Deriving Gauss's Law for Electric Flux via the Divergence Theorem from Vector Calculus

[@DrTrefor]

Note: I asked you to compute out the divergence of the electric field. But note the field is NOT defined at the origin, so the divergence theorem does not apply to the region inside the sphere centred at the origin. But that's ok, what we did in this video was the region between the small sphere and an arbitrary region and the divergence IS all zero there.

[@a.nelprober4971]

If anyone is confused about this problem: We can't use divergence theorem for the initial flux outside a sphere as you rightly say, we would get 0. The minor detail is that divergence theorem only holds when the field is continuously differentiable inside ALL of our boundary surface. Electric fields are not, as they fail the continuity property at the origin. When deriving that outward flux is independent of the boundary surface, we used the region BETWEEN boundary surfaces that did NOT contain the annoying origin, hence why we were allowed to use divergence theorem and set the divergence integral to 0. In this video, we exploit the fact that F*dsigma is equal everywhere on the surface of the sphere since the radial field is parallel to surface normal (special case). Thus, we can take the magnitude of the electric field out of the integral, and we are left with the surface integral of a sphere, which is just it's area. Thus multiply coulomb's law by area of sphere, and you get qenc/epsilon. Do not use divergence theorem!

[@ketaksingh5465]

Loved this series entirely....never learnt these things in such a meaningful way. Thank you so much

[@aryangupta7022]

I will be starting the Discrete Maths course today! I just want to say You are Legend and a very funny teacher. Thank you😊❤

[@DrTrefor]

Good luck and I hope to hear from you at the end:)

[@mnada72]

I hope there is a chance you can prove the rest of Maxwell's laws as an application on vector calculus.

[@mohamedirshaathm32123]

Same idea as i even thought being fascinated to know the physical meaning of maxwell's equations ,please do consider this sir

[@MrJames-nx3un]

a beautiful thing of mathematics, thank you professor for this lecture

[@liuwaiyip1671]

Very good explanation.

[@mohamedirshaathm32123]

Same idea as i even thought being fascinated to know the physical meaning of maxwell's equations ,please do consider this sir @Muhammad Nada

[@andrerossa8553]

you are the best! tks a lot!!

[@Rebecca-di7mt]

we started from the assumption that for this field divergence is field, but that isn't true right? isn't the divergence of E=ro/epsilonnot?

[@carultch]

The divergence is the sum of the partial derivatives of each vector field component, each with respect to the same spatial variable of the component. Analogous to a dot product. The fact that it equals rho/epsilon0, is the statement that Gauss's law makes. This isn't how we define the divergence of an electric field, but rather what it happens to equal in the statement of this law.

[@carultch]

The divergence of the field in his example, is zero everywhere except at the point charge as the source of the field. If you distribute the charge to a uniform sphere of charge, you would have a field in the form of F=x*ihat + y*jhat+ z*khat, inside the sphere. This would have a divergence, in this case equal to 3. When scaled by the appropriate constants, the divergence would be rho/epsilon0.

[@momen8839]

in electrostatic textbooks why they write that divergence of electric field equal charge density over permittivity however it equal zero?

[@DrTrefor]

The divergence is a local property while flux across a boundary is a global one, so one can be zero and the other not.

[@momen8839]

@@DrTrefor can you give me a source that explain this in details?

[@carultch]

@@momen8839 The divergence of an electric field equals zero in free space, or in a region of space with a distribution of charges that adds up to zero. In a region of space with a net charge density, the divergence equals charge density over free space permittivity.

[@softwarephil1709]

I’ve followed the other videos in this series, but this one confuses me. The first equation of Maxwell’s equations is Div(E)=q/epsilon0. I don’t know how to reconcile that with the premise of this video that Div(F)=0. Is Maxwell’s equation computing the flux through a bounding surface? I understand concentric bounding surfaces have the same total flux, but shouldn’t that be >0 if there’s a charge enclosed?

[@a.nelprober4971]

Your source of confusion is the same as what I had. We can't use divergence theorem for the initial flux outside a sphere as you rightly say, we would get 0. The minor detail is that divergence theorem only holds when the field is continuously differentiable inside ALL of our boundary surface. Electric fields are not, as they fail the continuity property at the origin. When deriving that outward flux is independent of the boundary surface, we used the region BETWEEN boundary surfaces that did NOT contain the annoying origin, hence why we were allowed to use divergence theorem and set the divergence integral to 0. In this case, we exploit the fact that F*sigma is equal everywhere on the surface of the sphere since the radial field is parallel to surface normal (special case). Thus, we can take the magnitude of the electric field out of the integral, and we are left with the surface integral of a sphere, which is just it's area. Thus multiply coulomb's law by area of sphere, and you get qenc/epsilon. Do not use divergence theorem!

[@weihungchew8228]

The divergence of electric field is zero, but I wonder why the differential form of gauss law states that electric field is not zero but equal to charge density over permittivity?

[@DrTrefor]

It's the difference between a local property (divergence) and the global property (flux across a surface). One can be zero and the other nonzero, and vice versa.

[@robmarks6800]

But isnt the flux through a surface equal to the tripple integral of divergence by the div Thm? If thats the case, the tripple integral is zero and the flux should be zero?

[@jsn1900]

I know this is late but i hope this will be useful for people in the future For a field with zero divergence, you may imagine an incompressible fluid flowing according to the field, because for any region the volume of fluid flowing in must equal to the volume of fluid flowing out. Therefore, outflow for inner sphere (A) = inflow for region between sphere and outer surface (B) = outflow for entire region (C). That's the geometric meaning of the proof in the last video in the playlist. Yes the divergence everywhere except the origin is zero, so if you calculate the "real" flux for a closed surface (C) around a charge it will be zero as well. However, in the video he is calculating the "net" flux for the region between the outer surface and the sphere (same as B), not the entire region, so it can be non-zero as long as C=A-B=0, hope the equations make sense to you But, the most important thing is that the sphere can be made as small as you like, but the net flux remains the same, so the flux for the entire region can be thought of as the limit of B as the radius of the inner sphere goes to zero. Personally I think this is a consequence of the field being undefined at the origin and causes all sorts of funny things, but that's my theory :)

[@matteovissani1071]

The divergence of F is 4*pi*delta(r), right?

[@DrTrefor]

The divergence is zero, but I think you are referring to flux across a small sphere.

[@sumitkumarsahoo7601]

A great idea, very intuitive, I never learnt it before, thank you

[@robmarks6800]

When I compute divergence of F i just get 3/r^3. What am I missing?

[@DrTrefor]

oh, don't forget the r has an x and y and z in it!

[@mathadventuress]

How do I convert I j k to the unit vectors in spherical/cylindrical coordinates? 😔

[@joanneal-adham2068]

Does anyone know of a book which has the integral and differential form equivalency proof for Gauss' electrostatics law? much appreciated :)

[@a.nelprober4971]

Idk but most of this course is coming from thomas calc

[@jhonnycamacho5262]

The divergence isn't 3(1/r³)?

[@carultch]

r isn't a constant, so you'd have to account for how r changes with x, y, and z, as you take the derivatives of divergence.

[@poonamhr]

It's my birthday today.

[@DrTrefor]

well hopefully this video is the best present you could ever hope for:D

[@sumitkumarsahoo7601]

well, happy birthday

[@tharunchougoni8293]

happy birthday After 2years

[@ElR1517]

I hate this class. You make me hate this class slightly less.

[@mathadventuress]

I’m a physics major. Can you be my professor? Mine are a joke.

[@DrTrefor]

I like jokes:D

[@mathadventuress]

@@DrTrefor legit made me lol!