The electric field of a plane electromagnetic wave is given by

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is

A point source of electromagnetic radiation has an average power output of $800\,W$ . The maximum value of electric field at a distance $3.5\,m$ from the source will be…..$V/m$

There exists a uniform magnetic and electric field of magnitude $1\, T$ and $1\, V/m$ respectively along positive $y-$ axis. A charged particle of mass $1\,kg$ and of charge $1\, C$ is having velocity $1\, m/sec$ along $x-$ axis and is at origin at $t = 0.$ Then the co-ordinates of particle at time $\pi$ seconds will be :-

The electric fields of two plane electromagnetic plane waves in vacuum are given by

$\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \hat{\mathrm{j}} \cos (\omega \mathrm{t}-\mathrm{kx})$ and

$\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{ky})$

At $t=0,$ a particle of charge $q$ is at origin with a velocity $\overrightarrow{\mathrm{v}}=0.8 \mathrm{c} \hat{\mathrm{j}}$ ($c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is 

If ${\varepsilon _0}$ and ${\mu _0}$ are respectively, the electric permittivity and the magnetic permeability of free space. $\varepsilon $ and $\mu $ the corresponding quantities in a medium, the refractive index of the medium is