Recent content by Solibelus

Hi, I'm trying to derive the displacement of an object with proper acceleration (constant) α. t is our laboratory time. The goal: x(t) = {1 \over α} [ \sqrt{1+α^2t^2} -1 ] How to approach this? :( Note for anyone who's wondering: no, this is not homework...

Well disagreeing with you might well be a big mistake. If there's an error in my way of thinking, I'd be more than glad to hear your correction, unless you've given up on trying to explain this topic to me! :)

Thanks for the help so far! I'm afraid I'll have to leave this hanging until Sunday since I've got another test in a completely different subject. However my plan now that I know Q is to approximate E using dipole expansion in the reference frame where the bar is moving and then apply Lorentz...

Sorry for being away, had an exam yesterday and some family events. I do care, because if the galaxy is receding from Earth at close to v=0.99c, length and time differs whether I'm measuring from Earth or from that galaxy. Now, I'm standing on Earth and we have that receding galaxy. If the...

According to the transformations, E should grow by factor γ, but in addition it will have the 0.99γBsinβ component. B will of course also depend on that 0.99... Gotta say that I haven't plugged in any numbers to actually get a feeling for the significance here.

Alright, I see, so I must have made a mistake there with the signs, because it does make sense why I have two identical results for E. The signs just need to be opposite I guess... Now for Q, if I understood you correctly, I just need to use Coulomb's law? If so, I get this: Q = {v \over...

I'm not that easily shocked... OK, so I think I understand your reasoning. Let me try again! Now according to what you're saying, ∇ x E = -∂B/∂t + ∇ x (v x B) B is constant and I wish to work in CGS (in hope that I'm doing it correctly). So, applying stokes theorem: \oint \vec E.d \vec l =...

That "standard formula" - is it not simply Faraday's law after applying Stoke's theorem? Does it even apply here, given that I don't have a closed circuit? What's the ∫dl loop here? If this is not what you had in mind, then I don't know of any "Blv law"... it's just that during the course I took...

So I'm guessing that either the cross-section or the density are missing from the original question. I think I'll just introduce this piece of data and solve using dipole approximation as usual. Thanks!

There is no cross section given though. I don't think it'd be correct to just assume it unless it cancels out somewhere during the actual calculations. I guess I can derive the charge density from Gauss' law, but E's gradient is zero, is it not? Maybe the E field outside for some reason exists...

Homework Statement We're working in the right-handed Cartesian coordinate system. Unit system is CGS. A conducting bar of length L is placed along the x axis. Center of mass at x=0 when t=0. It's moving with constant velocity V in the +y direction. There's a uniform magnetic field B, such that...

Look into Torricelli's law for the velocity. The volume discharged is just dV/dt. What exactly confuses you here? If it's a really big stand pipe, you can maybe ignore the rate at which the water decreases when it flows out of the tiny hole :) The first formula you posted there is...

I think that it is indeed correct, and you can also imagine the field lines. The symmetry will break due to the charge distribution. However, finding the actual charge distribution in case 2 is rather difficult I reckon... If it was a conducting shell, then it'd act as Faraday's cage, where...

Yeah but my original claim is that things aren't so simple. According to the guys who posted above, they indeed aren't. Meaning that B is really constant only if I ignore higher orders as WannabeNewton explained. Maybe I misunderstood you?

Yes, your first post indeed provided a good explanation and this post strengthens it. Thank you. My job will be to do some proper reading and exercises, in addition to examining the information given by UltrafastPED and vanhees71. It seems as though our Professor did not provided us with the...