Answer
The charge on each capacitor is $~6.0\times 10^{-5}~C$
$\Delta V_1 = 5.0~V$
$\Delta V_2 = 15~V$
$\Delta V_3 = 10~V$
Work Step by Step
We can find the equivalent capacitance:
$\frac{1}{C} = \frac{1}{12\times 10^{-6}~F} + \frac{1}{4\times 10^{-6}~F}+ \frac{1}{6\times 10^{-6}~F}$
$\frac{1}{C} = \frac{1}{12\times 10^{-6}~F} + \frac{3}{12\times 10^{-6}~F}+ \frac{2}{12\times 10^{-6}~F}$
$\frac{1}{C} = \frac{6}{12\times 10^{-6}~F}$
$C = \frac{12\times 10^{-6}~F}{6}$
$C = 2\times 10^{-6}~F$
$C = 2~\mu F$
We can find the charge $Q$ on each capacitor:
$Q = C~V$
$Q = (2\times 10^{-6}~F)(30~V)$
$Q = 6.0\times 10^{-5}~C$
We can find the potential difference across $C_1$:
$\Delta V_1 = \frac{Q}{C_1}$
$\Delta V_1 = \frac{6.0\times 10^{-5}~C}{12\times 10^{-6}~F}$
$\Delta V_1 = 5.0~V$
We can find the potential difference across $C_2$:
$\Delta V_2 = \frac{Q}{C_2}$
$\Delta V_2 = \frac{6.0\times 10^{-5}~C}{4\times 10^{-6}~F}$
$\Delta V_2 = 15~V$
We can find the potential difference across $C_3$:
$\Delta V_3 = \frac{Q}{C_3}$
$\Delta V_3 = \frac{6.0\times 10^{-5}~C}{6\times 10^{-6}~F}$
$\Delta V_3 = 10~V$
