Recent content by Biest

But how do i take the \hbar, -\hbar 0 condition into account?

Hi, I have a quick homework problem because I am confused. So a wave function with l = 1 in the state: |\psi> = \frac{1}{\sqrt{14}} \[ \left( \begin{array}{ccc} 1 \\ 2 \\ 3i \end{array} \right)\] and i have to find the probability to be in state \hbar, -\hbar, 0 in L_z , so i...

Hi, So for an harmonic oscillator we need to to find the average value for x^4, so <n|x^4|n> . We split it up to \sum_m |<n|x^2|n>|^2 and recognize that only m = n+2, m=n and m = n-2 can be used. We find that m=n \frac{\hbar}{2m\omega}<n|\hat{A}\hat{A^\dagger}|n> m= n+2...

That there are multiple possible positions for the image charges in a sphere. they are conserved even though they are positioned differently. You also should be be so strict on the conservation of charges when it comes to images charges as the actual position of charges is completely different...

Look at Example 3.2 and Exercise 3.8 they show the different options

Do consider that he said PLANE not sphere... there is a difference

rethink where you set V = 0 and consider example 3.8

You are right, but that question asks for more

Just find Q total in terms of V0 and then use that to find a ration between V and V0, you will see that it is not insane, but simply a limit case

Hi, So I have a lab question for which I have to drive a formula to find the optical cavity length of a laser, L . I have results from testing the fringe patterns when I move one of the mirrors in a Michelson interferometer. It shows the typical \frac{\sin x}{x} pattern. To find the cavity...

read the question again. you don't even have to know the areas... it is simply asking for a ratio between the two

Kittel is notorious to be hard... like jackson in E&M the first chapter in that one already threw me off when trying to use it as a reference recommended by my prof... will string theorists ever learn the fact that there are mortals around them... You might want to try...

The boundary condition is changed this will change the outcome when calculating the field outside

Now will you calm down... I have other things to do as well... It clearly shows what the x and y coordinates are... cos t = x and sin t = y. so if you want sin t = -.5 you might want to look in that area.

https://www.physicsforums.com/showpost.php?p=1613771&postcount=10