Recent content by PullMeOut

yes it is thin and surface isn't closed. and yes i did take the boundary cond. wrong. so ? i have made some calculation and i am not sure i did it right, you see i am not good at emt calculations. i have found potential and then electric field inside shell for a point, then run some...

i solved it, and no there is no bottom of the hemisphere. i used V(r,\theta)= Al rl Pl (cos\theta) and since i have V(R,\theta) = V0 only term i have is A0 and i find that A0=V0. so Vin=V0 i found. do you think it make sense to have the same potential everywhere ?

ok thank you very much , i will try to solve it with bndry cond. now

ok then i will try to use separation of variables, i know the general solution for sphere shell. it should be like (Al . r^l + Bl. r ^(l+1) ) . Pl(cos) , but can I use this for hemisphere or should i find a special solution for hemisphere.

i have a hemispherical shell with a radius of R ,and which has V=V0 potential on it. the main problem is i have to find the potential inside the shell for every (r,\theta,\phi) points. 1. I think the potential will depent on only the distance from origin, there won't be any \theta or \phi...

are you sure about \gamma and its derivative? there must be something wrong about it

well thank you for your help. now i will give it a try.

if i do that i should calculate the series from zero to infinity. when will i know that i should stop?

Homework Statement This is the integration i have to solve I=\int x^{2}In(1-exp(-ax))dx integration is from zero to infinity The Attempt at a Solution I know that it should be solved with integration by parts so u=In(1-exp(-ax)) du=[a exp(-ax)] / [1-exp(-ax)] dv=x^{2}dx...

you should use the \Psi(x,t) while calculating expectation value it should include t parameter

i think we can't use 10^k, becaouse it should be the same with dx , so it colud be x^k, but we solved the problem.you can check it out if you want and it is really an easy way. but thanks for your time

I have found it 2Bx=y dx2B=dy then put this into the integral. it becomes like \int \frac{4B^{3} y^{3}exp(-y) dy}{16B^{4}} then it equals to \frac{3!}{4B} with the help of the integral ı talk about in the previous post well, i saw your message after i solved it, but thanks anyway...

http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator there is a solution in this link. maybe it could help you solve the problem. energy is shown with this equation for harmonic oscillator. E =\hbar.w(n+(1/2))

I know that \int x^{n} exp(-x) dx =n! i have to turn my integral into this format and solve the integral.so what to do next?

this is the integral , and it's between (0,inf) \int 4B^{3} x.exp(-2Bx).dx how can i solve this? it seems easy but i couldn't figure it out :confused: