An isolated conduction sphere sphere whose radius `R=1` m has a charge `q=1/9nC.` The energy density at the surfasce of the sphere is

Options

A. `epsilon_0/2 J//m^3`
B. `epsilon_0J//m^3`
C. `2epsilon_0J//m^3`
D. `epsilon_0//3Jm^3`

Explanation

To find the energy density at the surface of an isolated conducting sphere with a given charge, we can follow these steps:

### Step 1: Identify the parameters
Given:
- Radius of the sphere, \( R = 1 \, \text{m} \)
- Charge on the sphere, \( q = \frac{1}{9} \, \text{nC} = \frac{1}{9} \times 10^{-9} \, \text{C} \)

### Step 2: Calculate the electric field at the surface of the sphere
The electric field \( E \) at the surface of a charged conducting sphere is given by the formula:
\[ E = \frac{k \cdot q}{R^2} \]
where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting the values:
\[ E = \frac{9 \times 10^9 \cdot \frac{1}{9} \times 10^{-9}}{(1)^2} \]
\[ E = \frac{9 \times 10^9}{9 \times 10^9} = 1 \, \text{N/C} \]

### Step 3: Calculate the energy density
The energy density \( u \) in an electric field is given by the formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Substituting \( E = 1 \, \text{N/C} \):
\[ u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot (1)^2 \]
\[ u = \frac{1}{2} \cdot 8.85 \times 10^{-12} = 4.425 \times 10^{-12} \, \text{J/m}^3 \]

### Final Answer
The energy density at the surface of the sphere is:
\[ u = \frac{\epsilon_0}{2} \, \text{J/m}^3 \]