For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
( c) energy stored in each capacitor .

Explanation

Capacitor b and c in parallel combination
`C_(p) =C_(b) + C_(c) = (6+2) mu F = 8 muF `

Capacitor a, `c_(p)` and d are in series combination , so the resulatant copacitance
`(1)/(C_(s))=(1)/(C_(a))+(1)/(C_(cp))+(1)/(C_(d))=(1)/(8) +(1)/(8) +(1)/(8) = (3)/(8)`
`C_(s)= (8)/(3) muF `

(a) Charge on each capacitor ,
Charge on capacitor a `Q_(a) =C_(s) V= (8)/(3) xx9`
`Q_(a)= 24muC`
Capacitor b and c in parallel
Charge on capacitor `b , Q_(b) =(6)/(3)xx9=18`
`Q_(b) = 18muC`
Charge on capacitor ` c , Q_(c) = (2)/(3)xx9=6 `
`Q_(c) = 6 muC`
(b) Potential difference across each capacitor `V= (q)/(C)`
Capacitor `C_(a), V_(a) =(q_(a))/(C_(a))= (24xx10^(-6))/(8xx10^(-6))=3V`
Capacitor `C_(b), V_(b)= (qb)/(C_(b))= (18xx10^(-6))/(6xx10^(-6))=3V`
Capacitor `C_(c), V_(c) = (q_(c))/(C_(c))= (6xx10^(-6))/(2xx10^(-6))= 3 V `
Capacitor `C_(d), V_(d)= (q_(d))_/(C_(d))= (24xx10^(-6))/(8xx10^(-6))=3V `
(c) Energy stores in a capacitor `U=(1)/(2) CV^(2)`
Energy in capacitor `C_(a), U_(a)=(1)/(2) C_(a)V_(a)^(2)=(1)/(2)xx8xx10^(-6) xx(3)^(2)`
`U_(a)= 36 muj`
Capacitor `C_(b), U_(b)= (1)/(2)C_(b)V_(b)^(2)=(1)/(2)xx6xx10^(-6)xx(3)^(2)`
`U_(a)=27 muj`
Capacitor `C_(c) , U_(c)=(1)/(2)C_(c)V_(c)^(2)=(1)/(2)xx2xx10^(-6)xx(3)^(2)`
`U_(a)=9muJ `
Capacitor `C_(d), U_(d)=(1)/(2)C_(d)V_(d)^(2)=(1)/(2)xx8xx10^(-6)xx(3)^(2)`
`U_(a)36 muJ`