Electricity & Magnetism - Is an electric field infinite? The inverse square law for an electric field is:

$$
E = \frac{Q}{4\pi\varepsilon_{0}r^2}
$$
Here: $$\frac{Q}{\varepsilon_{0}}$$
is the source strength of the charge. It is the point charge divided by the vacuum permittivity or electric constant, I would like very much to know what is meant by source strength as I can't find it anywhere on the internet. Coming to the point an electric field is also described as:
$$Ed = \frac{Fd}{Q} = \Delta V$$
This would mean that an electric field can act only over a certain distance. But according to the Inverse Square Law, the denominator is the surface area of a sphere and we can extend this radius to infinity and still have a value for the electric field. Does this mean that any electric field extends to infinity but its intensity diminishes with increasing length? If that is so, then an electric field is capable of applying infinite energy on any charged particle since from the above mentioned equation, if the distance over which the electric field acts is infinite, then the work done on any charged particle by the field is infinite, therefore the energy supplied by an electric field is infinite. This clashes directly with energy-mass conservation laws. Maybe I don't understand this concept properly, I was hoping someone would help me understand this better. 
 A: It goes out forever, but the total energy it imparts is finite. The reason is that when things fall off as the square of the distance, the sum is finite. For example:
$$ \sum_n {1\over n^2} = {1\over 1} + {1\over 4} + {1\over 9} + {1\over 16} + {1\over 25} + ... = {\pi^2\over 6} $$
This sum has a finite limit. Likewise the total energy you gain from moving a positive charge away from another positive charge from position R to infinity is the finite quantity
$$\int_r^{\infty} {Qq\over r^2} dr = {Qq\over r}$$
So there is no infinity. In two dimensions (or in one), the electric field falls off only like ${1\over r}$ so the potential energy is infinite, and objects thrown apart get infinite speed in the analogous two-dimensional situation.
A: I just want to add something besides Ron's answer, which in my opinion you should accept as "the answer".
The second formula which you quote does not apply to the field produced by a point charges. It is only true for a constant electric field. In general, the change in the potential energy when going from a point $\mathbf{r}_0$ in the space to infinity is
$$\Delta V = -\int_{\mathbf{r}_0}^\infty \mathbf{E}(\mathbf{r}) \cdot \text{d} \mathbf{r}$$.
which comes from the relation $\mathbf{E}(\mathbf{r})=-\nabla V(\mathbf{r})$ and integration in the space.
For your point charge in $\mathbb{R}^3$, you can integrate easily [using the the electric field is isotropic] to get 
$$\Delta V = V(\infty)-V(\mathbf{r}_0) = \dfrac{Q}{4\pi\varepsilon_0 r_{0}}$$
where $r_0=|\mathbf{r}_0|=\sqrt{x_0^2+y_0^2+z_0^2}$ and $V(\infty)=0$. 
Hence, even if the range of the electric field is infinite, the energy is always finite. Note that when $|\mathbf{r}_0| \rightarrow 0$ the potential energy blows up even though physically does not make any sense. To solve this you need to make use of quantum electrodynamics, the "quantum version" of electromagnetism.
A: The Landau Pole is not a problem for QED because at scales much smaller than it (the Planck scale, which is smaller than the Landau pole by 260 orders of magnitude) the (negative) gravitational self-energy of the particle will more than cancel out its electromagnetic self-energy. So string theory is not necessary in this case, just gravity.
