Questions: Two capacitors (C₁=6.00 μF and C₂=16.0 μF) are now connected in series and to a 9.00-V battery. (a) Find the equivalent capacitance of the combination. □ μF (b) Find the potential difference across each capacitor. v₁=□ V v₂=□ V (c) Find the charge on each capacitor. Q₁=□ μC Q₂=□ μC

Solution Steps

When capacitors are connected in series, the reciprocal of the equivalent capacitance $C_{\text{eq}}$ is the sum of the reciprocals of the individual capacitances:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substituting the given values:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{6.00 \, \mu \mathrm{F}} + \frac{1}{16.0 \, \mu \mathrm{F}}$$

$$\frac{1}{C_{\text{eq}}} = \frac{1}{6.00} + \frac{1}{16.0} = \frac{16 + 6}{96} = \frac{22}{96}$$

$$C_{\text{eq}} = \frac{96}{22} \approx 4.3636 \, \mu \mathrm{F}$$

The total voltage across the series combination is the sum of the voltages across each capacitor. The charge $Q$ on each capacitor is the same, and is given by:

$$Q = C_{\text{eq}} \times V = 4.3636 \, \mu \mathrm{F} \times 9.00 \, \mathrm{V} = 39.2724 \, \mu \mathrm{C}$$

The potential difference across each capacitor is given by:

$$V_1 = \frac{Q}{C_1} = \frac{39.2724 \, \mu \mathrm{C}}{6.00 \, \mu \mathrm{F}} \approx 6.5454 \, \mathrm{V}$$

$$V_2 = \frac{Q}{C_2} = \frac{39.2724 \, \mu \mathrm{C}}{16.0 \, \mu \mathrm{F}} \approx 2.4545 \, \mathrm{V}$$

Since the capacitors are in series, the charge on each capacitor is the same:

$$Q_1 = Q_2 = 39.2724 \, \mu \mathrm{C}$$

Final Answer