Define capacitance of a capacitor. Find the charge on each capacitor in the circuit shown in the figure.
Show Hint
To find the charge on each capacitor, first find the equivalent capacitance of the entire circuit, and then use the formula \( Q = C_{\text{eq}} \times V \).
Step 1: Definition of Capacitance.
The capacitance \( C \) of a capacitor is the ratio of the charge \( Q \) stored on one plate to the potential difference \( V \) between the plates. It is given by the formula:
\[
C = \frac{Q}{V}
\]
Where:
- \( C \) is the capacitance,
- \( Q \) is the charge on the capacitor,
- \( V \) is the potential difference across the capacitor.
Step 2: Analyze the Circuit.
Given the circuit:
- The capacitors are in a combination of series and parallel.
- Capacitors in series have an equivalent capacitance \( \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \).
- Capacitors in parallel have an equivalent capacitance \( C_{\text{eq}} = C_1 + C_2 \).
The circuit consists of capacitors \( 6 \, \mu\text{F}, 3 \, \mu\text{F}, 12 \, \mu\text{F}, 4 \, \mu\text{F} \) and a resistor \( 10 \, \Omega \) connected to a \( 20 \, \text{V} \) power supply.
Step 3: Find the Equivalent Capacitance of the Capacitors.
- The capacitors \( 6 \, \mu\text{F} \) and \( 3 \, \mu\text{F} \) are in series:
\[
\frac{1}{C_1} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2} \quad \Rightarrow C_1 = 2 \, \mu\text{F}
\]
- Now, the \( 2 \, \mu\text{F} \) capacitor is in parallel with the \( 12 \, \mu\text{F} \) capacitor:
\[
C_2 = 2 + 12 = 14 \, \mu\text{F}
\]
- The \( 14 \, \mu\text{F} \) capacitor is in series with the \( 4 \, \mu\text{F} \) capacitor:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{14} + \frac{1}{4} = \frac{1}{\frac{14 \times 4}{14 + 4}} = \frac{1}{\frac{56}{18}} \quad \Rightarrow C_{\text{eq}} = 3.21 \, \mu\text{F}
\]
Step 4: Find the Total Charge in the Circuit.
Using the formula \( Q = C_{\text{eq}} V \), where \( V = 20 \, \text{V} \), we can find the total charge:
\[
Q = 3.21 \times 20 = 64.2 \, \mu\text{C}
\]
Step 5: Find the Charge on Each Capacitor.
- The charge on each capacitor is the same in series combination, so the charge on \( 6 \, \mu\text{F} \) and \( 3 \, \mu\text{F} \) capacitors will be the same, \( Q_1 = 64.2 \, \mu\text{C} \).
- The charge on the parallel combination of \( 2 \, \mu\text{F} \) and \( 12 \, \mu\text{F} \) will be the same, \( Q_2 = 64.2 \, \mu\text{C} \).
Final Answer:
The charge on each capacitor is \( \boxed{64.2 \, \mu\text{C}} \).
