A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 $$\times$$ 10^$$-$$2 Am^$$-$$1 at a point, what will be the approximate magnitude of electric field intensity at that point?

(Given : Permeability of free space $$\mu$$₀ = 4$$\pi$$ $$\times$$ 10^$$-$$7 NA^$$-$$2, speed of light in vacuum c = 3 $$\times$$ 10⁸ ms^$$-$$1)

Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by

E = 20cos(2 $$\times$$ 10¹⁰ t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\mu$$_r = 1)

The magnetic field vector of an electromagnetic wave is given by $$B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)$$; where $$\widehat i,\widehat j$$ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q₁ of 4$$\pi$$ coulomb and q₂ of 2$$\pi$$ coulomb located at $$\left( {0,0,{\pi \over k}} \right)$$ and $$\left( {0,0,{{3\pi } \over k}} \right)$$, respectively, have the same velocity of 0.5 c $$\widehat i$$, (where c is the velocity of light). The ratio of the force acting on charge q₁ to q₂ is :-

Electric field in a plane electromagnetic wave is given by E = 50 sin(500x $$-$$ 10 $$\times$$ 10¹⁰ t) V/m The velocity of electromagnetic wave in this medium is :

(Given C = speed of light in vacuum)