If I assume orthonormality and the other 3 hybrids have some coefficients as negative, then am I right to conclude that the coefficients for this one are all +1/2?
No this is the whole problem. I keep thinking it must be entirely simple but I guess it's not. (My teacher has a way of omission to the point of ridiculous)
Assuming the 2s and 2p wavefunctions are normalized, determine the coefficients in the hybrid orbital:
Ψ(sp3) = aΨ(2s) + aΨ(2px) + aΨ(2py) + aΨ(2pz) (the other 3 hybrids have – signs for some of the coefficients.
I have no clue where to start. I know this is a tetrahedral hybrid orbital but...
So in order to cancel the uniform field exactly you would need the charge distribution to be even throughout the whole cube even the top and bottom.
Is the cube simply polarized because of the external field? Or is it conducting a charge?
I am envisioning something like the right image if the...
Problem statement, equations, and work done:
A perfectly conducting cube is placed in a uniform electric field in the x direction (see attached).
Step1 : Use Gauss's law to determine the electric field inside the cube.##\phi_E = 2E(r)A = \frac{q}{\epsilon_0} →...
That is what I have been thinking all along. But it sounds like some of these other people see something that I don't. 1/2 V seems like a logical by symmetry but can't R1 be arbitrarily anything here?
OK I think I get it now. Yea it seems misleading that the virtual image is the object for the second lens because the rays never make it there. I guess I was picturing something different as if rays eminated from the virtual image, which of course is impossible. That is why I was struggling with...