Problems Solving on Electric Flux | Booster Classes for IIT JEE Advanced Physics
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- čas přidán 9. 05. 2022
- Electric flux in electrostatics is an important chapter of Physics of class 12 is a critical topic on which questions are frequently asked in jee main & jee advanced and it is extremely useful topic of physics for board exams as well as all competitive examinations. In this live booster class on electric flux & Gauss's law, Ashish Arora Sir will cover several advance illustrations concepts on the topic for JEE Main & JEE Advanced Physics which are usful for IPhO aspirants also to fight well in International Physics Olympiad.
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Sir please take one free class on iit/ neet. I have genuine doubt. On Unacademy.
31:54ans is Q/6root5epsilont
31:54 answer is q/pi€ (sin inverse(1/5)
@@yourjeemate3590 since inverse 1/root 5 bro.
31:54ans is( 2Q/root6 epsilon)×since inverse 1/root5
Thank you sir
🌟
Sir thanks for you video very usefull.
In charge distribution, we can make a sphere of rad x, then we can use, kq/r2 + (field due to charge distribution in sphere) = kq/r3
VERY HELPFL SESSIONS THNQ SIR SO MUCH
absolute elegance :)
Sir at 35:45 if we use the differential form of Gauss Law (finding divergence of E-field), the value of ρ comes out to be a factor 3 less than your derived answer. Alternatively, I also calculated in the following way, and got the correct answer:
From integral form of Gauss Law, find Q_enclosed as a function of r, then use ρ = dQ_enc / dV. We get Q_enclosed(r) = q/r, and putting it in the expression we get ρ = -q/4πr^4 (note the missing factor of 3).
Field is uniform along x axis and hence div E = 2ax => ρ = 2aε₀x. dQ = ρdV = ρa²dx = 2a³ε₀x. Integrating from a to 2a we get Q(2a) - Q(a) = a³ε₀x² = 3a⁵ε₀. I'm getting the correct answer using the differential form of Gauss' law.
actually I was talking about the next question with E = Kq/r³
sir answers seems wrong as sir treated charge density constant (mathematically) in rhs !!!
use leibnitz
Divergence for spherical coordinates is different. Field is spherically symmetric therefore the divergence is 1/r^2 d/dr(r^2E)
Thank you sir, it's very important for us 👍
Thankyou sir
It's really helpful thanku so much sir to out from busy schedule ❤
Appreciable 🙌
Very genuine session
This was the best session on flux, that I have ever attended.
Thanks a lot sir❤
Mind blown
thank you sir it was a brilliant class !!
Sir please onion physics series ko pura kariye
Yes sir i have also that of through ball assuming it as a circle
bro I didn't understand why assuming that way is wrong ?
@27:08 Sir aapko innke comments ke chinta karne ki koi Jarurat nahi aisa Har kisi na kisi live class mein hota hi hai, aap bindaas padhaiye kyunki kuch live padhte hai,kuch live nahi padh pate due to some reasons toh vo recorded se padhte toh hai hi, aap ka content ki value ki Hum respect karte hai, vo waste na Jay uski poori koshis karenge, for me mujhe aapke kuch booster classes aur abhi jo chal rahe vo classes chahe short ho lekin 1 hour ideally lag hi Jaata hai samajte samajte Not telling ki you don't teach well lekin jab tak khud na samaj jau baar baar repeat karke padhne ki aadat si hai,
Isliye Sir aap nirash mat ho, We are with you 😊
SIR THANK YOU VERY MUCH 😊😊😊😊😊😊
For revision purpose : (me) 31:00 onwards
Physics Genius
Sir what's the difference between the earlier booster checklist and this
Sir ap to 🌌universe ka best teacher ho 🥰🥰🥰
Sir, will the chapter weightage change for JEE Mains as JAB is conducting it now?
Sir please🙏 continue booster classes
I think we need 4 cubes (2 half cubes at top and bottom and one on left side so flux through them will be q/24 epilson and then if u see there will be 16 equivalent faces through which flux will be passing so through one face flux will be q/24×16 = q/384 epilson
I am afraid it's wrong those 16 faces are not equivalent take caution
To be more precise the center's of those faces are not at the same distance from the point charge
To enclose cube 8 faces combine to form a single face of a big cube so it will be q/6*8 epsilon
Sir, Now please release a new block strategy as JAB has taken the authority of JEE now
Sir which pen do you use
Lecture starts at 5:37 , save time
full video is precious
@@DODO-gq4qu acha
Source
Ye mains ke liye bhi hai?
@@DODO-gq4qu dont be that guy , please.
You are superb continue to upload video
Video me ball se flux passing and coming me kya difference hai?
Yes sir main
Sir I have started my 12th a month ago and I am following those 5 volumes of physics galaxy, do I need to do advanced illustration book now or at last when I will complete all 5 books(I have completed 2 books of 11th and 1 book of 12th)
Abhi saath saath karlo
Have you completed the unsolved series of physics galaxy book and from which class you have started your JEE preparation
Easy h bhai ekdam krlo
Do it later
@@namanjain6144 bhai conceptual illustration hi to h kr lene do kaafi help milega
Sir jii double integration
4/a kq arctan(√3/10)
Cube vale ka answer
I ACCEPT IT
bro I didn't understand why assuming that way is wrong ?
i think the answer would be q/120 epilson
Flux for the charge present at face centre is q/20eo
Mera q/10eo
Sir please take frequent classes on unacademy also. Please please
Sir please make video on term-2 physics exam
Ans for the question 30:30 is --> q / 10 Eo
How
q/40e
30:47 Sir q/2πE⁰ tan^-1 (1/2)
35:54 By spherical symmetry the ans doesn't have a factor of 3🤔
bro we use that when density is constant
here its varying
@@AnandYadav-vc6yg yes you are right
Thankyou for the correction ☺️
Sir please make some videos for JEE 2023. TIPS AND LAUNCH CLASSES FOR ADVANCED ON APP IF POSSIBLE. PLEASE SIR🙏🏼
Sir 2nd que me flux ke liye jo apne angle liya h vo to only half ring hi cover krta h
Puri ring se flux ke liye to cos¢ ko double krna padega na
Sir plz restart Onion Physics 🙏
I didn't understand why projected area is not used at 23:00 part
31:10 sir is the answer q(3 + rt3 - rt6 - rt2) /12 E {where E is epsilum naught} ??
How did you solve this ?
I think.. that cube question can be solved by double integration...
First, taking an strip of thickess dx at a distance x from centre of face... then further taking an element on that strip... as electric field is not constant on that strip even...
And then it can be solved...
Sorry, bhai.. there is no feature of uploading images.. here..
Sahi sahi bilkul sahi...angle bi vary kar raha...aur strip position bi.....i was thinking about that only!!!...
What is the answer kindly tell
Can we do it by considering Cubes just like sir did for the corner case but this time the charge in the middle will be the corner to draw those cubes and we will use the fact that the area of the given face will still be the area of the face of the larger cube divided by 4 so we still get q/24e0
in that you can distribute total flux in ratio of per;endicular area of that opposite side to perpendicular area of oppositeside+4sides which we need to calculate flux. no need for integration although double integration is not in syllabusosite side
In flux through ball question - we can take solid angle from centre of sphere and between two Tangents and then thake q/eo/4pi 2pi(1-cos ø) R^2
yes sir hm vahi ans likhdete the
Ans to the cube question will be an inequality
φ>4KQ/√5 ,reason being,even if we will consider an elemental strip at x from edge ,field will not be constant at every point of the strip,we can calculate only at centre of that strip
True...
That can be calculated by further taking an element on the previous taken strip.. and then by double integration.. that can be solved...
@@prashantsingh1885 I also know that bro,but in double integration,you can’t solve one separately and then again integrate it
There a special methods to solve ∫∫ f(x)dx questions,it requires good knowledge of converging diverging series,eigen vector use
@@ujjwalsinha6405 Yeah bro.. but i think it's a way of solving that question..
Try deriving a formula for the solid angle of a Prism
Physics universe 🔥🔥🔥🔥
Physics metaverse🔥
@@ShivamSaini-sj8pd multiverse hota hai vo
@@vanmen1304 oh haan... Galti se mistake ho gyi😅
At 30:50 ans. will be q/10E0
Can you explain a little
I didn't get where was the charge placed
31:02 Ans is qsin^-1(1/5)/pi eph
How?
Bro it is 1/root5 not 1/5
Sir vo cube wala question ka answer kya (q×cos-¹((2 (root6))/5))/pi epsilon not hoga??
Flux through one surface of cube when charge is at another suface of same cube is = q divided by10 epsilon not (q÷10£)
Same answer bro 👍
No, there is no symmetry
😭😭😭
can anyone explain how to do those ring questions by integration approach
plz sir bring the pg application for desktop also
At 38:00 how E= kq/x³ , is it not E=kq/x²??
17:00
31:10 ans flux is 8kq(1-1/√2)
24:52
31:35 answer is 0.156q/Eo ?
30:05 it will be q/32E
E-Epsilon not
Sir I have feel that jab ham q. Solve karne k liye padhte h to ham q. Ko bas words me padh kar jitna samagh ata h utna Solve karte h Lekin ap ki Solving stategy alag h ap kaise concepts apply krte h please bataye
Mushe q. Solve hi nhi hote sirf simpal hi kar pati hu aur jab nhi hota to but depressed 😥 feel karti hu 🥺😭😭
❤️
Sir please take my doubt... in the calculation of electric field due to a ring of charge Q at a point on symmetrical axis passing through the centre is kqx/(x²+r²)^3/2 ... then what about the field at a point which is +- a distance above or below the axis of symmetry
not in jee syllabus i think
Flux=0
Bcz the angle between area vector and electric field will be 90°which in cos theta=0
You can calculate the field in the plane of the ring but only very slightly displaced from axis of symmetry.. at a general point in the plane will itself be mathematically tedious. Use gauss law and take gaussian surface as a small cylinder and you will find the required field to be passing through the curved part of the gaussian cylinder
@@pratyushsingh1211 thank you for your reply bro ... now I am in IITkgp doing my dream aerospace engineering 😁😁😁
Thanks sir 🙏 sita ram
Sir maths and chemistry please fast
For 31:00 question the charge will be giving equal flux to the top, bottom, left, right side of the cube but not the side asked in the question. We can get that by subtracting total flux and flux by the 4 faces. And the face containing thr charge will not get any flux. Am I right?
Yaa...but wo 4 faces ke through flux kaise nikaalenge ?
@@krishnathawani7613 wo four faces me flux equal hoga to total flux ko by 4 krke kr skte hai shayad
@@debajyotichatterjee1891 total flux by 4 kyu krenge...iska mtlb to yeh hua ki saamne waale face se 0 flux maan liya
I also thought this
But how will you find flux (equal) due to 4 adjacent faces ?
@@krishnathawani7613 Yes. By 4 nhi karenge...
Shayad isme jis face ke through flux nikalna hai , uspar area Lena padega using solid angle.
37:00 sir i think you have considered charge density constant while calculating the answer, thats why your answer is coming off by a factor of 3 (as calculated from differentiial form of gauss law )
5:30
33:00
31:32 i got q/10e
23:59 Yes sir mai bhi upar wala lkh deta hu😅
Yes i do this mistake
31:54ans is Q /6root5epsilont
please share the ans
Sir ka Hindi lecture better than English lectures
what is flux through infinite and rectangular plane in XY plane having breadth 2l .
q/16eo
@@tvh8354 how
30:46 Flux through required Surface = q/10e°
How? Please tell the process
If you distribute the the flux in 5 surfaces, then it will be wrong because, there is no symmetry
Keep one more cube of the same size attached with the face of cube containing positive charge q , then it will form a cuboid having charge q in the centre ,total flux will be q/e° ,Now since two cubes are joined internally so each cube will have 5 unique faces ,total faces of both cubes will be 10 ,so flux from each face will be q/10e° . ( Draw diagram it'll be more clear).
@@sameerbanarjee in the final wall there is no symmetry
@@sameerbanarjee we cant simply divide by symmetry as the last case does not have the faces in symmetry wrt the charge
Sir in my they don't taught to calculate the solid angle
30:44 is the ans (tan^-1(0.5)÷5^0.5÷pi)×q/epsilon? It is approx. 0.066q/epsilon
Plz check sir....
0.064 good 👍
How? Please explain the method
@@Jesus-cm6ln Go to Wikipedia and search solid angle in that one subtopic will be there where the formula for solid angle of a right rectangular pyramid is given use that to get omega and multiply by q/4pi*epsilon naught because flux per solid angle is (q/epsilon naught)/(4pi).
@@Soaring-Dragon for jee should we mug up that?
@@Soaring-Dragon is there any way to do without solid angle ?
Q/10€o at 31:02
Sir aapne hungama movie mein role kiya tha kya
Mai full demotivate hu 11 dhag se aati na pyq hote na kuch kuch backlog bhi hai future barbaad hai
30:40 is it {q/[2(epsilon not)]}multiplied by (root5)/[(root5)+4] ??????
Bas q/12E• na ?
31:00 flux will be q/12£°
Bhai tumne kese socha kyunki mera q/10 aya
This series is useful for neet aspirant??
physics is there in NEET
so should be useful ;-)
38:24 can we do this even by differential form of gauss law?
But only Olympiad aspirants know this, in jee we don't use it
@@Jesus-cm6ln our teacher told this as bonus concept
@@prathamshailani3184 yes, the teacher is a really good one 🙏. Pranam unko 🙏🙏
can someone explain how the field distribution is 2 dimensional for line charge?
Mtlb in the circular cross section
Wire k along koi Electric field nhi hoga.
Line wire se radially outwards niklega tabhi toh gauss law se cylindrical Gaussian surface lete hain
@@jatinbhatt7826 kyu nahi hoga bro. Line charge ka top point ko hum point charge man sakte hai to electric field upper part me bhi hoga.lekin bahut kam hoga .so ho Sakta hai neglet kardeta hoga .but I don't sure.
@@murlimanohar5567 bhai line charge m hmesha yhi assume kiya jata k charge length k along hi distributed hoga top pr nhi
rj sir op
Which series is this?
Onion 🧅
41:22 sir why cant we take a flat component?
Self note: 24:55
Pta nahi aaisa kyun lag raha hai ki In concept jee 2022 mai bhout question aainge
Video starts at 5:00
WHEN WILL THE PG MOBILE APP RELEASED ON APP STORE ??
Its already released
@@aswiniitm7 NO BUDDY ITS NOT
@@aswiniitm7 if so send link
It's already given in description
30:47 I got a little weird ans. I don't know if it is correct or not. Flux comes out to be
2q/pi epscilonnot (root5 -2/root5)
q/12 hoga bhai itna complex mt banao
Epscilnot
For cube face center pe q/2epcilnot hota h
So 1 surface pe by 6 ho jayega
@@lifeatamu4814 nhi hoga, sare surface symmetric nhi hai.
Sir can u please address this question 31:31 in the next video just by using gauss law please sir
You are the finest teacher respected sir 🥰😻
I'm not