if R=r then
energy=(k/2)*[(2n-1)2/22nr]
where k=one by 4 pie epsilon.......
i dont know how we type such things
Now as r is not equal to R the answer i am gettin is not even presentable, pls tell me how we type those integral symbols etc...
If u tell me how, i will be able 2 tell u the answer and how did manish sir present his solution for the capacitor problem u asked ,may be then i can effetively clear ur queries da .r u a member in orkut????
Manoj ,I am not a member of orkut.I also find difficulty in writing things invovling integration etc.If it is really difficult to write the answer ,then possible solution can be that you can explain the solution to this problem qualitatively.I think your answer is correct.
Are you joking , My ans is correct!!!!
If you are not a member of orkut then why dont you become one
Well for your question its like this girl,
All you got to do is to assume distributions and the equations invented by our loving ancestors will take care of the rest of the things.So in the 1st round the charge on S1 is Q.
Now after touching S2 , as there is a P.D between the 2 conducting spheres ,there now exists a charge transfer process to make the 2 spheres equipotential.(Here as the spheres are conducting it doesnt make any difference whether they are shells or solid spheres).Basically this concept is not potential concept but much of a field oriented one i.e charge does not flow as there is a P.D but it flows in order make the field inside the conductor zero which is a direct bi implication of the equipotential surfaces.
So now Assume a charge q on S2 equate their potentials
i.e kq/R =k(Q-q)/r which means Q/q=r+R/R where k is that useless one by 4 pie epsilon....................
Now comes fun time. You are now removing S1 ,charging it to q and again making it touch S2
So now kq1/R=k(Q-q0)/r
And Q+q1+q0=Q+q ( charge conservation)
Where q1=> Charge on S2 after the 2nd round
q0=> Charge transferred from S1
Now Continue till you observe a pattern in the charges on S2 which depends on n (Believe me you will need lot of paper & ink)
& finally self energy of a shell of charge q & radius R is
(k/2)(q/R)
Basically thats the reason i introduced 2 variables in my Eqn i.e q1 & q0 ,well it is a bit hard to observe this sequence thats the reason for takin r=R and observing things.Well i did have a tough time with observing the pattern because I am as stupid as a boy as you are as a girl.
Well you are a smart girl you should be able to observe it, because if you dont then something is wrong with you.
And become a member of orkut Please....
& dont tell me you didnt understand because ill kill you.It took a lot of time for typing all this because of the nobel prize winning advice you gave me for presenting my slution QUALITATIVELY.
You are great .Thanks a lot for trying out this problem. I will try to observe the pattern and will tell you if I could observe it. Again thanks a lot.This problem came in IIT 98.
The job is done. It's forming a G.P.
Charge in first round=QR/R+r
Charge in second round= QR/R+r + QR2/R+r
Charge in third round=QR/r+R + QR2/R+r + QR3/R+r
similarly charge in nth round ,QN= QR/R+r +QR2/R+r +.............+ QRn/R+r
therefore energy stored in sphere is 1/2 [QN2/C] .QN ,WE CAN CALCULATE AND C=kR,WHERE K= 4 PIE EPSILON.
QN= QR/R+r +QR2/(R+r)2 +.............+ QRn/(R+r)n
As it is R/r+R times the original charge added to original charge.
Your approach is absolutely correct.
My ans comes:
1/2Q22 /4(pie)(epsilon) R
where Q2=Q(R/r)[1-(R/r+R)n]
Hope you understood.
Bye and study well.
Hey todwal thats not her approach ,its mine!!
Anyways its a simple gp ,i messed up the whole thing and got a series named hermitian series god am i mad or what??