Recent content by RodolfoM

I tried to find the charge distribution using the given potential but couldn't produce the correct result. Also, Gauss's Law doesn't help, as the electric flux is 0 but we don't have any symmetry. Can someone please shine a light on this? Thanks in advance..

Yeah, I've been thinking about this problem. What do you think about the following solution? There will be a redistribution of the charge on sphere B, let's name ##q_{B_{3}}## the outer charge (radius ##R_{3}##) and ##q_{B_{2}}## the inner charge (radius ##R_{2}##). The inside sphere A will have...

My first attempt at solving this was to calculate the induced charge in A by making ##V=0##: ##\frac{q_{A}}{4πεR_{1}}+\frac{q_{B}}{4πεR_{3}}=0## ##q_{A}=-q_{B}\frac{R_{1}}{R_{3}}## But that's not the answer. Any help is welcome!

It makes sense! I'm still at the bargaining stage but after plugging some numbers I'm heading towards acceptance. The result is really impressive, and it only happens because of the symmetry of the problem. Consider that the left rocket is moving towards right with a speed of ##\beta c##, and...

Shouldn't I obtain some expression such as when ##\beta=1## I'd get ##\Delta t=0##?

In a fixed inertial frame of reference (distant stars), the events would occur at the same place, let's say, at the origin of the reference frame. In this frame, both rockets will have their lenghts contracted: $$L'=L\sqrt{1-\beta^2}$$ As one of the rockets moves to the right with speed ##\beta...

Thank you! But when I use the relative speed, some weird answer shows up. Let's say the rockets are moving not with ##0.8c##, but with ##\beta c##. The relative velocity will be $$v_{rel}=\frac{\beta c+\beta c}{1+ \frac{\beta c\cdot\beta c}{c^2}}=\frac{2\beta c}{1+\beta^2}$$ In this reference...

Firstly, I calculated the relative speed between the two rockets, finding v=1.6c/1.64. Then, I applied the length contraction: the length of the moving rocket will be 0.6L due to this phenomena, so the total distance traveled by one rocket, with respect to the the other, will be 1.6L. Therefore...