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

Explanation

`C_P = C_1+C_2`
`C_P = 6+2=8muF`
`1/C_s = 1/C_1+1/C-2+1/C_3`
`1/8+1/8+1/8 = 3/8`
`C_s = 8/3muF`
Total capacitance
`C_s = 8/3xx10^-6F`
Total charge
`Q = C_sV`
`= 8/3 xx 10^-6xx9`
`Q = 24 muC`
(a) Cahrge on capacitors a is
`Q_a = 24 muC` ...(1)
Charge on capacitors b is
`Q_b = 24 xx6/8`
`= 18 muC ` ...(2)
Charge on capacitors c is
`Q_c = 24 xx 2/8`
`= 6 muC` ...(3)
Charge on capacitors d is
`Q_d = 24 xx 8/8`
`24 muC` ...(4)
(b) Potential differences across `C_a` is
`V_a = Q_a/C_a = 24/8 = 3V` ...(5)
`V_b = Q_b/C_b = 18/6 = 3V` ...(6)
`V_c = Q_c/C_c = 6/2 = 3V` ...(7)
`V_d = Q_d/C_d = 24/8 = 3V` ...(8)
(c) Energy stored in each capacitor
`U= 1/2 CV^2`
Energy stored in
`C_a = 1/2C_aV_a^2`
`U_a = 1/2 xx 8xx10^-6xx(3)^2`
`=36muJ` ...(9)
Energy stored in
`C_b = 1/2C_bV_b^2`
`U_b = 1/2 xx 6xx10^-6xx(3)^2`
`=27muJ` ...(10)
Energy stored in
`C_c = 1/2C_cV_c^2`
`U_c = 1/2 xx 2xx10^-6xx(3)^2`
`=9muJ` ...(11)
Energy stored in
`C_d = 1/2C_dV_d^2`
`U_d = 1/2 xx 8xx10^-6xx(3)^2`
`=36muJ`