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