The electric fields of two plane electromagnetic plane waves in vacuum are given by
$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$ and
$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
At t = 0, a particle of charge q is at origin with
a velocity $$\overrightarrow v = 0.8c\widehat j$$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :

A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $$\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT$$. The corresponding electric field $$\overrightarrow E $$ is (speed of light c = 3 × 10⁸ ms^–1)

The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$

At t = 0, a positively charged particle is at the point (x, y, z) = $$\left( {0,0,{\pi \over k}} \right)$$.
If its instantaneous velocity at (t = 0) is $${v_0}\widehat k$$ , the force acting on it due to the wave is :

If the magnetic field in a plane electromagnetic wave is given by
$$\overrightarrow B $$ = 3 $$ \times $$ 10^-8 sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat j$$ T, then what will be expression for electric field ?