Electric field dependence on distance How can it be proved that for a point charge, $E$ is proportional to $$1/r^2$$ using the concept of Electric field lines (or lines of force)? I tried to show that if field lines are close, then magnitude of Electric field is higher. But, I couldn't show the given dependence.  
 A: As such there is no real theoretical proof to the inverse square dependence of the electric field in classical electrodynamics. It is an experimental fact famously known as the Coulomb's law. When combined with the superposition principle, it gives us the Gauss's law of classical electrodynamics: 
$$\nabla \cdot\mathbf E = \frac{\rho}{\epsilon_0}.$$ 
But, one can also think of the Gauss' law as an experimental fact and from it, he/she can derive (with suitable physical assumptions) the inverse square dependence of the electric field of a charged sphere in the following manner: 
Let's take a spherical charge whose charge $Q$ is distributed spherically symmetrically within some radius $a$. Consider a surface centered at the center of the sphere and having a radius $R>a$. Now, one can argue that at each point of the spherical shell, the only direction, an electric field can have is either radially outward or radially inward. Also, if the electric field points radially inward at one of the points on the spherical shell then it should point radially inward at every other point on the spherical shell. Also, the magnitude of the electric field must be the same at each and every point of the considered spherical shell.
Thus, the integral $\displaystyle\iint_{S} \mathbf E\cdot \mathrm d{\mathbf A}$ (where $S$ denote the integration over the spherical surface) can also be written as $4\pi R^2E$ where $E$ is the magnitude of the electric field - taken positive if it points radially outward and negative if it points radially inward. (This is just a convention - you could alter it and still get the right physical direction of the electric field provided you use the vector calculus properly.) Now from the Gauss' law, this integral must be equal to the total charge inside the spherical surface divided by $\epsilon_0$. i.e. $Q/\epsilon_0$. Therefore, 
$$4\pi R^2E = \dfrac{Q}{\epsilon_0}$$ 
Or, $$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{R^2}\;.$$ 
Since, there was nothing special about the radius $R$ except for $R>a$, we can consider this formula to be true for any $R>a$. 
A: You can prove it using the concept of electric flux. For instance. If you surround a point charge with a sphere if r=1, or a sphere with r =10, you know that the electric flux ( field strength times area) must be the same. A sphere is easy because every point is equidistant to the charge.
A: This is a much more deeper question then it looks in first glance. The simple logic given by @Anthony B is not enough for proving the inverse square law. There are numerous experiments that have been done to verify this law. There is a collection of the experimental works in this review. 
In earlier days Cavendish and Coulomb have performed experiments with conducting hemispheres and torsion spring, which proved the inverse square law. 
Procs and deBrolgie have postulated that if the photons have rest mass then there will be deviations from inverse square law. However the estimates of the rest mass of photons are really low. 
If there is a deviation from inverse square law then there will be critial situation for the physics.
A: As Anthony B said,the number of field lines cutting any sphere surrounding a point charge is the same(because any field line which passes through a sphere of radius 1 also Passes through a sphere of radius 200) given that, the flux  = E 4pir^2 should be constant. That explains the 1/r^2 dependance theoretically
