The electric field associated with an electromagnetic wave in vacuum is given by $E = 40 \cos(kz - 6 imes 10

Question

The electric field associated with an electromagnetic wave in vacuum is given by $E = 40 \cos(kz - 6 	imes 10^8 t)$, where $E$, $z$, and $t$ are in volt/m, meter, and second respectively. The value of the wave vector $k$ would be:

Accepted Answer

The given wave equation is $E = 40 \cos(kz - 6 	imes 10^8 t)$. Comparing this with the standard equation of a plane electromagnetic wave propagating in the $z$-direction, $E = E_0 \cos(kz - \omega t)$, we identify the angular frequency $\omega = 6 	imes 10^8 	ext{ rad s}^{-1}$ .
The speed of light in vacuum ($c$) relates the angular frequency ($\omega$) and the magnitude of the wave vector ($k$) by the relation:
$$ c = \frac{\omega}{k} $$
where $c = 3 	imes 10^8 	ext{ m s}^{-1}$.
Rearranging for $k$:
$$ k = \frac{\omega}{c} = \frac{6 	imes 10^8}{3 	imes 10^8} = 2 	ext{ m}^{-1} $$

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