The potential of point $A$ is greater than that of point $B$ by $19 \, V$. What is the potential difference in volts across the $3 \, \mu F$ capacitor?

(B) Applying Kirchhoff's voltage law from point $A$ to point $B$ along the circuit:
$V_A - \frac{q}{2 \times 10^{-6}} - 8 + 15 - \frac{q}{3 \times 10^{-6}} - \frac{q}{4 \times 10^{-6}} = V_B$
Rearranging the terms:
$V_A - V_B + 7 = q \left( \frac{1}{2 \times 10^{-6}} + \frac{1}{3 \times 10^{-6}} + \frac{1}{4 \times 10^{-6}} \right)$
Given $V_A - V_B = 19 \, V$:
$19 + 7 = q \left( \frac{6 + 4 + 3}{12 \times 10^{-6}} \right)$
$26 = q \left( \frac{13}{12 \times 10^{-6}} \right)$
$q = \frac{26 \times 12 \times 10^{-6}}{13} = 24 \times 10^{-6} \, C = 24 \, \mu C$
The potential difference across the $3 \, \mu F$ capacitor is:
$V_{3\mu F} = \frac{q}{C} = \frac{24 \times 10^{-6}}{3 \times 10^{-6}} = 8 \, V$