The electric field associated with an electromagnetic wave in vacuum is given by `vecE=40cos(kz-6xx10^(8)t)hati,`
where E, z and t are in volt per meter, meter and second respectively. The value of wave vector k is

Options

A. 2 `m^(-1)`
B. `0.5m^(-1)`
C. `6m^(-1)`
D. `3m^(-1)`

Explanation

To find the value of the wave vector \( k \) from the given electric field equation of an electromagnetic wave, we can follow these steps:

### Step 1: Identify the given electric field equation
The electric field associated with the electromagnetic wave is given by:
\[ \vec{E} = 40 \cos(kz - 6 \times 10^8 t) \hat{i} \]
Here, \( E \) is in volts per meter, \( z \) is in meters, and \( t \) is in seconds.

### Step 2: Compare the given equation with the general form
The general form of the electric field for an electromagnetic wave can be expressed as:
\[ \vec{E} = E_0 \cos(kz - \omega t) \]
where:
- \( E_0 \) is the amplitude of the electric field,
- \( k \) is the wave vector,
- \( \omega \) is the angular frequency.

From the given equation, we can identify:
- \( E_0 = 40 \) V/m,
- \( \omega = 6 \times 10^8 \) rad/s.

### Step 3: Relate wave vector \( k \) to angular frequency \( \omega \)
The relationship between the wave vector \( k \), angular frequency \( \omega \), and the speed of light \( v \) is given by:
\[ v = \frac{\omega}{k} \]
For electromagnetic waves in a vacuum, the speed of light \( v \) is approximately:
\[ v = 3 \times 10^8 \text{ m/s} \]

### Step 4: Rearranging the equation to find \( k \)
We can rearrange the equation to solve for \( k \):
\[ k = \frac{\omega}{v} \]

### Step 5: Substitute the known values
Now, substituting the values of \( \omega \) and \( v \):
\[ k = \frac{6 \times 10^8}{3 \times 10^8} \]

### Step 6: Calculate the value of \( k \)
Calculating the above expression:
\[ k = 2 \text{ m}^{-1} \]

### Final Answer
Thus, the value of the wave vector \( k \) is:
\[ \boxed{2 \text{ m}^{-1}} \]

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