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Consider two concentric spherical metal shells of radii `'a'` and `bgt a` .The outer shell has charge `Q` but the inner shell has no charge.Now the inner shell is grounded ,This means that the inner shell will come at zero potential and that electric fields lines leave the outer shell and end on the inner shell.(a) Find the charge on the inner shell
Find the potential on outer sphere.

Text Solution

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When an object is connected to earth (grounded) its potential is reduced to zero,Let `q` be the charge on A after it is earthed as shown in fig as

The charge`q'` on A induces `-q`' on inner surface of `B` and `+q` on outer surface of `B` ,In equilibrium the charge distribution is as shown in fig.
Potential of inner space =Potential of on `A+` potential due to charge on `B=0`
`V_(A)=(q')/(4piepsilon_(0)a)=(q')/(4piepsilon_(0)b)=(Q+q')/(4piepsilon_(0)b)=0`
or `q'=-Q((a)/(b))`
This implies that a charge `+Q(a//b)` has been transfered to the earth leaving negative charge on `A`.Final charge distribution will be as shown in fig.

As `bgta` so charge on the outer surface to outer shell will be `(Q(b-a))/(b) gt0`
(b) Potential of outer surface `V_(B)`=Potential due to charge on `A+` potential dueto charge on `B`.
`V_(B)=V_(A,out)+V_(b,"both surface")=(1)/(4piepsilon_(0))(q)/(b)+(1)/(4piepsilon_(0))(Q)/(b)`
`=(1)/(4piepsilon_(0))((-Q(a)/(b)))/(b)+(1)/(4piepsilon_(0))(Q)/(b)=(Q(b-a))/(4piepsilon_(0)b^(2))`
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