Solution:
Solution
Given: The electric field associated with an electromagnetic wave in vacuum is given by $$\vec E =40 \cos(kz−6\times 10^8t)\hat i$$ , where E, z and t are in volt per meter, meter and second respectively.
To find the value of wave vector k
Solution:
We know electromagnetic wave eqution is
$$E=E_0\cos(kz-\omega t)$$
And given equation is
$$\vec E =40 \cos(kz−6\times 10^8t)\hat i$$
By comparing these two, we get
$$\omega=6\times10^8$$ and
$$E_0=40\hat i$$
we also know,
Speed of electromagnetic wave, $$v=\dfrac \omega k$$
where v is the speed of the light
Hence, $$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$$
is the required value