The electric field associated with an electromagnetic $(EM)$ wave in vacuum is given by $\vec{E} = \hat{i} 40 \cos (kz - 6 \times 10^{8} t)$,where $E$,$z$,and $t$ are in $V/m$,$m$,and $s$ respectively. The value of the wave vector $k$ is .... $m^{-1}$.

Assume a bulb of efficiency $2.5\%$ as a point source. The peak value of the electric field produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is approximately.....$V\,m^{-1}$.

$A$ point source of electromagnetic radiation has an average power output of $800\,W$. The maximum value of the electric field at a distance of $4.0\,m$ from the source is...$V/m$

Given below are two statements:
Statement $I$: $A$ time-varying electric field is a source of a changing magnetic field and vice-versa. Thus,a disturbance in an electric or magnetic field creates $EM$ waves.
Statement $II$: In a material medium,the $EM$ wave travels with speed $v = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
In the light of the above statements,choose the correct answer from the options given below:

An electromagnetic wave with frequency $\omega$ and wavelength $\lambda$ travels in the $+y$ direction. Its magnetic field is along the $+x$ axis. The vector equation for the associated electric field (of amplitude $E_0$) is:

Suppose that the intensity of a laser is $\left(\frac{315}{\pi}\right) \ W/m^2$. The $rms$ electric field,in units of $V/m$,associated with this source is close to the nearest integer. Given: $\epsilon_0 = 8.86 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $c = 3 \times 10^8 \ m/s$.