2=c##...

So, what about the other poster who said there could be no constant electric field? Well, I thought ##D## would be constant since ##D=\epsilon_0 E## which is supposed to be constant. I suppose it doesn't hold in this case? Where can I start then?

Makes sense to me, I suppose. Thanks anyway for the help!