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show me a full solution for 4c using E = 1/2 QV

To solve **question 4c** in the problem (using \( E = \frac{1}{2} QV \)) for the energy stored in each capacitor after the charge redistribution, we need to proceed step-by-step.

### Problem 4c:  
In Figure 5, a **4.7 μF capacitor** is charged to 12.0 V by connecting it to a 12.0 V battery. Afterward, this charged capacitor is connected to a **2.2 μF uncharged capacitor**, causing the charge to be shared between the two capacitors. We are tasked with calculating the **final energy stored** in each capacitor after the charge is shared.

### Solution:

#### Step 1: Initial charge and energy in the 4.7 μF capacitor
The initial charge on the 4.7 μF capacitor, \( Q_i \), is calculated using the formula:

\[
Q_i = C V
\]

Where:
- \( C = 4.7 \, \mu\text{F} = 4.7 \times 10^{-6} \, \text{F} \)
- \( V = 12.0 \, \text{V} \)

So,

\[
Q_i = 4.7 \times 10^{-6} \, \text{F} \times 12.0 \, \text{V} = 5.64 \times 10^{-5} \, \text{C}
\]

The initial energy stored in the 4.7 μF capacitor is:

\[
E_i = \frac{1}{2} C V^2
\]

Substituting the known values:

\[
E_i = \frac{1}{2} \times 4.7 \times 10^{-6} \, \text{F} \times (12.0 \, \text{V})^2 = 3.384 \times 10^{-4} \, \text{J}
\]

#### Step 2: Charge redistribution
When the switch is closed, the charge redistributes between the two capacitors. The total charge remains conserved, so the total charge after redistribution is the same as the initial charge on the 4.7 μF capacitor.

The total capacitance of the two capacitors in parallel is:

\[
C_{\text{total}} = 4.7 \, \mu\text{F} + 2.2 \, \mu\text{F} = 6.9 \, \mu\text{F} = 6.9 \times 10^{-6} \, \text{F}
\]

The final voltage across both capacitors, \( V_f \), is given by:

\[
V_f = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{5.64 \times 10^{-5} \, \text{C}}{6.9 \times 10^{-6} \, \text{F}} = 8.17 \, \text{V}
\]

#### Step 3: Final energy stored in each capacitor

The energy stored in each capacitor is given by:

\[
E = \frac{1}{2} C V_f^2
\]

For the **4.7 μF capacitor**:

\[
E_{4.7} = \frac{1}{2} \times 4.7 \times 10^{-6} \, \text{F} \times (8.17 \, \text{V})^2 = 1.57 \times 10^{-4} \, \text{J}
\]

For the **2.2 μF capacitor**:

\[
E_{2.2} = \frac{1}{2} \times 2.2 \times 10^{-6} \, \text{F} \times (8.17 \, \text{V})^2 = 7.34 \times 10^{-5} \, \text{J}
\]

#### Step 4: Total final energy

The total final energy stored in both capacitors is:

\[
E_{\text{final total}} = E_{4.7} + E_{2.2} = 1.57 \times 10^{-4} \, \text{J} + 7.34 \times 10^{-5} \, \text{J} = 2.30 \times 10^{-4} \, \text{J}
\]

#### Step 5: Energy lost
The initial energy stored in the 4.7 μF capacitor was \( 3.384 \times 10^{-4} \, \text{J} \), and the final total energy stored in both capacitors is \( 2.30 \times 10^{-4} \, \text{J} \).

The energy lost during the process is:

\[
\Delta E = E_{\text{initial}} - E_{\text{final total}} = 3.384 \times 10^{-4} \, \text{J} - 2.30 \times 10^{-4} \, \text{J} = 1.08 \times 10^{-4} \, \text{J}
\]

This lost energy is due to heat and other losses in the system, possibly in the connecting wires or during the redistribution of charge.

### Final Answer:
- **Final energy stored in the 4.7 μF capacitor**: \( 1.57 \times 10^{-4} \, \text{J} \)
- **Final energy stored in the 2.2 μF capacitor**: \( 7.34 \times 10^{-5} \, \text{J} \)
- **Energy lost**: \( 1.08 \times 10^{-4} \, \text{J} \)

Would you like more details or have any questions?

### Related Questions:
1. How does energy get lost during the charge redistribution process?
2. What would happen if the capacitances of the two capacitors were equal?
3. How does the final voltage relate to the total capacitance in parallel circuits?
4. What factors contribute to the energy loss during the charge redistribution?
5. How would the situation change if the initial voltage was higher?

### Tip:
Always remember that in a capacitor circuit, when capacitors are connected in parallel, the total capacitance is the sum of individual capacitances, and the charge is shared according to the capacitances.

Learn how to calculate the final energy stored in two capacitors after charge redistribution using the formula E = 1/2 QV. This detailed solution covers initial and final energy calculations for a 4.7 μF and a 2.2 μF capacitor connected in parallel after one is charged to 12V. Suitable for high school physics students.

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