Uniform radial electric field?

[maurits]

In textbooks or other texts that discuss electrodynamics at some point always the term 'uniform' is introduced to describe particularly simple (symmetric) charge distributions, electric fields, etc.

In trying to imagine what a uniform charge distribution - which, of course, is a mathematical idealization - looks like in nature, I quickly end up at an image of many charges (electrons) sitting next to each other at a certain distance. They may form a string, cover a surface or fill up a volume, thereby creating a uniform line, surface or volume charge density, respectively.

Using Gauss's Law to find a formulation for the electric field for the last case of a uniformly charged sphere, we see that for a sphere of radius R the field inside the sphere (r <= R) is


\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}\hat{r}


Outside the sphere (r > R), it is

\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}


Hence, inside the uniformly charged sphere E is proportional to r, outside it E goes as the inverse square of r. This is nicely sketched at the bottom of http://www.2classnotes.com/digital_notes.asp?p=Electric_Field_due_to_a_Unifor m_Sphere_of_Charge. Note that E points in the radial direction as indicated by the \hat{r}.

What if we consider a uniform electric field instead of a uniform charge distribution? A sketch similar to that linked to before should then consist of or contain a horizontal line.

A well-known example of a uniform electric field is the one produced by (between) two (infinite) parallel plates of opposite charge polarity. The electric field lines are perpendicular to the plates in this case.

As to investigate radial uniform electric fields, let's try to find the spherical charge distribution (of finite volume) that produces the field


\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\hat{r}


within that sphere (r <= R). Outside the sphere, where we assume the charge density to be zero, the electric field will be the same as we found before.

Applying Gauss's Law again, this time to find the charge distribution, we get


\rho(r) = \frac{Q}{2\pi R^2}\frac{1}{r}


valid for r <= R. It surprises me that the charge distribution goes as 1/r (please correct me if this is not the case at all) and several question popped up in my mind the moment I derived this result:
- As we approach the origin (r = 0) the charge density becomes higher and higher. What happens near or at the origin?
- Does a charge distribution like this exist in nature? Are we capable of making one in some way?
- Are any cases know in nature of uniform radial electric fields?

Thanks!

[Andrew Mason]

If the field is to be constant, the charge would have to increase in proportion to r^2. Since the volume increases in proportion to r^3, the charge density would have to decrease in proportion to 1/r. So that makes sense. The fact that this means charge density would have to increase without bound as you decrease r makes it very difficult or impossible to make.

AM

[granpa]

the charge density becomes infinite but the total amount of charge is still finite

the field at the origin would be zero

[dgOnPhys]

- As we approach the origin (r = 0) the charge density becomes higher and higher. What happens near or at the origin?
- Does a charge distribution like this exist in nature? Are we capable of making one in some way?
- Are any cases know in nature of uniform radial electric fields?

Thanks!

It is easier to see what's going on by calculating q(r) for r<R, this goes as r^2; the problem is really that density in a radial variable is not such an intuitive quantity

[maurits]

the charge density becomes infinite but the total amount of charge is still finite

the field at the origin would be zero

Ah, of course! This reminds me of Dirac's delta function, which, in the end, we always integrate. I remember the Dirac delta is also introduced in electrodynamics (cf. Griffiths' Introduction to Electrodynamics (3e) p. 50) to cater for the problem with the divergence of the E-field at r = 0.

It is easier to see what's going on by calculating q(r) for r<R, this goes as r^2; the problem is really that density in a radial variable is not such an intuitive quantity

Yes, my intuition indeed seems to falter, here. I guess it has to do with the problem you point out: how can we understand a density in a radial variable? Thanks! I'll also try to look into the approach concerning q(r).