In the above question, if plates `P_1 and P_2` are connected by a thin conducting wire, then the amount of heat produced will be

Options

A. `Q^2/(Aepsilon_0)d`
B. `(5Q^2)/(Aepsilon_0)d`
C. `(2Q^2)/(Aepsilon_0)d`
D. none of these

Explanation

To solve the problem regarding the amount of heat produced when plates \( P_1 \) and \( P_2 \) are connected by a thin conducting wire, we will follow these steps:

### Step 1: Understand the System
When two charged plates are connected by a conducting wire, they will reach the same electric potential. The charge will redistribute between the plates until equilibrium is reached.

### Step 2: Determine the Initial Charge on the Plates
Let’s denote the initial charges on plates \( P_1 \) and \( P_2 \) as \( Q_1 \) and \( Q_2 \) respectively. The potential difference between the plates can be expressed as:
\[ V = \frac{Q_1}{C_1} - \frac{Q_2}{C_2} \]
where \( C_1 \) and \( C_2 \) are the capacitances of plates \( P_1 \) and \( P_2 \).

### Step 3: Calculate the Final Charge Distribution
After connecting the plates with a wire, the total charge \( Q_t = Q_1 + Q_2 \) will redistribute between the two plates. The final charges on the plates can be expressed as:
\[ Q'_1 = C_1 V_f \quad \text{and} \quad Q'_2 = C_2 V_f \]
where \( V_f \) is the final common potential.

### Step 4: Calculate the Heat Produced
The heat produced \( Q_h \) due to the redistribution of charge can be calculated using the energy stored in the capacitors before and after the connection:
\[ Q_h = \left( \frac{1}{2} \frac{Q_1^2}{C_1} + \frac{1}{2} \frac{Q_2^2}{C_2} \right) - \left( \frac{1}{2} \frac{(Q_1 + Q_2)^2}{C_{eq}} \right) \]
where \( C_{eq} \) is the equivalent capacitance of the two capacitors in parallel:
\[ C_{eq} = C_1 + C_2 \]

### Step 5: Substitute and Simplify
Substituting the expressions for \( Q_h \) and simplifying will yield the final expression for the heat produced.

### Final Expression
\[ Q_h = \frac{1}{2} \left( \frac{Q_1^2}{C_1} + \frac{Q_2^2}{C_2} - \frac{(Q_1 + Q_2)^2}{C_{eq}} \right) \]