A sphere of radius `1 cm` has potential of `8000 V`, then energy density near its surface will be

Options

A. `64xx10^(5) J//m^(3)`
B. `8xx10^(3) J//m^(3)`
C. `32 J//m^(3)`
D. `2.83 J//m^(3)`

Answer

To find the energy density near the surface of a sphere with a given potential, we can follow these steps:

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

### Step 2: Determine the electric field (E) at the surface of the sphere
The electric field (E) at the surface of a charged sphere can be calculated using the formula:
\[ E = \frac{V}{R} \]
where:
- \( V \) is the potential (8000 V in this case).
- \( R \) is the radius of the sphere (1 cm = \( 0.01 \, \text{m} \)).

### Step 3: Substitute the values into the electric field formula
Convert the radius from centimeters to meters:
\[ R = 1 \, \text{cm} = 0.01 \, \text{m} \]
Now substitute the values into the electric field formula:
\[ E = \frac{8000 \, \text{V}}{0.01 \, \text{m}} = 800000 \, \text{V/m} \]

### Step 4: Calculate the energy density (u)
Now substitute \( E \) into the energy density formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
\[ u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (800000)^2 \]

### Step 5: Perform the calculations
Calculate \( E^2 \):
\[ E^2 = (800000)^2 = 640000000000000 \, \text{V}^2/\text{m}^2 \]
Now substitute this back into the energy density formula:
\[ u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (640000000000000) \]
\[ u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (6.4 \times 10^{14}) \]
\[ u = \frac{1}{2} \times 5.664 \times 10^{3} \]
\[ u = 2.832 \, \text{J/m}^3 \]

### Step 6: Final answer
Thus, the energy density near the surface of the sphere is approximately:
\[ u \approx 2.83 \, \text{J/m}^3 \]

### Conclusion
The energy density near the surface of the sphere is \( 2.83 \, \text{J/m}^3 \).

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