A capacitor of capacitance `C_0` is charged to potential `V_0`. Now it is connected to another uncharged capacitor of capacitance `C_0/2`. Calculate the heat loss in this process.

Options

A. `1/2 C_0 V_0^2`
B. `1/3 C_0 V_0^2`
C. `1/6 C_0 V_0^2`
D. `1/8 C_0 V_0^2`

Answer

To solve the problem of calculating the heat loss when a charged capacitor is connected to an uncharged capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- A capacitor of capacitance \( C_0 \) is charged to a potential \( V_0 \).
- The charge on this capacitor is given by:
\[ Q_0 = C_0 V_0 \]

2. **Connect to Another Capacitor**:
- The charged capacitor \( C_0 \) is connected to an uncharged capacitor of capacitance \( \frac{C_0}{2} \).
- After connection, the total charge \( Q_0 \) will redistribute between the two capacitors.

3. **Determine Final Voltage**:
- Let the final voltage across both capacitors after connection be \( V_{AB} \).
- The charge on the first capacitor (charged one) will be:
\[ Q_1 = C_0 V_{AB} \]
- The charge on the second capacitor (uncharged one) will be:
\[ Q_2 = \frac{C_0}{2} V_{AB} \]

4. **Conservation of Charge**:
- According to the conservation of charge:
\[ Q_1 + Q_2 = Q_0 \]
- Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ C_0 V_{AB} + \frac{C_0}{2} V_{AB} = C_0 V_0 \]
- Factor out \( V_{AB} \):
\[ V_{AB} \left( C_0 + \frac{C_0}{2} \right) = C_0 V_0 \]
- Simplifying gives:
\[ V_{AB} \left( \frac{3C_0}{2} \right) = C_0 V_0 \]
- Thus, the final voltage \( V_{AB} \) is:
\[ V_{AB} = \frac{2V_0}{3} \]

5. **Calculate Initial and Final Potential Energy**:
- The initial potential energy \( U_i \) of the charged capacitor is:
\[ U_i = \frac{1}{2} C_0 V_0^2 \]
- The final potential energy \( U_f \) of the system (both capacitors) is:
\[ U_f = \frac{1}{2} C_0 V_{AB}^2 + \frac{1}{2} \left( \frac{C_0}{2} \right) V_{AB}^2 \]
- Substituting \( V_{AB} = \frac{2V_0}{3} \):
\[ U_f = \frac{1}{2} C_0 \left( \frac{2V_0}{3} \right)^2 + \frac{1}{2} \left( \frac{C_0}{2} \right) \left( \frac{2V_0}{3} \right)^2 \]
- Calculating each term:
\[ U_f = \frac{1}{2} C_0 \cdot \frac{4V_0^2}{9} + \frac{1}{4} C_0 \cdot \frac{4V_0^2}{9} = \frac{2C_0 V_0^2}{9} + \frac{C_0 V_0^2}{9} = \frac{3C_0 V_0^2}{9} = \frac{C_0 V_0^2}{3} \]

6. **Calculate Heat Loss**:
- The heat loss \( Q \) during this process is the change in potential energy:
\[ Q = U_i - U_f = \frac{1}{2} C_0 V_0^2 - \frac{C_0 V_0^2}{3} \]
- Finding a common denominator (6):
\[ Q = \frac{3C_0 V_0^2}{6} - \frac{2C_0 V_0^2}{6} = \frac{C_0 V_0^2}{6} \]

### Final Answer:
The heat loss in this process is:
\[ \boxed{\frac{C_0 V_0^2}{6}} \]