Recent content by hidemi

Thanks a lot!

The degeneracy would be a total of three because for the energy to be 6*h^2/(8mL^2), there are three possibilities: (2,1,1,), (1,2,1) and (1,1,2). Is it the correct reasoning?

If the energy level is twice the energy of the ground state in three dimensional cubic box, then the energy would be 2*[(nx^2 + ny^2 + nz^2) *h^2 / (8mL^2)] = 2*[(1^2 + 1^2 + 1^2) *h^2 / (8mL^2)] = 6*h^2 / (8mL^2) Is this correct?

Energy of the One-dimensional box: ground state: En = (n^2*h^2) / (8mL^2), where n=1 twice the ground state: 2* En = 2 [(1^2*h^2) / (8mL^2)] Energy of the Three-dimensional box: En = (nx^2 + ny^2 + nz^2) *h^2 / (8mL^2) = 2 (1^2*h^2) / (8mL^2) As stated, twice the ground state energy of one...

Thanks for your remider :)

The correct answer can be obtained by the calculation as attached. However it can not be gotten by the following way. Why? F = -∇U = -[ 16x + 8x^3] = ma Since m = 0.2, a = -80x - 40x^3 V = -40x^2 - 10x^4 +C =5 c= 50 + 5 =55

Thanks for your hints. I have solved it.

According to the equation, the answer is B. Since the lecture didn't cover much about it, can someone explain this formula in a less physics way? Thanks.

Thanks for the hint. I got it!

Thanks for confirming. However, I am still unsure how to answer "What is the maximum electric field magnitude between the cylinders?" As I calculated previously, 700 = λ/(2πε) *1.4 E = λ/(2πεr) = 700/(1.4*r) = 500/(2*10^-3) I use 20*10^-3 meters to plug in 'r' and get the correct answer, but...

Is it correct?

I think I has proved the formula for the electric field as my attachment in post #3(as below): V=∫ (λ/2πε0r)⋅dr = (λ/2πε0r)∫ (1/r)⋅dr = (λ/2πε0r) ln r Could you give me some more hints if it is not you want me to do which I attached in post #3 (as above)?

700 = λ/(2πε) *1.4 E = λ/(2πεr) = 700/(1.4*r) = 500/(2*10^-3) I use 20*10^-3 meters to plug in 'r' and get the correct answer, but I"m not sure why.

I used a couple ways to do this question, but I got neither correct. Can someone help, please? Thank you. 1. E= V/r = 700 / (60*10^-3) = 11667 (very far from the given answer) 2. E = (-kQ/r)⋅ dr = kQ/r^2 = kQ/ [( 1/20/ 10^-3)^2 - (1/80/10^-3)^2] (For this method, I stuck...