you have helped me so much in ochem and physics2!!!! i just cant emphasize enough that you have explained these complex concepts so well and so organized!!!!! your mind is so clear when you are going through the example problems one by one with specific steps and explanation. You are such a responsible teacher!!!
for those confused why .3 radius is used instead of 1.5. its asking for the CHARGE inside guassian cylinder. that means you use .3 because that's the cylinder causing the charge. the only thing we care about is the small cylinder not the big one
When solving the "c" part, you can just use the formula : electric flux=electric field intensity x area of gauss's cylinder. It is much easier this way and the answer is also same
The thing I struggle the most with is deriving the equations in the first place. Your videos have definitely helped though! I'll continue to practice, and eventually, I will be better at it. If anyone has any advice, I would more than welcome it.
What is the electric field 1.5m away from the cylindrical conductor? Idk but I thought it means computing the electric field outside the cylindrical conductor and not inside it yet you used the radius of it as the small r ;-; I don't understand. Anyways, thanks for making my life easier :D we love you TOCT!
thanks for the video. but i have the problem at (a) the place we were finding total charges enclosed by gaussian. when didn't you use 1.5m as radius as the way they give us in the question. i am confused at that point.?
@@princeklutse4950 i think it is because the charge does not depend on the Gaussian cylinder. They are 2 unrelated things that r used in the qn. Not sure whether this is true though
because you are using the radius of the conducting cylinder and not the guassian area. Read part a once more, we need the charge on the conducting cylinder that's enclosed by the guassian area.
My question sir.. the electric flux that passes through the Gaussian surface . The electric field is perpendicular to the Gaussian surface so I thought it should be zero. Help me please
The gaussian (external) sphere would be calculated as 4pir^2 as the charge enclosed by any closed gaussian surface is E x epsilon x Surface area. You could then equate this to the charge stored in the conducting sphere by using 4/3pir^3 x volume density. Hope this helps :)
It would because the electric field lines from the rod would not be perpendicularly away from it. When the rod is infinite, vector addition of field lines from neighbouring charges result in this configuration but not when its finite length
Donal Moloney if it’s a shape enclosed within a shape you use volume of the shape that’s enclosed. A good example is if there is a sphere within a sphere. So you use volume of the inner sphere and then surface area of the sphere enclosing the small sphere
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i love you in ways that i cannot explain
Toluwalogo Alase exactly same..
Same. He got me through physics 1 and here we are again getting me through physics 2
you have helped me so much in ochem and physics2!!!! i just cant emphasize enough that you have explained these complex concepts so well and so organized!!!!! your mind is so clear when you are going through the example problems one by one with specific steps and explanation. You are such a responsible teacher!!!
He's indeed Good in his work , thank you sir
physics 2? physics got so good they had to make a sequel
for those confused why .3 radius is used instead of 1.5. its asking for the CHARGE inside guassian cylinder. that means you use .3 because that's the cylinder causing the charge. the only thing we care about is the small cylinder not the big one
wait i didn’t know that i wondering that too! thank you!!😊
i hope you make lots and lots of money from youtube, you deserve it. you along with Sal Khan have guided thousands of us through school. thank you
.Thanks man, with the help of you I could pass, Calculus, Physics1, and the general chemistry. You are the BEASTTTTT!!!!!!
When solving the "c" part, you can just use the formula : electric flux=electric field intensity x area of gauss's cylinder.
It is much easier this way and the answer is also same
You would be praised as a saint in ancient era
The thing I struggle the most with is deriving the equations in the first place. Your videos have definitely helped though! I'll continue to practice, and eventually, I will be better at it. If anyone has any advice, I would more than welcome it.
What I like most is that your calculator voice :D
Haha same
You don't know how thankful I am to this channel from saving me to this situation (modular learning). Keep making videos sir. I'm with you💜💯
why part a use r= 0.3, but part c use r= 1.5? when the question is both 1.5m from the conductor, thank you
0.3 is finding Qs of conductor
Thanks for this video, its helped me. Hugs from Brazil
Very helpful, subscribed
why are we using the lateral area of the cylinder,instead of using the surface area of the whole cylinder,SA=2(pi)R^2+2(pi)Rh to calculate the charge.
What is the electric field 1.5m away from the cylindrical conductor? Idk but I thought it means computing the electric field outside the cylindrical conductor and not inside it yet you used the radius of it as the small r ;-; I don't understand.
Anyways, thanks for making my life easier :D we love you TOCT!
Yeah you're right brother I just noticed. Just substitute the 0.3 with 1.5 since it asked for a gaussian surface with radius 1.5m
Did u figure this out?
At 4.40 min the radius of the guassian cylinder is 1.5m but u substituted radius as .3 meter
thanks for the video. but i have the problem at (a) the place we were finding total charges enclosed by gaussian. when didn't you use 1.5m as radius as the way they give us in the question. i am confused at that point.?
@Kevin Le wow...really
@Kevin Le but why
@@princeklutse4950 I guess we will never know Lmao
@@whackapigeon3583 haha yeah
@@princeklutse4950 i think it is because the charge does not depend on the Gaussian cylinder. They are 2 unrelated things that r used in the qn. Not sure whether this is true though
i have a question. the charge of the conductor in the cylindrical gaussian surface is a surface charge distribution or a volume charge distribution
Hearing your calculator buttons click is very asmr
I wish you were my professor :(
Port part a why isn’t 1.5 used instead of 30 cm?
because you are using the radius of the conducting cylinder and not the guassian area. Read part a once more, we need the charge on the conducting cylinder that's enclosed by the guassian area.
I don’t get it still
same TT@@adarshdodeja5173
My question sir.. the electric flux that passes through the Gaussian surface . The electric field is perpendicular to the Gaussian surface so I thought it should be zero. Help me please
commenting vid for reference
😵💫
Please is the formular for total charge =surface density×area or total charge=surface density/area;
WHY is it sometimes we use 4PIr^2(area) and sometimes 4/3PIr^3(volume)? for a sphere
The gaussian (external) sphere would be calculated as 4pir^2 as the charge enclosed by any closed gaussian surface is E x epsilon x Surface area. You could then equate this to the charge stored in the conducting sphere by using 4/3pir^3 x volume density. Hope this helps :)
would the problem be different if the rod was finite length?
yes it whould
It would because the electric field lines from the rod would not be perpendicularly away from it. When the rod is infinite, vector addition of field lines from neighbouring charges result in this configuration but not when its finite length
@@maaan8494 thanks dude you're a life saver. I was wondering about that from the get go.
sir how to draw the electric field lines for infinite charge length
is there any way to solve it using coulomb'w law?
Why you didn't do a reviw of the lesson befor examples as usall
7:19
Nobody ever mentioned there is a typo in the title?
Isnt the length for the Area in flux = E.A, supposed to be L2 = L1 + 1.5 ?
4 years later i'm sure u figured it out right lol
@@lecctron Bruh dont remind me
why is that if i use E=kq/r^2 for the electric field 1.5m from the cylinder,im getting a different answer
Leeroy Majoni maybe depends on what constant k you’re using versus the epsilon naught “free space” value he’s using
Why gaussian surface was taken as cylinder why not sphere
why do you use surface area sometimes and volume in other videos.
Donal Moloney if it’s a shape enclosed within a shape you use volume of the shape that’s enclosed. A good example is if there is a sphere within a sphere. So you use volume of the inner sphere and then surface area of the sphere enclosing the small sphere
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