Prove EM Waves Are Transverse In Nature Why we say that EM waves are transverse in nature? I have seen some proofs regarding my question but they all calculate flux through imaginary cube. Here is My REAL problem that I can't here imagine infinitesimal area for calculating flux because em line of force will intersect (perpendicular or not) surface at only one point so $E.ds$ will be zero so even flux through one surface of cube will always be zero. I am Bit Confused. I DON'T KNOW VECTOR CALCULUS BUT KNOW CALCULUS.
 A: 
Why we say that em waves are transverse in nature?

In a region empty of electric charge, we have, from Maxwell's equations:
$$\nabla \cdot \vec E = \nabla \cdot \vec B = 0$$
Since you don't yet know vector calculus, let's rewrite these divergence equations as so:
$$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 0 $$
$$\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0 $$
Now, assume an electromagnetic wave is propagating in the $z$ direction so that the space and time variation of the field components are of the form
$$\cos(kz - \omega t)$$
Since the spatial variation is zero in the $x$ and $y$ directions, our equations become
$$\frac{\partial E_z}{\partial z} = 0$$
$$\frac{\partial B_z}{\partial z} = 0$$
Which means that electric and magnetic field components in the $z$ direction, the direction of propagation, must be constant with respect to $z$.
In other words, only the electric and magnetic field components transverse to the direction of propagation vary with respect to $z$. i.e., the electromagnetic wave is transverse.

Addendum to address a comment:

Why spatial Variations are zero in both x and y directions.

We stipulated that the field components are of the form $\cos(kz - \omega t)$ which means the wave is propagating in the $z$ direction.
Clearly, the partial derivative of $\cos(kz - \omega t)$ with respect to $x$ and $y$ is zero.
A: You say the 

em line of force will intersect (if perpendicular) surface at only one point. 

This is not the case for a plane wave, which is the simplest case one usually considers. I think I know what's going on to make you think this. This may or may not be the issue that helps you.
Take this typical picture of an EM wave.

It is so very tempting, yet incorrect, to think that the E and B field are only defined (non-zero) on that axis. No. Here's a better picture that only shows one of the two fields.

It's called a plane wave because the E and B fields have the same value everywhere on some plane. So if you imagine this wave incident on some cube, the dot product $\vec{E}\cdot d\vec{S}$ has a value at more than just one point.
A: I am going to try to "unconfuse" you.
Let's start with a picture of ocean waves; as seen from the side, they look like "sine" waves (just like BMS's red EM wave); as seen from the top, the crests and troughs make lines (not points). The same thing happens with the E and B waves (make lines), and since they are perpendicular to each other, they form a plane (two lines make a plane). This is why the flux is evaluated over a surface area (not a point).    
A: An EM wave is generated by the oscillation of an electron. Near the electron we have the near field and here all the wave components are non-zero. Far away from the source we have the far field and it is in the form of a spherical surface wave advancing along the radius of a sphere with centre at the source. If we take a small section of this spherical advancing surface we have a plane wave. Due to symmetry all components normal to the propagation direction cancel each other and become zero. This leaves a wave with variations of the z propagating wave component only varying with time and distance along z- hence the equation given in the other answers. Note that this applies to all waves from a single point.
