A solid conducting sphere of radius `10 cm` is enclosed by a thin metallic shell of radius `20 cm`. A charge `q = 20 mu C` is given to the inner sphere is connected to the shell by a conducting wire.

Options

A. 12 J
B. 9 J
C. 24 J
D. zero

Explanation

To solve the problem, we need to determine the heat generated when the inner conducting sphere is connected to the outer metallic shell by a conducting wire. The heat generated can be calculated based on the change in potential energy of the system.

### Step-by-Step Solution:

**Step 1: Understand the Configuration**
- We have a solid conducting sphere (inner sphere) with a radius of 10 cm and a charge \( q = 20 \, \mu C \).
- This sphere is enclosed by a thin metallic shell (outer shell) with a radius of 20 cm.

**Step 2: Determine the Initial Potential of the Inner Sphere**
- The potential \( V \) of a charged conducting sphere is given by the formula:
\[ V = \frac{k \cdot q}{r} \]
where \( k \) is Coulomb's constant (\( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \)), \( q \) is the charge, and \( r \) is the radius of the sphere.

- For the inner sphere:
\[ V_{inner} = \frac{k \cdot q}{R_{inner}} = \frac{9 \times 10^9 \cdot 20 \times 10^{-6}}{0.1} = 1.8 \times 10^6 \, V \]

**Step 3: Determine the Potential of the Outer Shell**
- When the inner sphere is connected to the outer shell, charge will redistribute until both have the same potential.
- The potential of the outer shell (which is also a conducting surface) is given by:
\[ V_{outer} = \frac{k \cdot Q}{R_{outer}} \]
where \( Q \) is the total charge on the outer shell.

- Initially, the outer shell has no charge, so:
\[ V_{outer} = 0 \, V \]

**Step 4: Charge Redistribution**
- When the inner sphere is connected to the outer shell, charge will flow until both spheres reach the same potential.
- Let \( Q \) be the charge that moves to the outer shell.
- The total charge will be conserved:
\[ Q_{total} = q + Q = 20 \, \mu C + Q \]

- The final potential of both the inner sphere and the outer shell must be equal:
\[ V_{inner} = V_{outer} \]
\[ \frac{k \cdot (20 \times 10^{-6} + Q)}{0.2} = \frac{k \cdot (20 \times 10^{-6})}{0.1} \]

**Step 5: Solve for Q**
- Simplifying the equation:
\[ \frac{20 \times 10^{-6} + Q}{0.2} = \frac{20 \times 10^{-6}}{0.1} \]
\[ 20 \times 10^{-6} + Q = 40 \times 10^{-6} \]
\[ Q = 40 \times 10^{-6} - 20 \times 10^{-6} = 20 \times 10^{-6} \, C \]

**Step 6: Calculate the Change in Potential Energy**
- The change in potential energy \( \Delta U \) can be calculated using:
\[ \Delta U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance of the system.

- The capacitance \( C \) of the system (considering both spheres) can be calculated as:
\[ C = 4 \pi \epsilon_0 R \]
where \( R \) is the distance between the centers of the spheres.

- For our case:
\[ C = 4 \pi \epsilon_0 (0.2) = 4 \pi (8.85 \times 10^{-12}) (0.2) \]
\[ C \approx 2.23 \times 10^{-11} \, F \]

- Now, substituting into the potential energy formula:
\[ \Delta U = \frac{1}{2} \cdot 2.23 \times 10^{-11} \cdot (1.8 \times 10^6)^2 \]

**Step 7: Calculate the Heat Generated**
- The heat generated \( Q_{heat} \) is equal to the change in potential energy:
\[ Q_{heat} = \Delta U \]