Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer
    
So would I be correct in saying that an electrical field of whatever source strength acts over the same distance that is: $$\frac{\pi^2}{6}$$? –  Ram Sidharth May 5 '12 at 14:51
1  
Of course not! Just go over what an "electric field" is, and what work/energy is. The pi-squared business is pure mathematics to show you that infinite sums going as the inverse square are finite. –  Ron Maimon May 5 '12 at 15:09
add comment

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.

share|improve this answer
1  
Even in quantum electrodynamics, the infinity problems are only postponed to exponentially tiny distances. If you make the fine structure constant much bigger than 1, then the classical picture is correct and leads to a blow-up before the positrons smooth out the blow-up to logarithmic. The only full solution is in string theory, and perhaps in Argyres-Douglas type SUSY theories where strong coupling limits are well defined. –  Ron Maimon May 5 '12 at 19:52
    
Very interesting. I thought that renormalization procedures solved the problem of the self-energy of the electron in QED without ressorting to string thery or SUSY, but I'm not an expert in this field. Can you please give some references to have a look at and learn a little bit more? –  DaniH May 5 '12 at 20:38
    
The renormalization in QED does solve the problem perturbatively to all orders, and for weak coupling, it will work to all distance scales larger than a tiny, tiny value, called the "Landau pole", which is so small that the problem is completely solved for all practical purposes. But at large fine-structure constant, the problems would reappear, because the self-energy in the electron's field would be comparable to the electron's mass at a distance scale where the theory is still classical. This is not our world, so we don't have to worry about this. Argyres/Douglas is a classic. –  Ron Maimon May 5 '12 at 20:51
    
There is no good reference for Landau pole issues in field theory, they aren't fully sorted out with rigor. There are heuristic arguments in some books, but you just have to muddle through Wilsonian articles. The condensed matter literature is sometimes good on this, usually a little better than the high energy literture, which is too Feynman-diagram focused (for good reason, but it obscures this stuff). You could try Wilson's 1974 Rev. Mod. Phys. classic, and consider the Ising model coupling renormalization as a model for Landau poles. –  Ron Maimon May 5 '12 at 20:53
    
Wonderful, thanks! –  DaniH May 5 '12 at 21:31
show 1 more comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.