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

We begin with the Lorentz force formula for a charged particle in an electromagnetic field: $$\vec{F}=q\left(\vec{E}+\vec{v}\times\vec{B}\right).$$

Both waves contribute to the net electric and magnetic fields, so we first find each field at the given instant.

For the first wave we have $$\vec{E_1}=E_0\hat{j}\cos\!\left(\omega t-kx\right).$$ At $$t=0$$ and $$x=0$$, the argument of the cosine is zero, hence $$\cos 0 = 1$$ and we obtain $$\vec{E_1}(0)=E_0\hat{j}.$$

For a plane wave propagating along the $$+x$$-direction, the magnetic field is perpendicular to both the electric field and the direction of propagation, with magnitude $$B=\dfrac{E}{c}.$$ Since $$\hat{j}\times\hat{k}=\hat{i}$$ gives the $$+x$$ direction, the magnetic field of the first wave must be along $$\hat{k}.$$ Thus $$\vec{B_1}=\frac{E_0}{c}\hat{k}\cos\!\left(\omega t-kx\right).$$ Again at $$t=0,\,x=0$$, $$\vec{B_1}(0)=\dfrac{E_0}{c}\hat{k}.$$

For the second wave we have $$\vec{E_2}=E_0\hat{k}\cos\!\left(\omega t-ky\right).$$ At $$t=0,\,y=0$$ the cosine equals 1, giving $$\vec{E_2}(0)=E_0\hat{k}.$$

This wave propagates along the $$+y$$-direction. For propagation along $$+y$$, we need $$\vec{E}\times\vec{B}$$ to point along $$\hat{j}\,.$$ Since $$\hat{k}\times\hat{i}=\hat{j},$$ the magnetic field of the second wave must be along $$\hat{i}.$$ Therefore $$\vec{B_2}=\frac{E_0}{c}\hat{i}\cos\!\left(\omega t-ky\right),$$ and at $$t=0,\,y=0$$ we get $$\vec{B_2}(0)=\dfrac{E_0}{c}\hat{i}.$$

Now we add the individual fields to get the total electric and magnetic fields at the origin and at the given instant:

$$\vec{E}=\vec{E_1}(0)+\vec{E_2}(0)=E_0\hat{j}+E_0\hat{k}=E_0\left(\hat{j}+\hat{k}\right),$$ $$\vec{B}=\vec{B_1}(0)+\vec{B_2}(0)=\dfrac{E_0}{c}\hat{k}+\dfrac{E_0}{c}\hat{i}=\dfrac{E_0}{c}\left(\hat{i}+\hat{k}\right).$$

The particle’s velocity is given as $$\vec{v}=0.8c\,\hat{j}.$$

We now compute the cross product $$\vec{v}\times\vec{B}.$$ We substitute the vectors and factor out the constants:

$$\vec{v}\times\vec{B}=\left(0.8c\,\hat{j}\right)\times\left[\dfrac{E_0}{c}\left(\hat{i}+\hat{k}\right)\right] =0.8E_0\,\hat{j}\times\left(\hat{i}+\hat{k}\right).$$

Using the basic vector product identities $$\hat{j}\times\hat{i}=-\hat{k}\quad\text{and}\quad\hat{j}\times\hat{k}=\hat{i},$$ we expand:

$$\hat{j}\times\left(\hat{i}+\hat{k}\right)=\hat{j}\times\hat{i}+\hat{j}\times\hat{k}=-\hat{k}+\hat{i}=\hat{i}-\hat{k}.$$

Multiplying by the scalar factor $$0.8E_0$$ gives $$\vec{v}\times\vec{B}=0.8E_0\left(\hat{i}-\hat{k}\right).$$

We now have everything needed for the Lorentz force:

$$\vec{F}=q\left(\vec{E}+\vec{v}\times\vec{B}\right) =q\left[E_0\left(\hat{j}+\hat{k}\right)+0.8E_0\left(\hat{i}-\hat{k}\right)\right].$$

Combining the $$\hat{k}$$ components carefully, we write out each component explicitly:

$$\vec{F}=q\Bigl[\,E_0\left(0.8\hat{i}\right)+E_0\left(1\hat{j}\right)+E_0\left(1-0.8\right)\hat{k}\Bigr].$$

The $$\hat{k}$$ coefficient simplifies to $$1-0.8=0.2.$$ Hence,

$$\vec{F}=E_0q\left(0.8\hat{i}+\hat{j}+0.2\hat{k}\right).$$

Comparing with the given options, this matches Option D.

Hence, the correct answer is Option 4.