IIT-JEE Professors make problems from this book

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PS: This problem is derived from Introduction to Electrodynamics by David G Griffiths, a renowned book on Classical Electrodynamics, which is considered one of the most challenging texts in the field. Many IIT professors also use this book, to create questions for electrostatics.
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[@jeesimplified-subject]

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[@fretting3264]

it can also be seen in as when we take the charge from the cavity to infinity the only field lines that do get changed are the ones that were produced in the spherical region extending from r=a to r=b ( initially there were no field lines in this region but when the charge is taken to infinity they appear ). So, the change in energy of configuration is due to those field lines and it's field energy can be obtained by integrating 1/2(epsilon not)(E)^2 × elemental volume with limits from r = a to r = b.

[@jeesimplified-subject]

Yes, could be another approach

[@abcd-e8k2s]

My approach
W ext= diff in total energies
Initially total energy is integral 1/2(epsilon not)(E)^2 × elemental small volume limits of integral is a to b
And it is the ext work since finally it is at infinity so no energy

[@darklord2096]

we can do it by capacitor concept take first energy that and find dinal energy that is zero so initial is KQ^2[1/a-1/b]/2 so lost energy is potential change (formula used q^2/2C and c is 4pi epsilon not (a-b)/ab

[@navisleegaming7453]

galti se upload krdi kya

[@kaushalgupta5209]

Loved your circle wala vid that you solved using cosec rule. Great content.

[@shedininjapsi]

Ef=0 and Ei= P.E (Between -q and +q i.e -kqq/r2) + self enregy of -q + self enegry of +q (radius b wala) but no interaction energy for radius b wala as it will become zero

[@L_Ratio_01]

I used
U(final)-U(initial) = 0 - U(initial)
U = sum of interaction energies (used similar logic like this video for this, i.e bringing small charge from infinity to surface) + sum of self potential energies (use formula for the 2 shells)

[@themightye4833]

Maine just PG se electrostatics revision karke break lene ke liye yt aya toh yeh vid dekha. lmfao imagine my surprise to see the last example of pg that i solved just before taking the break to be the very problem of the video lol

[@ketan8943]

I also revised it from PG advanced illustrations and saw it here

[@Omprasadmunde]

Mene toh community post me hi answer bata diya tha 😊

[@AtharwaKumar]

Ab pata chala mere Phy waale sir ke notes ke ques kaha se belong karte hai💀

[@mohitgoel4894]

Which coaching?

[@AtharwaKumar]

@ FIITJEE

[@unstoppablegaming577]

​@AtharwaKumar😂

[@AtharwaKumar]

@@prantikrai2368 Ho gaya bhai... Ab Allen m hoon☠️

[@samratdixit8048]

My approach : clearly net work is equal to change in potential energy
So we basically have to find potential energy of a charged double spherical capacitor and I knew the formula of its capacitance i.e. c= 4 pi e0 K ab / b-a
Then I used potential energy = q^2 /2c
Took me less than 2 minutes ❤

[@shivx3295]

just wrote final and initial energy done under 1 minute

[@Omprasadmunde]

MY CRAZY APPROACH ( without integration)
Key concept :- WD = ∆U =(∆self PE + ∆Interaction PE) .
Generally we consider only ∆self PE =0 , But here as charge is taken out of the cavity inside , Charges that were induced on conductor are vanished . Therefore we see change in self PE of two conducting shells .
To write self PE , u can write it by Q^2 / 2C . Considering both shells as capacitor here C= 4(pi)(ephsalen not) ab/(b-a) .
And interaction PE as usual .
The answer comes out to be (KQ^2 )(1/a-1/b) /2 .

[@Singularityfn]

Self energy of shell directly KQ²/2R laga dona aur sign dhang se balance

[@Omprasadmunde]

@singularityfn no way , it can get to answer coz it's q^2 there's no sign change for inner or outer shell therefore it can never lead to the answer .

[@Singularityfn]

@@Omprasadmunde going by the derivation we would already have -sign outside for self energy due to -q charge as the formula given by me above is for positive charge

[@Omprasadmunde]

@@Singularityfn might be , but instead of derivating all that y not do it through capacitors POV

[@Pulkit-tx6ih]

Bro this can be directly done by self energy, acutally this was tge whole concept this I guess। Correct me if I am wrong।

[@SanayM-ro1ci]

Bhaiya for maths ke liye professor kya book use karte woh batao please

[@Dhaneesh_7]

MATHS D GRIFFITH 😂

[@rajat23ranjan]

@Dhaneesh iske coaching ke master ne ek book bol diya hoga ..😂😂😂 muh se nikal gaya hoga master ka usiko pure jee calender me yaad rakha hai banda..
d griffith konsa book hai pehle search krlo
aur coaching ke module kro pehle 😂😂😂😂😂 @Deveshh kya jo v

[@onlybooks-zu9cv]

Isme
W=P.Ei-PEf
Laga denge na
Final zero initial inner surface aur outer surface pe induced charge -q,+q ki wajah se
Ye sahi hai kya

[@techbrosharma9243]

bro ye same question ashish arora sir ne apne ek video me karwa rakha h 2 saal pehle

[@mahimagupta6378]

Bro any specific series in which Ashish sir taught ??

[@kushal327]

Bro stop using this ai voice

[@Dhaneesh_7]

Starting 😂😂😂

[@rawatutkarsh]

Bhai ye question to irodov me bhi hai na ?

[@japvir3694]

hn

[@dikshasbotique6172]

Kya bhari aawaz hai

[@IITekSapna]

AI Voice 😂😂

[@shauryasoni1589]

Ie irodov wala question 😅

[@rudraksh_bhawsar]

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