Problems Solving on Electric Flux | Booster Classes for IIT JEE Advanced Physics

[@physicsgalaxyworld]

Notes of live classes will be available on Physics Galaxy mobile app. Click to setup - bit.ly/PhysicsGalaxyApp Join PG Telegram for Test Series papers of JEE & NEET. Click - t.me/physicsgalaxyworld Click to order 700+ Advanced Illustration Book for JEE Advanced. Click - amzn.to/3qyIK8H

[@YTayan27Bdn]

Sir please take one free class on iit/ neet. I have genuine doubt. On Unacademy.

[@murlimanohar5567]

31:54ans is Q/6root5epsilont

[@yourjeemate3590]

31:54 answer is q/pi€ (sin inverse(1/5)

[@murlimanohar5567]

@@yourjeemate3590 since inverse 1/root 5 bro.

[@murlimanohar5567]

31:54ans is( 2Q/root6 epsilon)×since inverse 1/root5

[@sushant2664]

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).

[@harshuldesai8901]

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.

[@sushant2664]

actually I was talking about the next question with E = Kq/r³

[@futureself8034]

sir answers seems wrong as sir treated charge density constant (mathematically) in rhs !!!

[@r4music220]

use leibnitz

[@random22453]

Divergence for spherical coordinates is different. Field is spherically symmetric therefore the divergence is 1/r^2 d/dr(r^2E)

[@arnavsingh1843]

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

[@Soaring-Dragon]

I am afraid it's wrong those 16 faces are not equivalent take caution

[@Soaring-Dragon]

To be more precise the center's of those faces are not at the same distance from the point charge

[@abhinavpandey7046]

To enclose cube 8 faces combine to form a single face of a big cube so it will be q/6*8 epsilon

[@potu6534]

@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 😊

[@tanmayjain2198]

Ans for the question 30:30 is --> q / 10 Eo

[@ayushsahu.841]

How

[@sarthakkhatale]

Galat

[@edunitian7701]

Lecture starts at 5:37 , save time

[@DODO-gq4qu]

full video is precious

[@riyasharma3992]

@@DODO-gq4qu acha

[@shivratanyadav8307]

Source

[@Dr.dhumketuIITDholakpur]

Ye mains ke liye bhi hai?

[@itachu.]

@@DODO-gq4qu dont be that guy , please.

[@rudrapatel7240]

35:54 By spherical symmetry the ans doesn't have a factor of 3🤔

[@AnandYadav-vc6yg]

bro we use that when density is constant here its varying

[@rudrapatel7240]

@@AnandYadav-vc6yg yes you are right Thankyou for the correction ☺️

[@debajyotichatterjee1891]

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?

[@krishnathawani7613]

Yaa...but wo 4 faces ke through flux kaise nikaalenge ?

[@debajyotichatterjee1891]

@@krishnathawani7613 wo four faces me flux equal hoga to total flux ko by 4 krke kr skte hai shayad

[@krishnathawani7613]

@@debajyotichatterjee1891 total flux by 4 kyu krenge...iska mtlb to yeh hua ki saamne waale face se 0 flux maan liya

[@imPriyansh77]

I also thought this But how will you find flux (equal) due to 4 adjacent faces ?

[@imPriyansh77]

@@krishnathawani7613 Yes. By 4 nhi karenge... Shayad isme jis face ke through flux nikalna hai , uspar area Lena padega using solid angle.

[@sameerbanarjee]

30:46 Flux through required Surface = q/10e°

[@Jesus-cm6ln]

How? Please tell the process

[@Jesus-cm6ln]

If you distribute the the flux in 5 surfaces, then it will be wrong because, there is no symmetry

[@sameerbanarjee]

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).

[@Jesus-cm6ln]

@@sameerbanarjee in the final wall there is no symmetry

[@prajwaltiwari4530]

@@sameerbanarjee we cant simply divide by symmetry as the last case does not have the faces in symmetry wrt the charge

[@Apocalypsepioneer]

Sir please onion physics series ko pura kariye

[@prashantsingh1885]

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...

[@prashantsingh1885]

Sorry, bhai.. there is no feature of uploading images.. here..

[@prajojeet]

Sahi sahi bilkul sahi...angle bi vary kar raha...aur strip position bi.....i was thinking about that only!!!...

[@hj-tu7hw]

What is the answer kindly tell

[@parikshitkulkarni3551]

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

[@TeeeeeeejAs]

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

[@rajbir2662]

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)

[@atikshagarwal2649]

Abhi saath saath karlo

[@sanjudutta8771]

Have you completed the unsolved series of physics galaxy book and from which class you have started your JEE preparation

[@ThakurGovinnSingh]

Easy h bhai ekdam krlo

[@namanjain6144]

Do it later

[@ThakurGovinnSingh]

@@namanjain6144 bhai conceptual illustration hi to h kr lene do kaafi help milega

[@parth1023]

31:02 Ans is qsin^-1(1/5)/pi eph

[@Jesus-cm6ln]

How?

[@ishaangupta246]

Bro it is 1/root5 not 1/5

[@zayan5128]

I think it's 1/4 inside arcsin

[@ujjwalsinha6405]

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

[@sunnymaurya238]

True...

[@prashantsingh1885]

That can be calculated by further taking an element on the previous taken strip.. and then by double integration.. that can be solved...

[@ujjwalsinha6405]

@@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

[@prashantsingh1885]

@@ujjwalsinha6405 Yeah bro.. but i think it's a way of solving that question..

[@Soaring-Dragon]

Try deriving a formula for the solid angle of a Prism

[@mathx.SB_2004]

At 30:50 ans. will be q/10E0

[@roshaninayak257]

Can you explain a little I didn't get where was the charge placed

[@futureself8034]

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 )

[@mathsphysicstutor8998]

This was the best session on flux, that I have ever attended. Thanks a lot sir❤

[@vedansh9004]

31:10 sir is the answer q(3 + rt3 - rt6 - rt2) /12 E {where E is epsilum naught} ??

[@krishnathawani7613]

How did you solve this ?

[@neerajparashar2161]

At 31:26 we should take another cube adjacent to it, it become cuboid. So flux through that cuboid is q/€ . So flux through one cube is q/2€ . So flux through its one side is 1/6.(q/2€) = q/12€

[@prajwaltiwari4530]

But the flux (q/2$) is not equally distributed among all sides! thats why we cant simply divide by 6

[@neerajparashar2161]

@@prajwaltiwari4530 It Will be equally distributed. If you have doubt you can check how ashish sir done the previous question to it!

[@shashvatsrivastava2580]

@@neerajparashar2161 No it is not equally distributed...In the previous question the charge is in the centre of a big cube that's why it has symmetry all around but here is not symmetrical situation as charge is not at a position symmetrical to all the sides because it is a centre of cuboid

[@imanmolrishi9590]

I think it should be q/10€

[@neerajparashar2161]

@@shashvatsrivastava2580 okk. If you are correct, now how will you do the question?

[@faith9666]

Sir which pen do you use

[@ayushaggarwal906]

Mind blown

[@Raj-xt4fk]

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

[@arijitkumardas2613]

30:44 is the ans (tan^-1(0.5)÷5^0.5÷pi)×q/epsilon? It is approx. 0.066q/epsilon Plz check sir....

[@Soaring-Dragon]

0.064 good 👍

[@Jesus-cm6ln]

How? Please explain the method

[@Soaring-Dragon]

@@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).

[@Jesus-cm6ln]

@@Soaring-Dragon for jee should we mug up that?

[@8BitGamerYT1]

@@Soaring-Dragon is there any way to do without solid angle ?

[@8BitGamerYT1]

I didn't understand why projected area is not used at 23:00 part

[@sakshamsharma3156]

35:45 SIR I HAVE A DOUBT . IF WE SOLVE THIS QUESTION BY TAKING A SPHERE OF RADIUS EQUAL TO r THEN BY WRITING GAUSS LAW FOR THAT WE CAN WRITE E(4pir^2) = Q(i.i. net charge inside )/epsilon naught . NOW E IS PPROPORTIONAL TO r^3 SO LHS IS PROPORTIOAL TO 1/r . NOW IF WE DIFFERENTIATE BOTH SIDES , THEN AT THE RHS , WE CAN WRITE Q AS 4PIR^2 RHO . BY SOLVING FROM HERE, ANS COMES OUT TO BE DIFFERENT FROM THE ONE THAT YOU SOLVED IN THE LECTURE . PLZ HELP SIR

[@Physics-Dimensions]

In fact solving by taking sphere is right way,because by slip, sir missed a point that small area. S taken at inner surface will not be same as small area at outer surface of the thin shell. There it should be S+dS. Also note that S is proportional to X and S+ dS is proportional to X +dx . If this is done ,both way answer may be same.

[@shanzalsiddiqui105]

@@Physics-Dimensions the answer is coming out to be kqenot/x^4

[@sakshamsharma3156]

@@shanzalsiddiqui105 YES MY ANS IS ALSO THIS

[@Physics-Dimensions]

@@shanzalsiddiqui105 correct 👍🏻

[@BetterCallHardik]

@@shanzalsiddiqui105 I also got the same answer

[@user-de6tj8jl2o]

Where can I find booster checklist notes? Pls anyone answer

[@prathamshailani3184]

38:24 can we do this even by differential form of gauss law?

[@Jesus-cm6ln]

But only Olympiad aspirants know this, in jee we don't use it

[@prathamshailani3184]

@@Jesus-cm6ln our teacher told this as bonus concept

[@Jesus-cm6ln]

@@prathamshailani3184 yes, the teacher is a really good one 🙏. Pranam unko 🙏🙏

[@gsrarmy3616]

Yes sir i have also that of through ball assuming it as a circle

[@8BitGamerYT1]

bro I didn't understand why assuming that way is wrong ?

[@theseusswore]

31:00 q/10e. by joining two cubes with the charge in the centre of conjoined faces, we get q/e flux exiting through 10 faces hence q/10e through each face. correct or no?

[@prashant34049]

No

[@prashant34049]

In 8 faces out of 10 flux is same but 2 face at full length distance flux is different

[@theseusswore]

@@prashant34049 oh yes right... thank you

[@prashant34049]

But I m also not able to find it through scaler addition One way to do this is electric field calculation which is complicated

[@theseusswore]

@@prashant34049 there has to be a way to do it simply because it feels wrong to do it with simple but lengthy calculation

[@Spavan3011]

At 38:00 how E= kq/x³ , is it not E=kq/x²??

[@khushankmaheshwari5149]

30:47 Sir q/2πE⁰ tan^-1 (1/2)

[@anuragbasu4140]

Flux through one surface of cube when charge is at another suface of same cube is = q divided by10 epsilon not (q÷10£)

[@imanmolrishi9590]

Same answer bro 👍

[@Jesus-cm6ln]

No, there is no symmetry

[@JEE-ff5pp]

😭😭😭

[@sandeepdas2639]

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)

[@lifeatamu4814]

q/12 hoga bhai itna complex mt banao

[@lifeatamu4814]

Epscilnot

[@lifeatamu4814]

For cube face center pe q/2epcilnot hota h

[@lifeatamu4814]

So 1 surface pe by 6 ho jayega

[@arijitkumardas2613]

@@lifeatamu4814 nhi hoga, sare surface symmetric nhi hai.

[@thinkbetter5384]

Sir, will the chapter weightage change for JEE Mains as JAB is conducting it now?

[@saitama1830]

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

[@AnandYadav-vc6yg]

not in jee syllabus i think

[@mrindia5251]

Flux=0 Bcz the angle between area vector and electric field will be 90°which in cos theta=0

[@pratyushsingh1211]

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

[@saitama1830]

@@pratyushsingh1211 thank you for your reply bro ... now I am in IITkgp doing my dream aerospace engineering 😁😁😁

[@panda-77]

Thank you sir 🌟

[@AbhishekKumar-oz3jn]

VERY HELPFL SESSIONS THNQ SIR SO MUCH

[@lifeisstruggle.struggleisl5899]

Sir thanks for you video very usefull.

[@TeeeeeeejAs]

30:40 is it {q/[2(epsilon not)]}multiplied by (root5)/[(root5)+4] ??????

[@SrishtiSingh-ek1hf]

Bas q/12E• na ?

[@GauravVerma-xs4uv]

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

[@studentgrowth]

20:11 why putting minus if it is scalar , aur scalar ko toh koi direction nahi hota .

[@arijitkumardas2613]

Flux is E.S and while solving a problem we have to fix the direction of area vector. We cannot change it everytime, there fore flux due to one charge has to be taken +ve and another -ve since they have different directions of electric fields and area vector assumption is kept constant.

[@studentgrowth]

@@arijitkumardas2613 thanks

[@wdivyankop]

thank you sir it was a brilliant class !!

[@vedansh9004]

41:22 sir why cant we take a flat component?

[@shivyakukku4423]

It's really helpful thanku so much sir to out from busy schedule ❤

[@filzasiddiqui]

Sir, Now please release a new block strategy as JAB has taken the authority of JEE now

[@kingremy723]

Sir what's the difference between the earlier booster checklist and this

[@ThakurGovinnSingh]

30:05 it will be q/32E E-Epsilon not

[@nayan9617n]

Thank you sir, it's very important for us 👍

[@saranshvottery6507]

31:35 answer is 0.156q/Eo ?

[@15thHarmonic]

Self note: 24:55

[@JaskaranSingh-ls4ex]

Sir jii double integration 4/a kq arctan(√3/10) Cube vale ka answer

[@mystic3549]

absolute elegance :)

[@anant7774]

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

[@deeeparka]

31:00 q/10E° I think will be the answer

[@ahsenequbal7234]

Q/6r

[@virajmhetar7456]

30:46 The ans to the opposite face flux would q/E0 because Electric feild strength through that face would be q/E0*A and thus flux would be (q/E0*A)*A and hence q/E0

[@Grace_moon_]

Sir ap to 🌌universe ka best teacher ho 🥰🥰🥰

[@aashreykumar9886]

can someone explain how the field distribution is 2 dimensional for line charge?

[@ishanalam5773]

Mtlb in the circular cross section

[@jatinbhatt7826]

Wire k along koi Electric field nhi hoga.

[@theadrijsamanta]

Line wire se radially outwards niklega tabhi toh gauss law se cylindrical Gaussian surface lete hain

[@murlimanohar5567]

@@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.

[@jatinbhatt7826]

@@murlimanohar5567 bhai line charge m hmesha yhi assume kiya jata k charge length k along hi distributed hoga top pr nhi

[@mohsinurrahmankhan9860]

Sir ,will you start class 11 from beginning ?

[@kushagrapiano9036]

Go and see concept videos of sir

[@shubhgupta4895]

@@kushagrapiano9036 YEAH

[@Apocalypsepioneer]

@@kushagrapiano9036 bhaiya kya sir ne concepts video mai pura jee syllabus complete karaya hai and kya yeh jo advance illustration ke question hai kya woh advance level kai hai

[@Apocalypsepioneer]

@@shubhgupta4895 bhaiya kya sir ne concepts video mai pura jee syllabus complete karaya hai and kya yeh jo advance illustration ke question hai kya woh advance level kai hai

[@shubhgupta4895]

@@Apocalypsepioneer dude if u think of him as mains level then u r mad (no offence) he has mentored air 1 many times and 100 of under air 100 rank he is teaching from 25 years i guess and he has his own books which r at top in physics jee mains plus advance so never underestimate Also he is allen jaipur srpg mentor he is hod there ❤️

[@abhinandandeepakraka5090]

Sir plz restart Onion Physics 🙏

[@Benzene42]

Mai full demotivate hu 11 dhag se aati na pyq hote na kuch kuch backlog bhi hai future barbaad hai

[@ReenaGuptaPCB]

5:30

[@dreadeddude6969]

Sir at 39:50 for the question can we take a Gaussian surface that is cuboidal and placed just like the cylinder with one face as the sheet? Then we can say flux equal to q/6e0?

[@VBM375]

Obviously not

[@J_Prince24]

Appreciable 🙌

[@vanmen1304]

Physics universe 🔥🔥🔥🔥

[@ShivamSaini-sj8pd]

Physics metaverse🔥

[@vanmen1304]

@@ShivamSaini-sj8pd multiverse hota hai vo

[@ShivamSaini-sj8pd]

@@vanmen1304 oh haan... Galti se mistake ho gyi😅

[@ridingrockband6968]

23:59 Yes sir mai bhi upar wala lkh deta hu😅

[@potu6534]

SIR THANK YOU VERY MUCH 😊😊😊😊😊😊

[@potu6534]

For revision purpose : (me) 31:00 onwards

[@murlimanohar5567]

31:54ans is Q /6root5epsilont

[@piyushpandey4688]

please share the ans

[@bricania1792]

Video starts at 5:00

[@harshivshah9376]

35:54 Sir ans is coming different if we take element as a sphere... Pls help

[@imPriyansh77]

Haa ,ek 3 ka factor numerator mai nhi aarha (iff we take spherical surface)

[@harshivshah9376]

@@imPriyansh77 yaa.. Ig we are making a conceptual mistake somewhere

[@UttamKumar-ez6nz]

Physics Genius

[@invincible9240]

Sir ,my ans for the question u gave at 31:31 is (q/pi €o )tan^-1(1/2sqrt6) .I was not able to do it by any elementary technique ,I had to use double integral to calculate the flux value .Could u please address this q ,using some elementary technique like symmetry arguments etc

[@SilentRaven77]

Can confirm, this answer is correct (Also solved by double integration)

[@arulvel96]

@@SilentRaven77 isnt q/10ε0 ? by symmentry

[@SilentRaven77]

@@arulvel96 Would you please elaborate a bit more on how you derived this answer?

[@arulvel96]

@@SilentRaven77 @Harsh Singh the charge is placed on centre of a side, so we are now placing a symmentric cube of same dimension to make the charge enclosed. The net flux for the new shape is now : q/ε0 Flux through each face of new shape is : q/5ε0 (since flux can't pass through the surface on which it's kept) This is for the new shape, so for the old shape(ie cube) is q/10ε0 This is how my teacher from my coaching instute taught us

[@SilentRaven77]

@@arulvel96 Okay so the problem is, when you add another cube to the side, you end up with a cuboidal gaussian surface. You’re using cubical symmetry arguments on a cuboidal surface. Think about it, does equal flux passes through all the faces of a cuboid? (Atleast q/10e0 is a better answer than q/2e0 so there’s that) I’ll be happy to clarify any further doubts/questions you have.

[@invincible9240]

Sir can u please address this question 31:31 in the next video just by using gauss law please sir

[@PwHighlights1]

what is flux through infinite and rectangular plane in XY plane having breadth 2l .

[@tvh8354]

q/16eo

[@PwHighlights1]

@@tvh8354 how

[@rocklee21236]

Sir vo cube wala question ka answer kya (q×cos-¹((2 (root6))/5))/pi epsilon not hoga??

[@tanishkgupta7237]

24:52

[@awesomechemistry7844]

WHEN WILL THE PG MOBILE APP RELEASED ON APP STORE ??

[@aswiniitm7]

Its already released

[@awesomechemistry7844]

@@aswiniitm7 NO BUDDY ITS NOT

[@awesomechemistry7844]

@@aswiniitm7 if so send link

[@krishnakatiyar8307]

It's already given in description

[@arnavborse5944]

Thankyou sir

[@yashagrawal8592]

30:44 I am getting Q/2.rt(5).π.ϵ0 by integration

[@sunnymaurya238]

But you cannot even integrate through the surface because due to charge the EF will be non uniform?

[@yashagrawal8592]

@@sunnymaurya238 yeah I didn't considered EF variation accross the elemental strip 😅, so it's wrong :(

[@akshatagrawal1294]

Flux for the charge present at face centre is q/20eo

[@tvh8354]

Mera q/10eo

[@FrAbhay.]

can anyone explain how to do those ring questions by integration approach

[@pramodsharma75158]

I ACCEPT IT

[@8BitGamerYT1]

bro I didn't understand why assuming that way is wrong ?

[@ThakurGovinnSingh]

Very genuine session

[@siddhantghildiyal2285]

31:10 ans flux is 8kq(1-1/√2)

[@devansh.gupta_]

17:00

[@tushitchatterjee9943]

30:40 my answer: Q/10 epsilon0

[@imPriyansh77]

How ?

[@bs6567]

31:32 i got q/10e

[@Bunnyyyy001]

Q/10€o at 31:02

[@HimanshuYadav-b1x]

Sir in my they don't taught to calculate the solid angle

[@shauryak5990]

Sir please take frequent classes on unacademy also. Please please

[@Dr.dhumketuIITDholakpur]

Pta nahi aaisa kyun lag raha hai ki In concept jee 2022 mai bhout question aainge

[@pratikgourav3654]

Sir please make video on term-2 physics exam

[@lobhnakurrey014]

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 🥺😭😭

[@vibhakudra]

You are superb continue to upload video

[@ishaansrivastava5376]

Video me ball se flux passing and coming me kya difference hai?

[@hasibreza6123]

Sir please🙏 continue booster classes

[@parth1023]

38:23 Ans jo he usko put karke dekhne pe galat aa raha he sir ka

[@Jesus-cm6ln]

Why? Please clarify

[@parth1023]

@@Jesus-cm6ln try to put it actually in the actual equations

[@arnav962]

i think the answer would be q/120 epilson

[@kapiljoshi970]

Sir ka Hindi lecture better than English lectures

[@dhruv_mseth]

Sir please make some videos for JEE 2023. TIPS AND LAUNCH CLASSES FOR ADVANCED ON APP IF POSSIBLE. PLEASE SIR🙏🏼

[@nishantbansal4785]

plz sir bring the pg application for desktop also

[@amanchandra6721]

31:00 flux will be q/12£°

[@bs6567]

Bhai tumne kese socha kyunki mera q/10 aya