In the arrangement shwn in Fig.4.96, plate (B) given a charge equal to `60`
`muC`, The ratio `d_(1)//d_(2)` is 2. Then
`q_(1)="___"`
`q_(2)="_____"`
`q_(3)="_____"`
`q_(4)="______"`
`q_(5)="_______"`
`q_(6)="_______"`
.

Explanation

Since charges on outer surfaces will be half of the total charge of all the three plates, So `q_(1)=q_(6)=60//2=30 muC`.
.
`q_(3)=q_(2), q_(4)=-q_(5)`
`rArr V_(AB)=V_(CB)`
or `(q-(2))/C_(1)=q_(5)/C_(2)`
`q_(2)d_(1)=q_(5)d_(2) rArrq_(2)2d_(2)=q_(5)d_(2)`
`q_(5)=2q_(2) rArrq_(4)=2q_(3)`
and `q_(3)+q_(4)=60`
`q_(3)=20 muC rArrq_(4)=40 muC`
And `q_(2) = -20 muC rArr q_(5) = -40 muC`