Electrostatics 26: Electrostatic Boundary Conditions
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- čas přidán 13. 09. 2024
- In this video I continue with my series of tutorial videos on Electrostatics. These videos follow on from my tutorial series on 'Vector Calculus for Electromagnetism' and are in preparation for my tutorial series on 'From Electromagnetism to Optics'.
Specifically in this video I derive the boundary conditions on the electrostatic field. This series is pitched at undergraduate level.
I hope it's of use!! Thank you for watching and I hope that this matches your requirements.
Please feel free to provide feedback via comments and share with your friends.
See my website for more: www.universityphysicstutorials.com
Adam , I think what you mean is "Stoke's Theorem" instead of "Divergence Theorem" (06:05). You can add a comment on that. But hey your videos are awesome.
Yup
You're very welcome, thanks for the feedback!
Hi Adam. I found your old videos of very limited use as they seemed to be quick demonstrations with little explanation. This video however is very explanatory and extremely useful, thanks!
Adam , videos are really great. I want to correct smt, In the 6 equations, first one should be ▲^2V= - ρ/ ε0 Because
E= - ▼V and ▼E = ρ/ ε0
▼E = -▼^2 V
So , ▼^2V= - ρ/ ε0
Very Helpfull. thanks for posting
Thanks for explaining
What exactly is meant by above surface and below surface.is this surface of pillbox represented by A or at boundary?
Good job🖖
Great series. Question: can you point me to the equations, and any useful results (eg w/ spharmonics), for the potential felt at points on the surface of a sphere, from sources on the surface of the same sphere. ( this is actually for a non-physics application, but the problem formulation seems to be similar to electrostatics). Thanks!
7:01 "Rehab " lol we needed it
"The net flux is the difference between what leaves the surface and what enters it." It looks like both Eup and Edown leave the surface.
No, they are coming through the surface. It doesn't matter where the flux came from, only where it's going. The flux is going through the surface.It's not being generated on the surface but rather the surface is inside an electric field. In this case (perpendicular to the surface) there are only 2 places it can go - down or up. So their difference is the next flux in that direction.
Adam Beatty I agree, but when you said, "The net flux is the difference between what leaves the surface and what enters it.." it seemed like you were implying that that is the source of the negative sign for (Eabove - Ebelow). Maybe you didn't mean this, but I would say the negative sign comes from the fact that the vector (da) on the bottom of the surface points downward.
kevinmm20 The electric flux is not affected by charges that are not within the closed surface, the net electric field, E, in the Gauss' Law equation, can be affected by charges that lie outside the closed surface.
@@kevinmm20 I also found this very confusing
I want your complete lecture series
HELPFUL
Helpfull
suuuuuuper
I am bit confused about what leaves or enters when you are calculating flux if we see figure both components E(UP AND E(down) leaving surface
Me too. Because E.da for both surfaces are positive
Yes, I am also confused about this
According to me, both the flux are positive so they should be added and not subtracted.
@@princymalhotra3434 We actually assume we don't know the direction of E(up) and E(down) as you can see from the definitions of them, there are no vector signs on them so don't think about their directions. What we are looking for is "the difference between them" occurred by the charged surface. Since the total flux on a Gaussian surface is not affected by external sources we can calculate the net flux from the amount of charges inside of it, which is essentially the same thing what we would get from the difference between E(up) and E(down) but only if the Gaussian surface is infinitely close to the charged surface because if it wasn't, an external source could create flux on the side surfaces, therefore changing the partial fluxes on the Gaussian surface. The final result would not change but calculating it would become impossibly hard since there would be 4 additional surfaces hence 15 different paths external field lines could follow from one surface to another, with two surfaces there is only one path to follow and it is from the down to up or from up to bottom, a single field line could only follow one at a time so it is much easier to calculate the net flux.
i am having problem with ur hand writing
Seriously? I've had to read so much worse.