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. `35.2xx10^(-10)J//m^(3)`
B. `35.2xx10^(-11)J//m^(3)`
C. `35.2xx10^(-12)J//m^(3)`
D. `35.2xx10^(-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 directly needed for this calculation).
- Amplitude of electric field \( E = 4 \, \text{V/m} \).
- Permittivity of free space \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

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 \( E^2 \):**
\[ E^2 = 4^2 = 16 \]

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

5. **Perform the multiplication:**
\[ u = \frac{1}{2} \times 140.8 \times 10^{-12} \]
\[ u = 70.4 \times 10^{-12} \, \text{J/m}^3 \]

6. **Final calculation:**
\[ u = 7.04 \times 10^{-11} \, \text{J/m}^3 \]

### Conclusion:
The average energy density of the electric field is \( 7.04 \times 10^{-11} \, \text{J/m}^3 \).