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(?) we did use the normal vectors for each surface, note for the spherical upper volume we converted A to the spherical coordinate system, then found the flux through the half sphere Is there a timestamp where you feel there is a typo/error?

I actually didn't plug in either just left m and g as variables in the final solution. The Normal Force ended up being a percentage of each at different points

@@redSTEM I needed this reply before my physics exam😂😂. But thanks for the video it helped

cuz at both 1 and 2 you calculated for the case r<b

[TYPO!!] YES, at 6:02 it should read b=< R < Ri the inequality is written down incorrectly there with the leq sign facing the wrong way. That is a typo that was also copy pasted into 16:32 this is a mistake. Thank you for bringing this to my attention. **Please note the inequality should be b=< R < Ri** for part 2 and this region

@@redSTEM then the integral limits are from b to R right?

@@yunustalhaerzurumlu6547 czcams.com/video/SKLKTP-wBVk/video.html here is a very similar simplified example

@@yunustalhaerzurumlu6547 for part 2, no because you are now outside the full charged volume of b…. all of the charged volume will now be contributing to the E-field… the integral bounds are just for the enclosed charge. The max enclosed is just zero to b. The R on the left side represents the variable R distance we are from our center. In part 2 because R is in between b and Ri … we have all the charged volume of b enclosed, the integral bounds in part 2 will still be 0 to b, with all of the 0 to b the volume of charge contributing to the E field while we are some “R” distance away

Hm. Not sure what section you are asking about because each surface here had a surface normal vector…. Maybe you could indicate a time stamp for your question? This response might be slightly off topic or more than what you’re looking for. Note: Each dS differential surface has a vector normal direction. I.e. a vector normal to that surface that lines up with a coordinate system unit vector (r hat, theta hat, phi hat) We have been trying to find flux through types of surfaces where we can use Cart, Cyl, Sph directions to represent surfaces. Looking at a surface in these coordinate systems we can use the coordinate factors as normal vectors we just have to be careful with our sign conventions. Note that we are taking a dot product with a DS with a surface and unit vector. The dot product is going to tell us how much of a vector field is going to flow through a surface exactly in the Normal Vector direction (anything that’s not in the normal Vector direction it’s going to cancel out.) Taking the dot product of a Vector Field in a unit vector direction, will just give us how much of the vector field is acting in that direction. We don’t need to normalize the normal vectors because we are using the coordinate system cardinal directions that already line up really nicely with dS surfaces and our unit vectors are already normalized here.

@@redSTEM i think the last sentence was what i was looking for. @16:20 Ive been doing alot of these problems and ive noticed that sometimes we have to make the normal vector unit and sometimes we don't. im jsut confused to why that is. so if we're using a different coordinate system, we won't have to normalize the normal vector? sort of ? lol

@@tsalt Ohh I see. Ok. Yes it just depends on the direction you need a unit vector in. I.e. when we’re trying to find flux through a surface, we need the unit vector that’s normal to that surface… in this case when I say normal I mean at a 90 degree angle to a surface. Normalized is a little different. (the unit normal (90 degrees to) is so we can get the Flux vector “flow” that’s coming through in that same 90* direction. The dot product lets you figure out how much of one vectors in the direction of another, we dot that A vector with a surface area, and we need that that 90 degree through the surface normalized unit direction to figure out the flux/flow through that surface area) When it lines up perfectly with the coordinate system already we don’t need to find a ‘normalized’ unit vector the unit direction is pointing one way, (note normalized means broken down into components, like little fractions of each unit that show how to point in the direction you want, it’s way easier if you just have it all in one unit direction) If we know the unit vector direction lines up perfectly with just 1 direction. The unit direction at a right angle to the dS surface area is just r hat, z hat etc. It’s kind of like when you have a vector with only a y component, the normalized unit vector is just y hat? We can tell by inspection The surfaces we’ve looking at have just worked out cleanly where the unit normal vectors to the dS surface, dS(normal unit vector) have just been dS(z hat) dS(r hat) etc. If they didn’t line up cleanly with the coordinate system you were using then that’s when you may need to ‘normalize your’ unit vectors to get a unit vector that shows flux pieces broken up in the right coordinate directions. TLDR; You don’t need to normalize here if the normal vector is just in one coordinate system direction

@@redSTEM ok thank you very much for the help i will let that soak in, im actually suprised you answered since it was a video posted a long time ago. I am writing my final in about 2 hours hahah have a good day!

@@tsalt goodluck!

@@helloredspry Ahh thank you! Also, I never understood the argument that because of symmetry we immediately know that there is no electric field in the phi or theta directions. Are you able to explain this?