The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains `4mu C` charge, its radius will be:

Options

A. 32 mm
B. 16 mm
C. 28 mm
D. 20 mm

Answer

To find the radius of a metal sphere given the energy stored in its electric field and the charge it contains, we can use the formula for the energy stored in the electric field of a charged sphere:

\[ U = \frac{3}{5} \frac{Q^2}{4 \pi \epsilon_0 R} \]

Where:
- \( U \) is the energy stored (in joules),
- \( Q \) is the charge (in coulombs),
- \( R \) is the radius of the sphere (in meters),
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step-by-step Solution:

1. **Identify the given values:**
- Energy \( U = 4.5 \, \text{J} \)
- Charge \( Q = 4 \, \mu C = 4 \times 10^{-6} \, \text{C} \)

2. **Substitute the known values into the formula:**
\[ 4.5 = \frac{3}{5} \frac{(4 \times 10^{-6})^2}{4 \pi \epsilon_0 R} \]

3. **Calculate \( Q^2 \):**
\[ Q^2 = (4 \times 10^{-6})^2 = 16 \times 10^{-12} \, \text{C}^2 \]

4. **Substitute \( Q^2 \) into the equation:**
\[ 4.5 = \frac{3}{5} \frac{16 \times 10^{-12}}{4 \pi \epsilon_0 R} \]

5. **Use the value of \( \epsilon_0 \):**
\[ \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \]
Thus, \( 4 \pi \epsilon_0 \approx 4 \pi (8.85 \times 10^{-12}) \approx 1.112 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \).

6. **Rearranging the equation to solve for \( R \):**
\[ R = \frac{3 \times 16 \times 10^{-12}}{5 \times 4.5 \times 1.112 \times 10^{-10}} \]

7. **Calculate the numerator:**
\[ 3 \times 16 \times 10^{-12} = 48 \times 10^{-12} \]

8. **Calculate the denominator:**
\[ 5 \times 4.5 \times 1.112 \times 10^{-10} = 24.99 \times 10^{-10} \approx 2.499 \times 10^{-9} \]

9. **Now substitute back to find \( R \):**
\[ R = \frac{48 \times 10^{-12}}{2.499 \times 10^{-9}} \approx 0.0192 \, \text{m} = 19.2 \, \text{mm} \]

10. **Final answer:**
The radius of the sphere is approximately \( 19.2 \, \text{mm} \) or rounded to \( 20 \, \text{mm} \).