A capacitor of capacitance C has charge Q. it is connected to an identical capacitor through a resistance. The heat produced in the resistance is

Options

A. `(Q^2)/(2C)`
B. `(Q^2)/(4C)`
C. `(Q^2)/(8C)`
D. dependent on the value of the resistance

Explanation

To solve the problem of finding the heat produced in the resistor when a charged capacitor is connected to an identical capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Condition:**
- We have a capacitor with capacitance \( C \) and charge \( Q \).
- The initial energy stored in this capacitor can be calculated using the formula for energy stored in a capacitor:
\[ U_i = \frac{1}{2} \frac{Q^2}{C} \]

2. **Connect to an Identical Capacitor:**
- The charged capacitor is connected to another identical capacitor (also with capacitance \( C \)) through a resistor.
- Initially, the second capacitor is uncharged.

3. **Charge Redistribution:**
- When the two capacitors are connected, charge will redistribute between them until they reach the same potential.
- Let the final charge on each capacitor be \( Q_f \). Since they are identical, the total charge \( Q \) will be equally divided:
\[ Q_f = \frac{Q}{2} \]

4. **Calculate Final Energy:**
- The energy stored in each capacitor after redistribution is:
\[ U_f = \frac{1}{2} \frac{Q_f^2}{C} = \frac{1}{2} \frac{\left(\frac{Q}{2}\right)^2}{C} = \frac{1}{2} \frac{Q^2}{4C} = \frac{Q^2}{8C} \]
- Since there are two identical capacitors, the total final energy is:
\[ U_{f,total} = 2 \times U_f = 2 \times \frac{Q^2}{8C} = \frac{Q^2}{4C} \]

5. **Calculate Heat Produced:**
- The heat produced in the resistor is equal to the loss of energy during the process:
\[ Q_{heat} = U_i - U_{f,total} \]
- Substituting the values we calculated:
\[ Q_{heat} = \frac{1}{2} \frac{Q^2}{C} - \frac{Q^2}{4C} \]
- To simplify, we can factor out \( \frac{Q^2}{C} \):
\[ Q_{heat} = \frac{Q^2}{C} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{Q^2}{C} \left( \frac{2}{4} - \frac{1}{4} \right) = \frac{Q^2}{C} \cdot \frac{1}{4} \]
- Thus, the heat produced in the resistor is:
\[ Q_{heat} = \frac{Q^2}{4C} \]

### Final Answer:
The heat produced in the resistor is \( \frac{Q^2}{4C} \).

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