A plane electromagnetic wave moving through free space has an electric field (also referred to as the optical field) given by `E _(x) =0, E _(y) =0 and E _(z) =100 sin [ 8pi xx 10 ^(14) (t - (x)/( 3 xx 10 ^(8)))]V m ^(-1).`
The corrsponding flux density is

Options

A. `13.3 W m ^(-2)`
B. `11.4 W m ^(-2)`
C. `25.0 W m ^(-2)`
D. `20.4 W m ^(-2)`

Explanation

To find the corresponding flux density \( D \) for the given electric field of the electromagnetic wave, we can follow these steps:

### Step 1: Identify the Electric Field
The electric field is given as:
\[ E_z = 100 \sin \left( 8\pi \times 10^{14} \left( t - \frac{x}{3 \times 10^8} \right) \right) \, \text{V/m} \]
with \( E_x = 0 \) and \( E_y = 0 \).

### Step 2: Use the Relationship Between Electric Field and Flux Density
The flux density \( D \) is related to the electric field \( E \) by the equation:
\[ D = \epsilon_0 E \]
where \( \epsilon_0 \) is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \, \text{F/m} \).

### Step 3: Calculate the Flux Density
Substituting the values into the equation:
\[ D_z = \epsilon_0 E_z \]
\[ D_z = (8.85 \times 10^{-12} \, \text{F/m}) \times (100 \, \text{V/m}) \]
Calculating this gives:
\[ D_z = 8.85 \times 10^{-10} \, \text{C/m}^2 \]

### Step 4: Final Result
Thus, the corresponding flux density \( D \) is:
\[ D = 8.85 \times 10^{-10} \, \text{C/m}^2 \]