Consider an electromagnetic wave propagating along x-axis. Since parallelopiped OABCDEFG does not enclose any charge. Therefore from Gauss.s law
`oint vec(E ).vec(dS) = 0 or underset(ABCD)int vec(E ).vec(dS) + underset(OEFG)int vec(E ).vec(dS) + underset(BCFG)int vec(E ).vec(dS) + underset(OADE)int vec(E ).vec(dS) + underset(CDEF)int vec(E ).vec(dS) + underset(OABG)int vec(E ).vec(dS)= 0` ...(i)
`vec(E )` does not depend upon y and z. So, electric flux from surface BCFG and OEDA will be equal and opposite.
Hence, `underset(BCFG)int vec(E ).vec(dS) + underset(OADE)int vec(E ).vec(dS)= 0` ...(ii)
Similarly `underset(CDEF)int vec(E ).vec(dS) + underset(OABG)vec(E ).vec(dS) = 0` ...(iii)
Using equation (ii) and (iii), equation (i) becomes
`underset(ABCD)int vec(E).vec(dS) + underset(OEFG)int vec(E ).vec(dS)= 0`
`underset(ABCD)int E_(x) ds cos 0^(@) + underset(OERG)int E_(x) dS cos 180^(@)= 0`
Where `E_(x) and E._(x)` are the value of electric field at faces ABCD and OEFG. or `E_(x) underset(ABCD)int dS- E._(x) underset(OEFG)int dS= 0`
If S is the area of ABCD and OEFG, then `E_(x) S- E._(x) S= 0 or (E_(x)- E._(x))S= 0`
As S cannot be zero, therefore either `E_(x)- E._(x)= 0` i.e., `E_(x)= E._(x) or E_(x)= E._(x)= 0`
`E_(x)= E._(x)` implies that electric field is static. Static field cannot propagate as a wave. Hence `E_(x) ne E._(x)` then `E_(x)= E._(x)= 0` i.e., field along x-axis is zero. It means that field is perpendicular to x-axis. Similarly, we can show that the magnetic field has no component along x-axis (direction of propagation of wave). Since both electric and magnetic fields are perpendicular to the direction of propagation. Therefore, electromagnetic waves are transverse in nature.
