Recent content by Blanchdog

How can I tell the handedness?

I believe I figured it out, though I would love confirmation. Since ##cos(\delta) = cos(\frac{\pi}{2})=0## and ## A = B##, we end up with ## E_\alpha = E_{\alpha_\pm+\frac{pi}{2}}##. That means we have circularly polarized light! So of course ##\alpha## is undefined; a circle has no determined axes!

Case 1 worked out great, I found it to be linearly polarized light at an angle ##\alpha = \frac{\pi}{4}##, but Case 2 is giving me trouble. As best I can tell, ##\alpha## is undefined in case 2. How do I solve case 2?

Ah I see the issue, I have a typo. The expression I simplified it to should be $$B^2-4AC=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$ This expression is invariate under a cyclic permutation, and I can show 2 of the positive...

I'm not sure what a cyclic permutation of indices is, but I pretty much figured that my simplification had a mistake. Any chance you can help me find it?

I got as far as simplifying the expression to $$\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$ But that doesn't seem to be a form that is necessarily positive and satisfies the criteria of the homework statement. Little help...

Electric fields are always continuous, so this cannot be correct. However, if you were to bend the ends of those two lines together at the origin that would pretty much be it. Near the disk the point charge is affected by the disk's presence, it's only further away from the disk that the...

It isn't, its asymptotic. Essentially as you move further away from the missing disk, the electric field looks more and more just like the field from the charged plane, which is constant.

Sorry I deleted that because of a mistake, the fraction goes to 1 rather than zero. See post #18.

You are varying x by some amount, but we know that it will be perfectly mirrored on either side of zero so let's set zero as the lower bound for a plot of what's going on. As mentioned before, E(0) = 0 so that's an easy starting point for your graph. Next, what happens as x becomes large...

That looks right to me, ignoring the typo on the square root. It would not be a straight line however, as you can see by x appearing in the denominator. Instead, look for something that begins to look like the electric field of the infinite plane as x become large.

You're right, you're right, I misread (x, 0, 0) as being a point in the plane, not on the x axis. So to amend my earlier statements, at the point (0, 0, 0) the E field and force are both zero, but as the point charge is moved along the x-axis the E field and force acting on it become non zero.

Thanks, that helps a lot with understanding the math of what is going on. I've never encountered a geometric mean in relation to a 'matching' operation though, could you help me understand what the math is describing? As an aside: For two layers of AR coating you would want the second layer to...

At (x, 0, 0), the force ## F = qE## is indeed zero because ##E = 0## at that point. As one varies the value of x away from zero however, the E field (and thus the force) become non-zero.

In part a we are evaluating the electric field at a point with perfectly symmetric amounts of charge in each direction. So the E field is just zero.