Recent content by Alex G

Thanks gneil. I agree on both of those approaches. I was attempting that first one to try and avoid any integrals. I kept getting lost after determining which charge has experienced what. Eventually I was able to determine with that same idea that each point charge is receiving in terms of...

I am not quite seeing 28 distances, in fact I can only really imagine the 8 points here.

Homework Statement How much work is required to assemble eight identical point charges, each of magnitude q, at the corners of a cube of side s? Note: Assume a reference level of potential V = 0 at r = infinity. (Use any variable or symbol stated above along with the following as necessary...

Wow thank you both, that was a lot easier than I thought. I suppose I'm just uncertain when I should integrate and when I shouldn't.

Homework Statement A closed surface with dimensions a = b = 0.400 m and c = 0.800 m is located as shown in the figure below. The left edge of the closed surface is located at position x = a. The electric field throughout the region is nonuniform and given by E = (2.70 + 2.20 x2)i N/C, where x...

Wow, I guessed using the idea that the total flux should be 0 because of symmetry (not sure if that's an okay assumption) and also figured that since the normal of the flat surface is 180 degrees of the electric lines coming in that the answer would be the negative of the curved surface because...

Well the qenclosed/2Epsilon0 was correct for the curved surface, however the flat surface it does not apply for. I'm in full understanding of the curved surface and I know the point is in the center and symmetric. But why is the flat surface giving me a hard time here. Perhaps plane symmetry...

Ahhh yes it would, I see the significance of that now. So the lines to pass through are of equivalence since flux can be considered as the number of electric lines passing through an area? So both of them use the same equation then? It really through me off because I assumed the point to be...

Thanks for helping me sort that out so I have: qenclosed/Epsilon0*(1/2) However what can I say about the flat surface, I'm assuming that formula applies to the curved surface.

Ah that is qenclosed/Epsilon0 correct? So for a hemisphere would it be half that? Or am I getting ahead of myself?

If it was just a sphere would it be 4(pi)*kq?

Homework Statement A particle with charge Q is located immediately above the center of the flat face of a hemisphere of radius R as shown in the figure below. https://www.webassign.net/serpop/p19-33.gif (a) What is the electric flux through the curved surface? (Use any variable or...

Haha! YES! Thank you! And it all seemed to click much better than before! (Yes using Epsilon0 was the best way to go) Answer ended up at -232.658 N*m2/Cwhich makes much more sense. So I would probably use the integral definition of Gauss's Law when given an object with no value of the...

Yay! Thank you! (: Since q2 and q3 are the only affecting charges: (1.03*10-9-3.09*10-9)/ (1/4*(pi)*8.9876*109)? That appears to come out to -2.9183246*10-19, does that seem reasonable for fluxes? My teacher hasn't used any numbers in class /-:

Oh, I guess it isn't equal, my mistake, I got ahead of myself when I saw a sphere and assumed the constant radius meant constant E, but they are not centered. (: Ah, ha! Gauss's Law states that it must = qenclosed / Epsilon0 So, if Epsilon0 = (1/4(pi)k) Then is it: (the sum of the...