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

Answer

To solve the problem step by step, we will analyze the situation involving the solid conducting sphere and the thin metallic shell.

### Step 1: Understand the System
We have a solid conducting sphere of radius \( r_1 = 10 \, \text{cm} = 0.1 \, \text{m} \) with a charge \( q = 20 \, \mu C = 20 \times 10^{-6} \, C \). This sphere is enclosed by a thin metallic shell of radius \( r_2 = 20 \, \text{cm} = 0.2 \, \text{m} \). The conducting wire connects the inner sphere to the shell.

### Step 2: Determine the Charge Distribution
Since the inner sphere is connected to the shell by a conducting wire, the charge will redistribute itself. The total charge \( q \) will be shared between the inner sphere and the outer shell. However, since the outer shell is neutral initially, it will acquire a charge of \( -q \) to maintain overall neutrality.

### Step 3: Calculate the Potential Energy Change
The potential energy \( U \) of a charge \( q \) at a distance \( r \) from the center of a sphere is given by:
\[ U = \frac{1}{2} \cdot \frac{q^2}{4 \pi \epsilon_0 r} \]
We will calculate the potential energy for both the inner sphere and the outer shell.

1. **Potential Energy of the Inner Sphere:**
\[ U_1 = \frac{1}{2} \cdot \frac{q^2}{4 \pi \epsilon_0 r_1} \]

2. **Potential Energy of the Outer Shell:**
\[ U_2 = \frac{1}{2} \cdot \frac{q^2}{4 \pi \epsilon_0 r_2} \]

### Step 4: Substitute the Values
Substituting the values into the equations:
- For \( r_1 = 0.1 \, \text{m} \):
\[ U_1 = \frac{1}{2} \cdot \frac{(20 \times 10^{-6})^2}{4 \pi \epsilon_0 (0.1)} \]
- For \( r_2 = 0.2 \, \text{m} \):
\[ U_2 = \frac{1}{2} \cdot \frac{(20 \times 10^{-6})^2}{4 \pi \epsilon_0 (0.2)} \]

### Step 5: Calculate the Change in Potential Energy
The change in potential energy \( \Delta U \) is given by:
\[ \Delta U = U_1 - U_2 \]

### Step 6: Final Calculation
Substituting the values into the equation:
\[ \Delta U = \frac{1}{2} \cdot \frac{(20 \times 10^{-6})^2}{4 \pi \epsilon_0} \left( \frac{1}{0.1} - \frac{1}{0.2} \right) \]
Now, using \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ \Delta U = \frac{1}{2} \cdot \frac{(20 \times 10^{-6})^2}{4 \pi (8.85 \times 10^{-12})} \left( 10 - 5 \right) \]
\[ \Delta U = \frac{1}{2} \cdot \frac{(400 \times 10^{-12})}{4 \pi (8.85 \times 10^{-12})} \cdot 5 \]
Calculating this gives:
\[ \Delta U = 9 \, \text{J} \]

### Conclusion
The final answer is \( 9 \, \text{J} \).