Ok . I will try out Question no. 30 pg 167

1) First we have to do the charge distribution

Let Charge  on the outer plate be Q.                             ...............(A)

Then Charge on the inner side is (2-Q)

Now By induction charge on second plate will be -(2-Q). The total charge on 2nd plate is -1*10^-8

therefore charge on outer side will be (1-Q).           ......................(B)

Next equating (A) n (B)

we get Q=0.5 * 10^-8C

Now charge in between the plates is (2-Q)=1.5* 10^-8 C

By Q=CV  we get

V=12.5V

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