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 will follow these steps:

### Step 1: Calculate the initial charge on the charged capacitor
The initial charge \( Q_0 \) on the charged capacitor with capacitance \( C_0 \) and voltage \( V_0 \) is given by:
\[ Q_0 = C_0 V_0 \]

### Step 2: Calculate the initial energy stored in the charged capacitor
The initial energy \( E_i \) stored in the charged capacitor can be calculated using the formula for energy stored in a capacitor:
\[ E_i = \frac{1}{2} C_0 V_0^2 \]

### Step 3: Determine the final voltage after connecting the two capacitors
When the charged capacitor \( C_0 \) is connected to an uncharged capacitor \( C_0/2 \), the total capacitance \( C_f \) of the system becomes:
\[ C_f = C_0 + \frac{C_0}{2} = \frac{3C_0}{2} \]
The charge \( Q_0 \) remains constant, and the final voltage \( V_f \) across both capacitors can be found using the relationship:
\[ Q_0 = C_f V_f \]
Substituting \( Q_0 \) and \( C_f \):
\[ C_0 V_0 = \frac{3C_0}{2} V_f \]
Solving for \( V_f \):
\[ V_f = \frac{2V_0}{3} \]

### Step 4: Calculate the final energy stored in both capacitors
Now we calculate the energy stored in both capacitors after they are connected. The energy stored in the first capacitor \( C_0 \) is:
\[ E_{1} = \frac{1}{2} C_0 V_f^2 = \frac{1}{2} C_0 \left(\frac{2V_0}{3}\right)^2 = \frac{1}{2} C_0 \cdot \frac{4V_0^2}{9} = \frac{2C_0 V_0^2}{9} \]
The energy stored in the second capacitor \( C_0/2 \) is:
\[ E_{2} = \frac{1}{2} \left(\frac{C_0}{2}\right) V_f^2 = \frac{1}{2} \cdot \frac{C_0}{2} \cdot \left(\frac{2V_0}{3}\right)^2 = \frac{1}{2} \cdot \frac{C_0}{2} \cdot \frac{4V_0^2}{9} = \frac{C_0 V_0^2}{9} \]
Thus, the total final energy \( E_f \) is:
\[ E_f = E_{1} + E_{2} = \frac{2C_0 V_0^2}{9} + \frac{C_0 V_0^2}{9} = \frac{3C_0 V_0^2}{9} = \frac{C_0 V_0^2}{3} \]

### Step 5: Calculate the heat loss
The heat loss \( \Delta E \) during this process is the difference between the initial energy and the final energy:
\[ \Delta E = E_i - E_f = \frac{1}{2} C_0 V_0^2 - \frac{C_0 V_0^2}{3} \]
Finding a common denominator (which is 6):
\[ \Delta E = \frac{3C_0 V_0^2}{6} - \frac{2C_0 V_0^2}{6} = \frac{C_0 V_0^2}{6} \]

Thus, the heat loss in this process is:
\[ \Delta E = \frac{C_0 V_0^2}{6} \]

### Final Answer:
The heat loss in this process is \( \frac{C_0 V_0^2}{6} \).

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