You are the physics master teacher! I was searching for this exact problem, and that was very explanatory. I fully understand the concept behind it. Thanks a lot 😍

Thank you so much! Your explanation is so clear, you are such a great teacher!

great channel, love your explanations!

Thank you so much! Your explanation is very easy to understand

wow great explanation. after watching 5min i knew how to do it plus i got a refreshment of gauss. 10/10

Thank you for your clear explanation!😍

really nice explanation thank you

Which type of gussian surface we will take in order to find E field due to a hollow capsule like object which have both hemisphere and cylindrical symmetry?

Thanks a lot , finally i got it 👏❤️

Please I need a clue on how to estimate the magnetic field of a concentric squares with uneven turn space. Thanks

it is awesome and undersrstood it completely 😎

Thanks homie.

What is the field at the centre of the cavity? Can't make a gaussian surface there. There is no enclosed charge.

any jee apirants ??

what if the sphere was a conducting shere? , what will the electric field inside the cavity?

@huvarda

8 ай бұрын

just 0

At first glance I just thought well if I draw a Gaussian surface inside the cavity, the Qenclosed will be 0 so there can’t be an electric field inside the cavity. I still don’t understand how to disprove that statement? (aside from your mathematical solution)

@mousaha78

2 жыл бұрын

Initially it has been considered there is +ρ charge density same as the outer sphere inside the cavity i.e. considering the cavity is not there initially. As it's not the case in the question, we have to subtract the extra charge at the portion of the cavity so -ρ charge is taken at the portion of cavity. Which finally on addition give total 0 in the cavity. This is how the sum is done and when calculating individually using gauss theorem we get -charge inside the cavity.

@ibrahimb1030

Жыл бұрын

It just says that no field emerges from that cavity, but it doesn't mean there is no Electric field passing through cavity. In simpler case, if there is a point charge and you draw a circle in space near it which doesn't contain the charge itself, obviously it doesn't mean there is no field passing through it, the field lines entering the gaussian surface(circle) all exits. It is the same here.

@suleymanaghamoglanli4439

Жыл бұрын

@@mousaha78 Well, we already saw what you told from the video, the guy was asking for a way to disprove (aside from solution in the video) his inference which seems reasonable but is not correct.

what about the field inside the sphere but outside the cavity????????/

@blur966

8 ай бұрын

you got the answer?

but isnt the enclosed charge zero?

For ucy students: When you cheese your battle with the poor.

The most appalling, revolting and vertiginous video where no notion of a, j and r and the subtraction and addition of the electric field is explained; just free flinging of equations!

@PhysicsNinja

5 ай бұрын

Lol

@lightshadow2707

12 күн бұрын

Probably too late but a is the given constant in this case the small sphere has a radius of a and the big one has a radius of 2*a, j is a unit vector pointing in the y direction==(0,1,0) and r is the absolute value of the r vector which points from 0,0,0 to a point in your coordinate space