An electromagnetic wave of frequency $1\times10^{14}\, hertz$ is propagating along $z-$ axis. The amplitude of electric field is $4\, V/m$ . lf ${\varepsilon_0}=\, 8.8\times10^{-12}\, C^2/Nm^2$ , then average energy density of electric field will be:

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\,Hz$ and amplitude $48\,Vm ^{-1}$. Then the amplitude of oscillating magnetic field is : (Speed of light in free space $=3 \times 10^8\,m s ^{-1}$)

The electric field component of an electromagnetic wave in vaccum is given as $\vec E = 3\cos \,\left( {1.8y + 5.4 \times {{10}^8}\,t} \right)\hat i$ Its direction of propagation and wavelength is 

An antenna is placed in a dielectric medium of dielectric constant $6.25$. If the maximum size of that antenna is $5.0\, mm$. it can radiate a signal of minimum frequency of $GHz .$

(Given $\mu_{ r }=1$ for dielectric medium)

In a certain region uniform electric field $E$ and magnetic field $B$ are present in the opposite direction. At the instant $t = 0,$ a particle of mass $m$ carrying a charge $q$ is given velocity $v_o$ at an angle $\theta ,$ with the $y$ axis, in the $yz$ plane. The time after which the speed of the particle would be minimum is equal to

The magnetic field in a plane electromagnetic wave is $\mathrm{B}_y=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5\right.$ $\left.\times 10^{11} \mathrm{t}\right) \mathrm{T}$. The corresponding electric field will be