Recent content by Samuelriesterer

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. [SIZE=16px]I have no clue where to start. I know this is a tetrahedral hybrid...

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...

I'm not sure how to draw a field inside the cube if the net field inside is zero?

[SIZE=16px] Would this be: ##\sigma = E \epsilon_0##

So the field inside the cube would be: ##ϕ_E=2E(r)A=q/\epsilon_0 = 0 ## But what about the induced charge density? What is inducing it?

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} →...

More specifically, what happens and why when a magnet is brought close to a nonferrous charged object.

Just an update to those who were wondering. I asked my teacher and he said to just pick a value for R1 because yes, there are infinite possibilities.

[SIZE=16px] 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?

[SIZE=16px] What does this mean because a converging lens can still form a virtual object if the object is within the focal length.

Well the logical choice would be 1/2 V?

OK so: ##V_s (\frac{R_s}{R_s +R_1} - \frac{R_1}{R_a +R_1})=0## Then solve for R1?

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...