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...
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
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 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...
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)?
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...