An electromagnetic wave of frequency `1xx10^(14)` Hertz is propagating along z-axis. The amplitude of electric field is `4V//m`. If `epsilon_(0)=8.8xx10^(-12)C^(2)//N-m^(2)`, then average energy density of electric field will be:

Options

A. `35.2 xx 10^(-10) J// m^(3) `
B. ` 35.2 xx 10^(-11) J// m^(3) `
C. `35.2 xx 10^(-12) J//m^(3) `
D. `35.2 xx 10^(-13) J//m^(3)`

Explanation

To find the average energy density of the electric field in an electromagnetic wave, we can use the formula:

\[ u = \frac{1}{2} \epsilon_0 E^2 \]

where:
- \( u \) is the average energy density,
- \( \epsilon_0 \) is the permittivity of free space, and
- \( E \) is the amplitude of the electric field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency \( f = 1 \times 10^{14} \) Hz (not needed for this calculation).
- Amplitude of electric field \( E = 4 \, \text{V/m} \).
- Permittivity of free space \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

2. **Substitute the Values into the Formula:**
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
\[ u = \frac{1}{2} \times (8.8 \times 10^{-12}) \times (4)^2 \]

3. **Calculate \( E^2 \):**
\[ E^2 = 4^2 = 16 \, \text{(V/m)}^2 \]

4. **Substitute \( E^2 \) into the Equation:**
\[ u = \frac{1}{2} \times (8.8 \times 10^{-12}) \times 16 \]

5. **Perform the Multiplication:**
\[ u = \frac{1}{2} \times 140.8 \times 10^{-12} \, \text{J/m}^3 \]

6. **Calculate the Final Value:**
\[ u = 70.4 \times 10^{-12} \, \text{J/m}^3 \]

7. **Final Result:**
\[ u = 35.2 \times 10^{-12} \, \text{J/m}^3 \]

### Conclusion:
The average energy density of the electric field is \( 35.2 \times 10^{-12} \, \text{J/m}^3 \).