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 :

For $\overrightarrow{E_1}$  corresponding  $\overrightarrow{B_1}=\frac{E_0}{C}\widehat{k}\cos \left(\omega t-kx\right)$
For $\overrightarrow{E_2}$  corresponding   $\overrightarrow{B_2}=\frac{E_0}{C}\widehat{i}\cos \left(\omega t-ky\right)$
 $\begin{array}{l}Total force acting at t=0, x=0 is\text{}\\ {\overrightarrow{F}}_{net}=q{\overrightarrow{E}}_{net}+q\left(\overrightarrow{v}\times {\overrightarrow{B}}_{net}\right)\text{}\\ here\text{}{\overrightarrow{E}}_{net} ={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=E_0\widehat{j}+E_0\widehat{k}\text{}\\ and\\ {\overrightarrow{B}}_{net} ={\overrightarrow{B}}_1+{\overrightarrow{B}}_2=\frac{E_0}{C}\widehat{k}+\frac{E_0}{C}\widehat{i}\\ \Rightarrow \overrightarrow{F} =q\left(E_0\widehat{j}+E_0\widehat{k}\right)+q\left(0.8c\widehat{j}\times \left(\frac{E_0}{C}\widehat{k}+\frac{E_0}{C}\widehat{i}\right)\right)\\ \Rightarrow \overrightarrow{F} =q{E}_0\widehat{j}+q{E}_0\widehat{k}+0.8q{E}_0\widehat{i}-0.8q{E}_0\widehat{k}\text{}\\ \Rightarrow \overrightarrow{F}=q{E}_0\left(0.8\widehat{i}+\widehat{j}+0.2\widehat{k}\right)\end{array}$