Problem: Consider the circuit shown in (Figure 1). Assume E = 16 V. What is the charge on 4.0 μF capacitor? What is the charge on 6.0 μF capacitor? 

Equivalent capacitance for capacitors in series:

$\boxed{\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}}$

The charge stored on a capacitor:

$\boxed{\mathbf{Q}\mathbf{=}\mathbf{C}\mathbf{V}}$

The 4.0 μF and the 6.0 μF capacitors are in series:

C_4,6 = (1/C₁ + 1/C₂)^-1 = (1/4 + 1/6)^-1 = 2.4 μF(10^-6F/1μF) = 2.4 × 10^-6 F

The voltage across the two capacitors is 16 V.