Recent content by askcr9

My understanding has improved tremendously on this subject! I know the answer to part i and understand how to get it correctly and I'm pretty sure I can do part ii by myself now. It has been a privilege to receive help from such knowledgeable and experienced users on this website. Thanks for...

Alright alright. Things are making a little bit more sense now. A little bit. I have just a couple of questions relating to all of this. 1. So a Gaussian surface can never be infinite? I guess that'd make sense, right, on a purely common sensical level. But then again, to me, it makes zero...

@gneill Ok So then, back to part i, when I sum up all the faces on my Gaussian slab, I end up getting a 2 in the denominator of my fraction, so I get ρd/2ϵ0. Is that right or am I still wrong? (I mean, I'm guessing it's wrong, I think that's what Orodruin was saying in the post above yours...

The region inside the slab? Um, I'm sorry, I don't think I quite understand. I'm solving for the area outside the slab (in part i, at least). Why would I need to worry about what's inside the slab? All I would need to do is just sum up the areas of all of the faces of my Gaussian slab, right...

Oh yeah, it's like charge density / 2ϵ0. I had heard the result before, but I had never seen the derivation for it until I looked it up just now. I wasn't thinking about that anyway, I guess it's similar to this question. In the derivation I saw, they took into account both sides of the...

Ok, I get it, it's dumb for to think it'd be dependent on area. Every other Gauss' law problem I've ever done the answer has been dependent on area… but this is just an exception and the area is infinity so the answer in reality make sense, at least to me. When I said two different area values...

My enclosed charge is incorrect? Ok… I don't know why… Oh wait, I think I know the problem. It's because ρ as given in the problem is the charge density for the surface given in the problem. It isn't the charge density of my Gaussian surface. So Qen = ρAd, right? So my E would be ρd/ϵ0. But...

(and I would use the same exact Gaussian surface for part ii as well; x would have a different domain, though, as the problem itself says)

Yeah, sorry, I'm using the Gaussian surface which is exactly the same as the surface in the problem just with a thickness of 2x rather than d.

The first time I saw this question I had no idea how to do it (as you can see in the figure, I lost a lot of points :s) because I was confused on how to even approach it with area of the slab from all sides being infinity. Right? That's problematic, no? Today, I just tried the problem again for...