electrostatics

sorry......par mere method mein kya galat hai???

haan priyam.[4][4]..........mera bhi aa gaya tha raat ko[3][3]...............main to mathie ko chaalenge kar raha thaa ki meri galti nikale[6]..............mujhe mere answers par pura bharosa hai .....hamesha......[1]

are eureka...

u are right... apne par bharosa rakh...

put ρ=Q/(4piR³/3)

aa jaega... kyunki method sahi hai...

and mathiee.. u asked in terms of ρ so should have given answer in its term..

arre yaar..........tum log hc verma ke peeche pade ho........mere method mein galti batao.....

hc verma ki ebook nahin milti yaar....mere paas sachi mein hc verma nahin hai...........

arey e buk dhundle :P
waise bhi etre yahn koi tym pass ke liye kuch ni hai xcept 4 pool :P

plzzzzz batao....[12]

Good prblem[1]

U_sphere=Interaction energy+Self energy

Here we consider a ring of radius r and width dr , so Self energy associated with it can be neglected.

=> dU=V_sdQ

Here V_s =(4/3 Ï€r³Ï)/4πεr =ρr/3ε
and dQ=ρdV=ρ(4Ï€r²dr)

=>dU=4Ï€/3ε ρ²r⁴dr
Integrating with limits 0 to R
=>U=4πρ²/3ε R⁵/5

eureka, woh tho answer la liya na yaar.

any more doubts?

lol

woh x^2 n x^3 ko niche khech lao :P

haan ni karnege mein kuch aur soch ri thi sorry

here d sol

energy=Vdq
=(4/3 Π^x3)(4Π^x2σdx)

isse int karne se m gettin
(3/20q^2)/ΠεR

answer is 3Q²/20ΠεR

no replyzzz[2][2]

hey mathie...........reply to kar diya kar.........bata ye theek hai naa???????[12][12]