Class 12th – Gauss Theorem Proof | Electric Charges and Fields | Tutorials Point

Gauss Theorem Proof
Lecture By: Mr. Pradeep Kshetrapal, Tutorials Point India Private Limited

[Sparkletta]

You’re the most brilliant physics teacher I’ve ever come across

[juniasmukhebi]

You make it so easy for me to understand concepts. Thank you so much

[thecount]

this is not a proof, this is verification, instead should be proved for any arbitrary surface.

[krishshiv]

Ur 💯 a fabulous teacher... thanks 😊 sir

[anikhilsannat]

For concept strengthening, you are no. 1

[panchamibr]

This is really helping a lot and thank you so much sir 😊

[DyNoBeAtS]

It's flawless awesome
Such a nice explanation

[anish]

U R Classes are my strength for me to learn physics

[googleuser]

Sir you didn't solved in general as it is applicable to all 3-d surfaces to be charge inside you only derived for sphere.

[adityagarkoti]

Appreciated☺😊😀😁....but sir, the proof you did, basically was not a proof, it is just an explanation😃

[Ashutoshe]

nice and lovely Application 👏hats off sir

[Physicschutkimai]

What about the irregular closed surface?
What is the origin of a 4 pi solid angle

[mohdarifansari]

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I become your fan after watching this video
Once again thank you so much

[ravikant]

Aapke batane ka tarika bahut bahut badhiya hai sir
Thankyou

[smittyflufferson]

Your proof is kind of circular because you derived it using coulombs law as a lemma which is derrive from Gauss's law

[JayCoolChannel]

YO, it's only the special case where it's a sphere! Where can I find the general proof for any closed surfaces containing the charge q ?

[bhindasingh]

Thnku so much sir ... It is very helpful for me🤗

[mayankjaiswal]

Thank you sir I'm greatful of your till my last breath

[arunbammidi]

Thank you so much sir...
I understood very well gauss therom

[anupamagoswami]

Good explanation. But to say that the theorem cannot be applied to a dipole just because the field falls off as 1/r3 is a wrong argument. Then by extension of said argument, it should not be applied to an infinitely long wire (where field falls off as 1/r) nor to an infinite plane sheet (where the field doesn't weaken with distance), but the Gauss theorem has been applied in the above two cases and the results have been experimentally verified.

So, the true reason why the Gauss law cannot be applied to a dipole, is a lack of symmetry. Not only is the total charged enclosed by the Gaussian surface zero (a dipole is a +ve and - ve charge after all, which would add to zero), but there is no symmetry whatsoever. No spherical symmetry, no cylindrical symmetry, no plane symmetry. You need symmetry to be able to pull the E out of the integral. Without symmetry, it's a tedious exercise and the magic of the Gauss equation is subdued.

The Gauss law here, simply put, is useless.