Recent content by Terrycho

I found this explanation, The contact potential is the difference between the work functions of the cathode and anote, since they are oppositely directed in the electric field, that is, the electric field has to work against the cathode potential but is helped in the case of the anode. Thus we...

It does look like the first peak is at 4.9V. However, when I did the experiment, the first peak was not 4.9V. It was larger than that. It seems like this plot also has the first peak is larger than 4.9V. https://foothill.edu/psme/marasco/4dlabs/4dlab8.html

In the experiment, I know that the spacing between successive valleys gives the excitation energy to be somewhere around 4.9eV. However, when you look at the plot, you can see that the spacing from zero to the first peak is much longer than any other spacings between two successive peaks. I was...

That was fast and very clear. Haha Thanks!

Hello! I have been recently studying Quantum mechanics alone and I've just got this question. If the potential function V(x) is an even function, then the time-independent wave function can always be taken to be either even or odd. However, I found one case that this theorem is not applied...

Thanks! I guess I was considering some useless stuff haha

So to let A_θ(r,θ,ϕ)=A_ϕ(r,θ,ϕ)=0, does a field have to be spherically symmetric ,or as you said, you can assign any function?

Thank you! So a vector field can be defined as either $$\vec A(r, \theta, \phi) = \vec r = r \hat r$$ or $$\vec A(r, \theta, \phi) = A_r(r, \theta, \phi) \hat r + A_{\theta}(r, \theta, \phi) \hat \theta + A_{\phi}(r, \theta, \phi) \hat \phi $$. Is that right? Also then, from my understanding...

I know the divergence of any position vectors in spherical coordinates is just simply 3, which represents their dimension. But there's a little thing that confuses me. The vector field of A is written as follows, , and the divergence of a vector field A in spherical coordinates are written as...

So with that boundary condition I specified, F(t)=2t meets all the characteristic curves, which are the hyperbole, right? Then how do you express F(t)=2t? Is it like y=2t, which is parallel to x-axis for each constant t?

I mean, this is another question of mine too. φ(x,y)=F(t)=F(x^2/2 - y^2/2 ), this is the general solution of the partial differential equation irrespective of the choice of the function F. Then, I set the boundary condition as φ(2y,-y)=3y^2. While doing this, I got F(t)=2t, where t=(3y^2)/2...

what I heard was the boundary conditions should meet all the characteristic curves only once. Is this one wrong?

Consider the following linear first-order PDE, Find the solution φ(x,y) by choosing a suitable boundary condition for the case f(x,y)=y and g(x,y)=x. --------------------------------------------------------------------------- The equation above is the PDE I have to solve and I denoted the...

Everything is okay besides the current gain of this circuit, which is "g" there The formula of current gain is I_out/I_in If I call the current flows to R_e I_2, according to KCL, the following formula must be satisfied. I_in + (β+1)I_b = I_2 Therefore, to know I_in, we should use the...

Sorry for all the confusing arrows... I tried to express those symbols as vectors, so I put the arrows on the upper line with lots of space but they just ignored the space and put all the arrows together... If you let me know how to compute vectors, I will edit the post right away! Thanks!