Two capacitors of $6\ \mu$F and $3\ \mu$F are connected in series across a $12$ V battery. What is the equivalent capacitance and the charge on each?

**Equivalent capacitance (series).** $\dfrac{1}{C_{eq}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1 + 2}{6} = \dfrac{1}{2} \Rightarrow C_{eq} = 2\ \mu$F. **Charge.** In series the same charge sits on every capacitor: $Q = C_{eq} V = 2 \times 12 = 24\ \mu$C. Option D uses the *parallel* formula ($C_{eq} = 9$). Option B halves the charge incorrectly.