Gauss law of electricity | Electrostatics | Physics | Khan Academy

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[@VG__]

4:10 theta is 0, E is constant 7:20 Thankyou🌟 8:40 sphere offcentre 9:50 I am blown away😮🧐🤔u always pull my brain chords and activate them to think_ there's a rush of neurons😄 10:30 imagine that charge is shooting out bullets 11:55 😮ohkkkk 13:00 more charges 13:45

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[@vivekgupta3692]

Summary Of The Video.... *Gauss's Theorem* * Electric Flux is independent of radius. * If we increase or decrease the radius of sphere , the electric Flux Remain the same. * If we move the charge from the centre anywhere in the sphere then in that case also the flux doens't change . * If charge is positive then we get a positive flux * But if the charge is negative then we get a negative flux. * If we take an irregular surface in that case also flux remain the same. * If there is a hole in the sphere then the flux decreases because in that case some flux (Electric field lines) may leak through the sphere. That's why we always consider a *Closed Surface* * If we increase the charge inside a surface then the flux increases because of increase in Number of electric field lines. * Charge located outside the sphere doesn't contribute on flux. * Only those charges will contribute in flux which are located inside the sphere.

[@vidit2700]

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[@olaoluwaomopariola4867]

Please can you explain why in some textbooks, electric flux is Φ = ∫ E dA and in other textbooks electric flux is Ψ= Q, and in this video too? Which one is correct and which is wrong?

[@tranquilmelodies9034]

I can sense and feel the enthusiasm and passion in this lesson I really appreciate this video it gives me a more defined version of Gauss's Law

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Now I know why electric flux through asymmetrical surfaces is also the same as symmetrical. Huge thanks to Khan academy.

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Excellent....For myself,the bullet analogy did the trick.

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Understanding at the peak level....

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3:26 dA is perpendicular to the sphere.

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Greatly explained👏🏻....

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6:01 what if the area is not sphere? Will the equation change?🫣🫣

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Excellent explanation sir!! Thank you

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[@Darwish_Habiba]

11:37 well, I have a question. It would be 12bullets/min also but the resistivity is changed and flux's depended on it. I mean the resistivity of sphere is different from the air

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[@dumb_tan2539]

we know that electric field lines per unit area is directly proportional to the electric field strength ( as that's how we define EFL)..so EFL α ES... We define this quantity (where the E and S are along each other) = E•S as electric flux... so electric flux is basically EFL but in a physical quantity form.... So we can imagine it like those isotropical EFLS. So it's just some lines coming out and the best thing is, no matter what or how many closed surfaces are there covering it and from whichever orientation, all the bullets which are emitted from the charge centre, if they follow the inverse square law, we know all the overlapping distorted looking closed surfaces have the same no. of lines coming out. Also even if it was something like inverse cube law, we could get the formula using integration!!!! Happy learning :D

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[@CE113378]

Good video, but the electric flux is equal to the charge enclosed by the Gaussian surface. It is the integral of the electric flux density D, not the integral of the electric field intensity E. D = eps*E. In fact, the flux depends only on the charge enclosed. The electric field intensity depends on the charge enclosed AND the permittivity of the space (eps).

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[@yash1152]

10:31 12:25 even though i understood gauss's law earlier, still i couldnt apply it in questions earlier. But this bullet's analogy really helped me to apply it to non-trivial situations with ease. So, thanks a lot for giving this analogy/memorisation tool.

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[@philoso377]

Nice video and presentation. Following page 12:00 what will be the model behave when the sphere has an opening?

[@pewpew7]

Awesome!! I have a question... What if radius is approaching zero(r~0) i.e radius is comparable to the charge size What happens when radius of Gaussian surface (sphere) r < charge *radius I mean Electric flux inside the charge 😯

[@randomstuff6799]

The surface you must understand is imaginary. If the radius were approaching zero, the surface would still be existing around the charge, but now the imagination would be that it is shrinked yet there, and since it is there the bullet analogy would also work and the flux will remain same. When the radius reaches zero our imagined surface has vanished, so there is no flux to calculate. But again you could always consider (/imagine) a second surface, and about which you can again go on calculating flux in the exact same way. So, the conclusion would be, flux is always considered/calculated with respect to certain given space(usually area), be it imaginary or physical.

[@pewpew7]

@@randomstuff6799 got it! But what about flux inside a charge? Incase of a 🧲 magnet Flux lines(magnetic field lines) Go from S pole to N pole inside the magnet... So gauss's law int(B.ds)=0 what about flux inside a charge?

[@randomstuff6799]

Consider a hollow spherical conductor situated at a distance almost equivalent to infinity, so naturally from the point of reference (infinity), the sphere would seem to be a point charge. Now if that is the charge you are talking about, you must understand there is no flux to be considered inside the charge as there are no electric field lines existing inside it(electric field lines do not exist inside a uniformly charged spherical shell). If the sphere was not hollow but solid, the field lines would be directed outwards(as the direction of an electric field is given by the direction of its experiencing force). So, there is no existence of flux inside a charge. Mathematically zero.

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I really like the way you put these concepts into perspective. Great work.