Recent content by Bashkir

Ignoring the 2.12! After taking some time back to think, my change in area per strip is dxdy, and I hold DX constant while considering the contribution from it, and integrate dy/r from 0 to xa/b. That gets me pretty close, and then just integrating dx from 0 to x gives me the correct answer.

These are more mathematics than they are problems with understanding of physics. 1. Homework Statement The problems are 2.18 and 2.12 which are as follows, 2.18A hollow circular cylinder, of radius a, and length b, with open ends, has a total charge Q uniformly distributed over its surface...

Thank you very much. That looks perfect.

Thank you for your detailed response! How close in mathematical difficult would you say that Griffith's and Fetter and Walecka are?

I currently have completed a two-semester course in physics where University Physics by Ronald Reece was used. This two-semester course was your basic introduction: the first semester comprised of Newton's laws and their extension to energy, momentum etc., and the final semester was a light...

Are you positive that that is the correct answer? Each section of the ring has some infinitesimal mass that is proportional to some infinitesimal length. Mathematically, dM = \frac{M}{2πa}dl\\ dM = \frac{M}{2πa}adø If this substitution is not obvious to you, all I did was say that the...

Remember that induced currents are caused by a changing magnetic flux. Knowing that, do any of the solutions present a situation where there isn't a change in flux or there is no flux at all?

That is the width of the band gap as I stated in the last post. The fermi level for undoped silicon lies directly inbetween the conduction band and the valence band. I have also provided you with the equation that you need to find the fermi level.

Working the full derivation from Newton's second law we can say, m\frac{dv}{dt} = eE - \frac{mv}{τ} The steady-state drift velocity implies we can write the Ohm's Law, J=nev=\frac{ne^2τ}{m}E=σE If there is no scattering term, Ohm's law is replaced by an accelerative supercurrent...

Well, if we are talking about intrinsic silicon (undoped silicon) then the fermi energy is directly in the center of the band gap. This means that the concentration of holes and electrons is equal. Doping the silicon will move the fermi energy towards the valence band or the conduction band...

Introductory college course can be a little misleading. At my school there are 4 different "introductory" physics classes. There is physics for non-science majors (no math, all concepts), physics for life sciences (algebra based physics), physics for engineers (calculus based applications of...

You are approaching this well, but is easier to why this result comes if you examine it like this. The x-components of the electric field add, so we say that, E_t = E_1 + E_2\\ E_1 = \frac{1}{4\pi\epsilon_o}\frac{Q}{r^2}sin\theta Which is the same as E_2 because both value of sines...

1) Well, when an electron emits and absorbs a photon all it is doing is absorbing and re-emitting energy again. For a "radiationless de-excitation" as they are called, meaning they don't radiate a photon, the energy is just transferred to something else. This is normally done through some...

In addition to what Sunil has stated, there was also an algebra mistake at the end preventing you from getting the correct answer. \cos\theta = \frac{v_x}{v_i} to (cos\theta)v_x = v_i instead of v_i =\frac{v_x}{cos\theta} An easier way to do this problem it to remember that...

It happens by photon-induced chemical damage and covalent modification. As I'm sure you might know, florescence is caused by the absorption of a photon, then the admittance of a florescent photon. Well, pretend we have two electrons in the ground state, one with a spin of +\frac{1}{2} and the...