The relative permittivity of distilled water is $81$. The velocity of light in it is $v \times 10^{7} \text{ m/s}$. Find the value of $v$. (Given $\mu_{r} = 1$ and speed of light $c = 3 \times 10^{8} \text{ m/s}$) (in $.33$)

Even though an electric field $E$ exerts a force $qE$ on a charged particle, the electric field of an $EM$ wave does not contribute to the radiation pressure (but transfers energy). Explain.

For a given electromagnetic wave,the magnitude of the electric field is $6.6 \,V \,m^{-1}$ at a point in space. The magnitude of the magnetic field at this point is . . . . . . $T$.

Suppose that the electric field part of an electromagnetic wave in vacuum is $E = \{(3.1 \; N/C) \cos [(1.8 \; rad/m) y + (5.4 \times 10^{6} \; rad/s) t] \} \hat{i}$.
$(a)$ What is the direction of propagation?
$(b)$ What is the wavelength $\lambda$?
$(c)$ What is the frequency $\nu$?
$(d)$ What is the amplitude of the magnetic field part of the wave?
$(e)$ Write an expression for the magnetic field part of the wave.

An electromagnetic wave is propagating in vacuum along $-\hat{j}$ direction. The magnetic field of the wave is given by $\vec{B} = (2 \times 10^{-8}) \cos [\pi \times 10^{15}(t + \frac{y}{c})] \hat{k} \text{ T}$. The electric field $\vec{E}$ of this wave is $(c = \text{speed of light})$

The frequency of an electromagnetic wave is $25 \ MHz$. If the magnitude of the electric field at a point at any instant is $6.3 \ V/m$,then the magnitude of the magnetic field at that point is ..... $Wb/m^2$.