Equivalent capacitance for capacitors in series:

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

The charge stored on a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

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

C_{4,6} = (1/C_{1} + 1/C_{2})^{-1} = (1/4 + 1/6)^{-1} = 2.4 μF(10^{-6}F/1μF) = 2.4 × 10^{-6} F

The voltage across the two capacitors is 16 V.

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?