A battery of `10V` is connected to a capacitor of capacity `0. lF`. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charges are distributed equally on these two capacitors, find the total energy stored in the two capacitors. Find the ratio of final energy to initial energy stored in capacitors.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial charge on the capacitor
The initial charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \times V \]
where:
- \( C = 0.1 \, \text{F} \) (capacitance)
- \( V = 10 \, \text{V} \) (voltage)

Substituting the values:
\[ Q = 0.1 \, \text{F} \times 10 \, \text{V} = 1 \, \text{C} \]

### Step 2: Calculate the initial energy stored in the capacitor
The initial energy \( U_{\text{initial}} \) stored in the capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U_{\text{initial}} = \frac{1}{2} \times 0.1 \, \text{F} \times (10 \, \text{V})^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \]

### Step 3: Connect the charged capacitor to an uncharged capacitor
When the charged capacitor is connected to an uncharged capacitor, the total capacitance becomes:
\[ C_{\text{total}} = C_1 + C_2 = 0.1 \, \text{F} + 0.1 \, \text{F} = 0.2 \, \text{F} \]
The charge \( Q \) remains the same, but the voltage across both capacitors will change.

### Step 4: Calculate the final voltage across the capacitors
The final voltage \( V_f \) across the capacitors can be found using the formula:
\[ V_f = \frac{Q}{C_{\text{total}}} \]
Substituting the values:
\[ V_f = \frac{1 \, \text{C}}{0.2 \, \text{F}} = 5 \, \text{V} \]

### Step 5: Calculate the final energy stored in the capacitors
The final energy \( U_{\text{final}} \) stored in the two capacitors can be calculated using:
\[ U_{\text{final}} = \frac{1}{2} C_{\text{total}} V_f^2 \]
Substituting the values:
\[ U_{\text{final}} = \frac{1}{2} \times 0.2 \, \text{F} \times (5 \, \text{V})^2 = \frac{1}{2} \times 0.2 \times 25 = 2.5 \, \text{J} \]

### Step 6: Calculate the total energy stored in both capacitors
The total energy stored in both capacitors is:
\[ U_{\text{total}} = U_{\text{initial}} + U_{\text{final}} = 5 \, \text{J} + 2.5 \, \text{J} = 7.5 \, \text{J} \]

### Step 7: Calculate the ratio of final energy to initial energy
The ratio of the final energy to the initial energy is:
\[ \text{Ratio} = \frac{U_{\text{final}}}{U_{\text{initial}}} = \frac{2.5 \, \text{J}}{5 \, \text{J}} = \frac{1}{2} \]

### Final Answers:
- Total energy stored in the two capacitors: **7.5 J**
- Ratio of final energy to initial energy: **1:2**