capacitance

1. C1 is connected directly to the 24v supply, its potential difference thus equals to 24 v.

Since C2 and C3 are connected in series, hence V2 + V3 = 24, where V2 and V3 are the potential differences across C2 and C3 respectively
But potential difference = charge/capacitance
thus, Q/C2 + Q/C3 = 24, where Q is the charge stored in each capacitor
i.e. Q = 77.54 uC
Therfore, V2 = 77.54/6 v = 12.9 v
V3 = 77.54/7 v = 11.1 v


2. Charge on C1 = 24 x 1 uC = 24 uC
Since C2 and C3 are connected in series, the equivalent capacitance C' is,
C' = 2x2.4/(2+2.4) uF = 1.091 uF
Let Q1 and Q' be the final charges on C1 and C'
Hence, Q1 + Q' = 24
i.e. Q' = 24-Q1
Since the voltage across C1 and C' must be equal,
Q1/1 = Q'/1.091
i.e. Q1/1 = (24-Q1)/1.091
solve for Q1 gives Q1 = 11.48 uC
hence Q' = (24-11.48) uC = 12.52 uC
voltage across C1 = 11.48/1 v = 11.48 v

Since C2 and C3 are connected in series, charges on each of them are equal, hence Q2 = Q3 = 12.52 uC

Voltage across C2 = 12.52/2 v = 6.26 v
voltage across C3 = 12.52/2.4 v = 5.22 v
Therefore, the charges on each capacitors C1, C2 and C3 are respectively 11.48 uC, 12.52 uC, 12.52 uC. The respective voltages are 11.48 v, 6.26 v and 5.22 v