very easy problem which im not able to solve....(any suprises there?)a solid conducting sphere of having uniform charge Q is surrounded by an uncharged conducting shell..let Potential difference between the surface of the solid sphere and the outer surface of the shell be V....now if the shell is given a charge -3Q, the new PD is_

1)V

2)2V

3)-V

4)-2V

@KARNA POST THE SOLN

BUT I THINK RADIUS IS NEEDED TO SOLVE THE QUESTION....

let the small sfere has radius R!.....and d one surrounding has radius R2

now in d first case (i.e +Q is given to R1).......-Q will be induced in the outer surface of R2 and+q in its outer surface.............

potential of outrer spfere(i.e R2)=kQ/R2

potential of inner spfere(i.e R1)=kQ/R1

HENCE DIFFERNCE V=KQ(1R1-1/R2)

now in d SECIND case (i.e -3Q is given to R1).......+3Q will be induced in the outer surface of R2 and -3Q in its outer surface.............

potential of outrer spfere(i.e R2)=k-3Q/R2

potential of inner spfere(i.e R1)=k-3Q/R1

HENCE DIFFERNCE V'=-3KQ(1R1-1/R2)=-3V

charge is given to the shell, dudes....

anyway, heres what i did...

a charge -Q is induced on the inner surface of the shell and hence a positive charge Q is induced on the other surface.

in the second situation, the the charge induced on the outer surface of the shell is -2Q

initial potential of solid sphere=KQ/R-KQ/r+KQ/r=KQ/R

initial potential of the outer surface of the shell=KQ/r-KQ/r+KQ/R

so PD=0

now final V of the sphere=KQ/R-3KQ/r

final V of the outer surface of the shell=-3KQ/r+KQ/R

PD=0 again

this is also consistent with the fact that the potentia at surface  of a hollow isolated conductor is the same as that in the interior points.

...

but im not convinced though answer is given as V......

let charge on the inner shell = q

potential difference = kq (r2 - r1) / (r1r2) .........am going to prove this in next 5 minutes

implies ...potential difference is directly proportional to charge on inner sphere .. (beautiful)

it is independent on the charge of outer sphere..

case ! ...inner sphere has charge Q ...potential difference is V 

case 2 ..somethings happen ..but still charge on inner sphere is Q ...

outer sphere has charges  - 2Q (outer) and -Q ( inner) resp,. ...

here ..we can see ..the charge on inner sphere remains unchanged . ...hence potential difference is the same i.e. V.

Proof :

let inner sphere ( q1 , r1) ...and outer (q2 , r2)

potential of inner sphere     V1 = kq1/r1 + kq2/r2

potential of outer sphere = kq1/r2 + kq2/r2

potential differemce =  kq1/r1 + kq2/r2 - kq1/r2 - kq2/r2

                                    =kq1(r2 - r1) / (r1r2) .......

potential difference  depends only on q1            .tadaa..!!!!

@tantri ..ur mistake ...initial potential difference is not zero ..

how can it be 0 ?

potential at inner sphere's surface = kQ/r .......and at outer sphere = kQ/R .(r not equal to R )

another mistake ....final potential difference isnt zero either.

potential at inner sphere's surface = kQ/r -3kQ/R ........and at outer sphere = kQ/R - 3kQ/R

...initial potential difference =  kQ/r -kQ/R = V

fianl potential difference =  kQ/r -3kQ/R - kQ/R + 3kQ/R

                                          =  kQ/r -kQ/R

                                          = V

darn ..i did it all over again...

 yea the answer is V

let radius of the solid sphere be a and that of the shell be b.

initially potential ofsolid sphere = kq/a and that of shell was kq/b

so PD = kq/a - kq/b = V (given)

Now when -3q is given to the shell

potential of solid sphere = kq/a - k3q/b

and that of shell = kq/b - k3q/b

hence PD = kq/a - kq/b = V

Hence the answer is V

hey its damn easy.........electric field inside d chrgd shell is alwz zero.......so wat eva b d charge u put on d outer sphere.....E inside wil be zero...hence making V=constant......which is here inturn d VD