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 the electromagnetic wave, we can follow these steps:

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

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

### Step 3: Identify the parameters from the given equation
From the comparison of the two equations, we can identify:
- The amplitude \( E_0 = 40 \) V/m,
- The angular frequency \( \omega = 6 \times 10^8 \) rad/s.

### Step 4: Relate the wave vector \( k \) to the angular frequency \( \omega \)
The wave vector \( k \) is related to the angular frequency \( \omega \) by the equation:
\[ k = \frac{\omega}{c} \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.

### Step 5: Substitute the values into the equation
Substituting the values of \( \omega \) and \( c \):
\[ k = \frac{6 \times 10^8}{3 \times 10^8} \]

### Step 6: Calculate the value of \( k \)
Calculating the above expression:
\[ k = \frac{6}{3} = 2 \text{ m}^{-1} \]

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

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